Dr Dave's Math-Science Blog

LESSON 11

2010-08-23T17:40:00.002-07:00

PHYS 1100 LESSON 10 FOR
MONDAY, AUGUST 23, 2010

I Introduction

II Logistics – Running Grades

III Turn in Assignments Due and Return of Papers

IV Review of Lesson 9: Waves and Fluids
A. Waves
1. Length, l
2. Frequency, n
88 Mhz – 108 MHz FM
3.8 m
530 KHz – 1710 KHz
Light, v = c = 300,000,000 m/s
= pitch in sound
Sound, v = 340 m/s
3. Period, P = 1/n
4. Amplitude, A
5. Speed v = l n
6. Types
7. Other

B. Fluid Dynamics
1. Definition of “fluid”
2. And, defn of “dynamics”
3. Density; r = M/V; g/cm3.
4. Pressure; P = F/A; Pascal; 1.0 N/m2.
Pascal; 1013 mB = ATM; 14.7 lb/in2. Mm of Hg; inches of Hg; ATM
5. Bernoulli’s Law – pressure differences between surfaces
6. Archimede’s Principle – displacement of fluids by masses
And more….

V Lesson 10: Density, Archimedes, Bernoulli
Examples of Aviation: airplanes, gliders, helicopters, and similar.

VI Laboratory Exercise 10: Archimedes, Liquids, and Solids

VII HWK Assignment 4: Read Chapters 15-16 and do Problems 1, 3, 9, 13, 19, 25, 29, 31, 39, 53, 63, 65, and 67 on pp 530-534 and do Problems 1, 3, 9, 13, 19, 25, 29, 31, 39, 53, 63, 65, and 67 on pp 566-569; Due 8/26

VIII Essay 4: Albert Einstein, due 8/26

LAB 10: ARCHIMEDES, SOLIDS, LIQUIDS (for Monday, August 23)

2010-08-23T17:40:00.001-07:00

Physics Lab 10: Archimedes, Solids, and Liquids Name _____________
Monday, August 23, 2010 Dr Dave Menke, Instructor

I Title: Solids and Liquids

II Purpose: To observe the effects of solids in a liquid environment

III Equipment and Supplies:
500-mL beaker, Large clear glass, or similar
Water
A metallic cylindrical weight much denser than water
A pair of tongs
Paper Towels to clean up your mess
Metric Ruler
Paper Clip(s)
Straw
“Sharpie” style pen

IV Procedure
1. Determine the diameter, d, of the beaker or glass in cm. Record.
2. Determine the area, A, of the base of the beaker or glass (A = p r2, where r = ½ d). Record.
3. Determine the height/depth of the beaker or glass, h, in cm. Record.
4. Calculate the maximum volume of the beaker or glass in cm3. Record. [Do this by multiplying the height/depth that you found in #3 (h), with the area of the base of the glass or beaker, A, that you found in #2].
5. Fill beaker about ¾ full with water (measure ¾ h from the bottom of the beaker, record, and mark the outside of the beaker/glass with Sharpie Pen).
6. Be sure that the marked water line is level with the water.
7. Use a straw to remove some of the water, via capillary action (don’t use it to drink the water). This means to lower the straw into the water almost all the way to the bottom, place your finger over the top of the straw, and slowly remove it. (The water should stay in the straw, unless it is defective. DON'T THROW THE WATER AWAY THAT IS IN THE STRAW).
8. Note the new water line. Measure how far the water line dropped in millimeters. Convert to centimeters. Record.
9. Return the water in the straw to the beaker and put away the straw. Make sure the water line is the same as before; if not, then add a few more milliliters of water to make it so.
10. Carefully place ONE paper clip on the water’s surface – if you are good at it, the paper clip will float. How many times did you have to do this to make it work? Record.
11. How many paper clips can you put on the water before it breaks the surface tension? Record.
12. Explain what allows the paper clip(s) to float.
13. Remove the paper clip(s) and put them away. Make sure the water line is back to where you started.
14. Select a metal (lead) weight. Calculate its volume, in cm3. Record. (Do this by measuring the diameter of the base, then get the Area of the base in cm2, then measure the height, and multiply the area times the height.)
15. Slowly place (with tongs) the dense lead weight into the water; note the new water line. Measure how far the water line increased, in millimeters, above the water line. Convert to centimeters. Record.
16. Find the increased volume of the water from the addition of the weight in cm3. Do this by multiplying what you found in #15 with the Area that you found in #2. Record.
17. Compare what you just found in #16 with what you calculated in #14 (they should be the same.) Explain.

V Data & Observations
A. Diameter of the beaker or glass d = ______ centimeters
B. Area of the bottom of the beaker or glass = ___________cm2.
C. Height/depth of the beaker or glass, h = ______ centimeters
D. Volume of the beaker or glass, Vb = ____________ cm3.
E. The distance from the bottom of the beaker/glass to ¾ h = ________ cm.
F. Distance that the water dropped from capillary removal ______ cm
G. Number of attempts to make the paper clip float: _____ times
H. Number of paper clips before breaking surface _______ paper clips
I. Explanation of floating paper clips:
J. Volume of lead weight, Vc = ___________ cm3.
K. Distance that the water increased from submersion of dense weight: ______ cm.
L. Increased volume of the water due to the submersion of the weight: ___________cm3.
M. Comparison between #14 and #16.

VI Results:

VII Error Analysis
A. Qualitative Only
1. Personal
2. Systematic
3. Random
B. Quantitative: NA

VIII Questions
1. Why can the straw hold water, even if the bottom is open and gravity is pulling on it?
2. What can you conclude about denser materials in water?
3. In one paragraph of a few sentences, explain Archimedes' Principle.

END

SOLUTION SET 4 - COMPLETE

2010-08-22T08:56:00.000-07:00

Chapter 15: Problems 1, 3, 9, 13, 19, 25, 29, 31, 39, on pp 530-534.

1. Find the weight (in Newtons) of the air in your physics class. Assume that the room is a cube of 4 meters on a side, and that the density of air is rair = 1.29 kg/m3.
Solution:
Volume 4 x 4 x 4 = 64 m3 ; rair = 1.29 kg/m 3. The density, r, is defined to be:
r = mass/volume.

So, the mass = r vol = (1.29)(64) = 82.56 kg. Finally, the “weight” in Newtons is its force in gravity: F = m g = (82.56)(9.8) = 809 Newtons.~ 8.1 x 102 N.

3. You buy a gold ring at a pawn shop, and run a “test” to determine if it's pure gold. The ring's mass is m = 0.014 grams, and has a volume of 0.022 cm3. The density of gold is rAu = 19.3 kg/m3.
Solution: Density is mass divided by volume, and it this case, r = M/V = (0.014g/0.0022 cm3). = 6.6 g/cm3. However, since rAu = 19.3 kg/m3. You can be sure that the ring is NOT solid gold.

9. Crutches will often have rubber tips at the bottom, so that contact with a flooring surface will decrease scratching of the floor, increase friction, and its broader tip spreads the pressure out over a larger area. Assume the radius of the tip of the cane to be 1.2 cm, and that the radius of the rubber tip is 2.5 cm. What is the ratio of the pressure of the cane alone, P1, to the pressure with the rubber tip, P2: P1/P2.
Solution:
Pressure = Force/Area
Area p r2
Area 1 = p r2
Area 2 = p (2.5/1.2)2r2 =p (2.08)2r2
Thus, Area 2 is 4.34 greater than Area 1, and thus, the pressure in Area 2 is 1/(4.34) as great; in other words, is 0.23 the pressure. P1/P2. = 4.34.

13. Two drinking glasses, G1 and G2, are filled with water to the same level of 5.0 cm. The diameter of the base of G1 is twice the diameter of the base of G2, but otherwise the two glasses are identical.
a. Is the weight of the water in G1 greater, less, or the same as weight in G2?
b. Is the water pressure at the bottom of G1 greater, less, or the same as the water pressure at the bottom of G2?
Solution: There is more water in G1 than G2, but the same area
a. G1 has a greater weight.
b. Water pressure is identical.

19. A circular wine barrel is shown in figure 15-26 on page 530. The barrel has a diameter of 75 cm. If a net upward force exceeds 643 Newtons it will burst. A tube of 1.0 cm in diameter extends into the barrel through a hole in the top. Initially the barrel is filled to the top and the tube is empty above that level. What weight of water must be poured into the tube in order to burst the barrel? And why would anyone want to do that anyway?
Solution:
Pressure = force/area
Density = mass/volume
Force = m g = 643 N (in barrel)
W = weight of water
Force density = r g
PE = mgh
PEdensity = r g h
Pbarrel = Ptube
F/Abarrel = W/Atube
The weight of the water must be poured into the tube, such that the above holds true:

W = [(Atube)/(Abarrel)] F

Abarrel = p (0.375)2 = 0.4415625 m2 . (Barrel’s area)
Atube = p (0.005)2 = 0.0000785 m2 . (Tube’s area ). So,

W = [(Atube)/(Abarrel)] F = [(0.0000785)/(0.4415625)] (643 N) = [0.000178](643) = 0.114 N.

Or, W = 0.11 N.~ 1.1 x 10-1 N.

25. Referring to example 15-4 on page 507 in the book, suppose that some vegetable oil has been added to both sides of the U-tube. Not YouTube. On the right side of the tube the depth of oil is 5.0 cm as before. On the left side of the tube the depth is 3.0 cm. Find the difference in fluid level between the two sides of the tube.
Solution:
We know the density of the oil and the water, respectively:

roil = 9.2 x 102 kg/m3.
rwater = 103 kg/m3.

The pressure point at “C” is given by,

PC = Pat + rogh1 + rwgx

and the pressure point at “D” is given by,

PD = Pat + rogh2 + rwgy.

Now we must find the difference in fluid level between the two sides of the tube:

(y + h2) – (x + h1)

At equilibrium, PC = PD, or, Pat + rogh1 + rwgx = Pat + rogh2 + rwgy..This leads us to:

rogh1 + rwgx = rogh2 + rwgy. Because the atmospheric pressure, Pat, is on both sides.

An ordinary orangutan can also see that the acceleration of gravity, g, is in each term, so we can divide that out, and now end up with:

roh1 + rwx = roh2 + rwy. We can rewrite this as rw (x - y) = ro(h2 – h1).

(h2 – h1) = [(rw)/(ro)](x – y). Or we can invert it to be: (x – y) = [(ro)/(rw)](h2 – h1)].

Now, multiply both sides by negative one ( - 1), then add (h2 – h1) to both sides: Then we will have:

[(h2 – h1) - (x – y)] = (h2 – h1) - [(ro)/(rw)](h2 – h1)..And that can be factored to look like this:

(y + h2) - (x + h1) = (h2 – h1)[1 - (ro)/(rw)]. And this equals (5 cm – 3 cm)[1 - (9.2 x 102 kg/m3)/(103 kg/m3)] = (2 cm)[1 – 0.92] = (2 cm)(0.08) = 0.16 cm.~ 1.6 x 10-2 m.

29. On Wednesday, August 15, 1934, William Beebe and Otis Barton made history by descending in the Bathysphere, a steel sphere 4.75 feet in diameter, to 3028 feet below sea level.
a. as the Bathysphere was lowered, was the buoyant force exerted on it at a depth of 10 feet greater, less, or equal to it at 50 feet?
b. Choose the best explanation from below.
I Magic
II Miracle
III Mayonnaise.
Solution:
a. Same
b. III

31. A fish called Wanda is carrying a pebble in its mouth swims with a small constant velocity in a small bowl. When the fish drops the pebble to the bottom of the bowl, does the water level rise, fall or stay the same?
Solution: Stays the same.

39. A hydrometer is a device for measuring fluid density. It's constructed as shown in Figure 15-30 on page 532. If the hydrometer samples fluid 1, the small float inside the tube is submerged to level 1. When fluid 2 is sampled the float is submerged to level 2. Is the density of fluid1 greater, less, or the same as fluid 2's density? [This is how auto mechanics test the antifreeze level in your car engine. Since antifreeze (ethylene glycol) is denser than water, the higher the density of coolant in your radiator the more antifreeze protection the engine has.]
Solution:
How would I know? I'm not an auto mechanic. He he he.

A hydrometer is an instrument used to measure the relative density of liquids to water, which has a density of r = 103 kg/m3. It comes from two words, hydro, meaning “water,” and metros meaning “to measure.”
Hydrometers are typically made of glass, which, itself is a liquid. The device consists of a cylindrical stem and a bulb weighted with the element mercury (not the planet or the car or the Roman god). This allows the stem and bulb to float upright. The liquid to be tested (such as honey or antifreeze) is poured into a tall jar, and the hydrometer is gently lowered into the liquid – the honey or antifreeze, etc., until it floats freely. The point at which the surface of the liquid touches the stem of the hydrometer is noted. Hydrometers usually contain a paper scale inside the stem, so that the relative density can be read directly. The scales may be of different types, as hydrometer units have not yet been standardized by the SIU.
Hydrometers may be calibrated for different uses, such as a lactometer for measuring the density (creaminess) of milk, a saccharometer for measuring the density of sugar in a liquid, or an alcoholometer for measuring higher levels of alcohol in liquor and other distilled spirits.

Chapter 16: Problems 1, 3, 9, 13, 19, 25, 29, 31, 39, on pp 566-567

1. The official record for lowest temperature ever recorded on Earth was set at Vostok, Antarctica, on July 21, 1983. The temperature on that day fell to T = ( – 89.2° C), which is well below the temperature of dry ice (frozen CO2). Give this temperature in degrees Fahrenheit.
Solution: We use the “famous” equation that °F = (9/5) °C + 32°. So, let's “plug and chug” and that will be it: °F = (9/5) °C + 32° = (9/5)(-89.2°C) + 32° = (1.8)(-89.2°C) + 32° = (- 160.56°) + 32° =
- 128.56 °F.~ - 129 °F.

3.Normal body temperature for humans is T = 98.6° F. Convert this temperature to degrees:
a. Celsius (aka Centrigrade)
b. Kelvin
Solution: We use the “famous” equation that °C = (5/9)(°F – 32°).
a. °C = (5/9)(°F – 32°) = (5/9)(98.6° – 32°). = (0.556)(98.6° – 32°). = (0.556)(66.6) = 37°C .
b. Using °K = °C + 273, then °C + 273 = 37 + 273 = 310°K.

9. A world record for the greatest change in temperature was set in Spearfish, North Dakota on January 22, 1943. At 7:30 AM, the temperature was T = ( - 4° F); two minutes later, at 7:32 AM, the temperature was T =+ 45° F. Calculate the mean rate of temperature change, per second, using Celsius or Kelvin scales (they use the same units); Hint: find DT.
Solution: We use the “famous” equation that °C = (5/9)(°F – 32°). At the lowest temperature, °C = (5/9)(- 4° – 32°). = (0.5555)( - 36) = - 20°C. At the highest temperature, °C = (5/9)(45° – 32°). = (0.5555)(13) = 7.2°C. Then, DT. = [7.2 - ( -20)] = 27.2 °C. in two minutes. Or, 27.2 / (120 sec) =
+ 0.227 °C/s.

13. Referring to Table 16 - 1 on page 545, which would be more accurate for an all-season, outdoor use: a tape measure made of steel or one made of aluminum?
Solution: Steel (iron) because it will shrink and expand less.

19. It is desired to slip an Aluminum ring over a steel bar as shown in Figure 16-14 on page 567. At T = 10.00° C, the inside diameter of the Aluminum ring is da = 4.000 cm, and the diameter of the steel rod is ds = 4.040 cm. The steel bar maintains a constant temperature of Ts = 10.0° C.
a. In order for the ring to slip over the bar, should the ring be heated, or cooled? Explain.
b. Find the temperature, Ta, of the ring at which it fits over the over the bar.
c. Who is the Lord of the Rings?
Solution: In this case, we must consult the coefficients of linear expansion, or, a. If the aluminum expands at a faster rate than the steel as we heat them, then we should heat the aluminum. Otherwise, cool it.
a. a(Al) = 24 x 10- 6 per degree K; and a(Steel) = 12 x 10- 6 per degree K. This means that Aluminum expands faster, so we heat the aluminum.
b. The Aluminum ring will NOT fit over the Steel bar at this temperature (10° C), so we need it to expand to exceed the diameter of the steel, i.e. Da > ds. So, Da > 4.040 cm, such that Da = (4.000 cm )+ Dd. According to the linear expansion formula, DL = aL0DT. So, for this problem, Dd = a(da DT). And, of course, Dd = ds – da = 0.040 cm; DT = Tf – Ti = Tf - 10.00° C. Where We are looking for Tf. So, let's re-write this as:

(Dd)/ (ada ) = DT = (Tf – Ti). And now, we can re-arrange that to get Tf.

Tf = Ti + (Dd)/ (ada ) = 10.00° C + (0.040 cm) / [(24 x 10- 6)(4.000 cm)] =

10.00° C + (0.040 cm) / [(96 x 10- 6)] = 10.00° C + (4.17 x 102 °C) = 427°C.

c. Exactly.

25. An aluminum saucepan with a diameter of d = 25.0 cm and a height of h = 6.0 cm, is filled “to the brim” with H2O. The initial temperature of the pan and water is T = 19° C. But then the pan with the water in it is placed on a range where it is heated to T = 88° C.
a. Will the water overflow from the pan, or will it decrease? Explain.
b. Calculate the volume of water that either overflows or decreases.
c. Go make some popcorn. Turn off the stove.
Solution: In this case, we must consult the coefficients of linear expansion for Aluminum, or, a. and volumetric expansion for water, or b. If the aluminum expands at a faster rate than the water as we heat them, then the water level will go down. Otherwise, it will overflow. We will use both DL = aL0DT for Aluminum, and DV = bV0DT for water.
a. a(Al) = 24 x 10- 6 per degree K; and b(water) = 0.21 x 10- 3 per degree K. This means that the water will expand faster in volume than the Aluminum will in linearity. The pot will overflow. By 24 cm3 .
b. The volume of the pan (and thus, the water) is the area of the pot, which is the area of the circle of the pan, multiplied by its height, or, (h)(p r2) where r = ½ d. Thus, the volume is:

V = (h)(p r2) = (6.0 cm)(p)(12.5)2 = (6.0 cm)(3.14)(156.25) = (6.0 cm)(3.14)(156.25) = 2945 cm3. So, this is the “original” volume, or V0. DT = 88° C – 19° C = 69° C.

Thus, DV(water) = bV0DT = (0.21 x 10- 3 )(2945)(69) = 42.7 cm3.

But the Aluminum pot will expand, too, by:

DV(Al) = 3aV0DT = (3)(2.4 x 10- 5)(2.945 x 103)(6.9 x 101) =

(3)(2.4)(2.945)(6.9)(10- 5)(103)(101) = 146.31 x 10- 1. = 14.63 cm3.

Finally, overall change of volume is [DV(water) - DV(Al)] = (42.7) - (14.63) = 28 cm3.

c. Okay.

29. Consider the apparatus the Joule used in his experiments on the mechanical equivalent of heat, shown in Figure 16-8 on page 550. Suppose both blocks have a mass of m1 = m2 = 0.95 kg and that they fall through a distance of h = 0.48 meter.
a. Find the expected rise in temperature, DT, of the water, given that 6200 Joules are needed for every increase of DT = 1.0° C. Give your answer in degrees Celsius (°C).
b. Would the temperature rise in degrees Fahrenheit (°F) be greater than, or less than, the result in part (a)? Explain.
c. Find the rise in temperature, +DT, in Fahrenheit degrees (°F).
Solution: In each case, the stored gravitational energy is PE = m g h = (0.95 kg)(9.8 m/s2)(0.48 m) = 4.4688 J. Since there are two of them, then the PE = 2(4.4688) = 8.9376 J.
a. If all the PE were converted to heat (which it is not), then 8.9376 J of heat went into the water. And if 6,200 J are needed for +1.0° C, then the rise is +[(8.9376) / (6200)](1.0° C) = + 0.00144° C which is negligible.
b. Greater.
c. Using the “famous” relationship: we find that each degree rise in Celsius is the same as each 1.8° rise in Fahrenheit. So, +DT(Fahrenheit) = (1.8)(0.00144) = + 0.002595° F.

31. Two objects are made of the same material, but have different temperatures, T1 and T2. Object 1 has a mass of “m” while object two has a mass of “2m.” If the objects are brought into thermal contact,
a. Is the temperature change of object 1, DT1, greater, lesser, or equal to the temperature change of object 2, DT2?
b. Choose the best explanation from among the following three choices:
I The larger object gives up more heat and therefore its temperature change is greatest
II The heat given up by one object is taken up by the other. Since the objects have the same heat capacity the temperature changes are the same
III One object loses heat of magnitude Q the other gains heat of magnitude Q. With the same magnitude of heat involved the smaller object has the greater temperature change.
Solution: DQ = m C DT, this is the change of heat energy relative to Temperature. In this case, Q is heat energy; m is the mass; C is the specific heat capacity, and you know T.
a. The smaller one.
b. III

39. If 2200 Joules of heat are added to a 190-gram object, its temperature increases by +DT = 12° C.
a. Find the heat capacity of the object.
b. Find the specific heat of the object.
Solution: We already know that DQ = m C DT, so...
a. DQ = m C DT, and we have DQ, m, and DT. We need only to find C: Let's re-write to isolate C:

C = (DQ) / (m)(DT) = (2200 J) / [(0.190 kg)(12)] = (2200 J) / (2.28) ~ 965 Joules / kg-deg = 9.65 x 102 kg-deg.

b. The same, 9.65 x 102 kg-deg..

END

HOMEWORK SET 4

2010-08-20T06:35:00.000-07:00

Chapter 15: Problems 1, 3, 9, 13, 19, 25, 29, 31, 39, on pp 530-534.

1. Find the weight (in Newtons) of the air in your physics class. Assume that the room is a cube of 4 meters on a side, and that the density of air is rair = 1.29 kg/m3.

3. You buy a gold ring at a pawn shop, and run a “test” to determine if it's pure gold. The ring's mass is m = 0.014 grams, and has a volume of 0.022 cm3. The density of gold is rAu = 19.3 kg/m3.

9. Crutches will often have rubber tips at the bottom, so that contact with a flooring surface will decrease scratching of the floor, increase friction, and its broader tip spreads the pressure out over a larger area. Assume the radius of the tip of the cane to be 1.2 cm, and that the radius of the rubber tip is 2.5 cm. What is the ratio of the pressure of the cane alone, P1, to the pressure with the rubber tip, P2: P1/P2.

13. Two drinking glasses, G1 and G2, are filled with water to the same level of 5.0 cm. The diameter of the base of G1 is twice the diameter of the base of G2, but otherwise the two glasses are identical.
a. Is the weight of the water in G1 greater, less, or the same as weight in G2?
b. Is the water pressure at the bottom of G1 greater, less, or the same as the water pressure at the bottom of G2?

19. A circular wine barrel is shown in figure 15-26 on page 530. The barrel has a diameter of 75 cm. If a net upward force exceeds 643 Newtons it will burst. A tube of 1.0 cm in diameter extends into the barrel through a hole in the top. Initially the barrel is filled to the top and the tube is empty above that level. What weight of water must be poured into the tube in order to burst the barrel? And why would anyone want to do that anyway?

25. Referring to example 15-4 on page 507 in the book, suppose that some vegetable oil has been added to both sides of the U-tube. Not YouTube. On the right side of the tube the depth of oil is 5.0 cm as before. On the left side of the tube the depth is 3.0 cm. Find the difference in fluid level between the two sides of the tube.

29. On Wednesday, August 15, 1934, William Beebe and Otis Barton made history by descending in the Bathysphere, a steel sphere 4.75 feet in diameter, to 3028 feet below sea level.
a. as the Bathysphere was lowered, was the buoyant force exerted on it at a depth of 10 feet greater, less, or equal to it at 50 feet?
b. Choose the best explanation from below.
I Magic
II Miracle
III Mayonnaise.

31. A fish called Wanda is carrying a pebble in its mouth swims with a small constant velocity in a small bowl. When the fish drops the pebble to the bottom of the bowl, does the water level rise, fall or stay the same?
39. A hydrometer is a device for measuring fluid density. It's constructed as shown in Figure 15-30 on page 532. If the hydrometer samples fluid 1, the small float inside the tube is submerged to level 1. When fluid 2 is sampled the float is submerged to level 2. Is the density of fluid1 greater, less, or the same as fluid 2's density? [This is how auto mechanics test the antifreeze level in your car engine. Since antifreeze (ethylene glycol) is denser than water, the higher the density of coolant in your radiator the more antifreeze protection the engine has.]

Chapter 16: Problems 1, 3, 9, 13, 19, 25, 29, 31, 39, on pp 566-567.

1. The official record for lowest temperature ever recorded on Earth was set at Vostok, Antarctica, on July 21, 1983. The temperature on that day fell to T = ( – 89.2° C), which is well below the temperature of dry ice (frozen CO2). Give this temperature in degrees Fahrenheit.

3.Normal body temperature for humans is T = 98.6° F. Convert this temperature to degrees:
a. Celsius (aka Centrigrade)
b. Kelvin

9. A world record for the greatest change in temperature was set in Spearfish, North Dakota on January 22, 1943. At 7:30 AM, the temperature was T = ( - 4° F); two minutes later, at 7:32 AM, the temperature was T =+ 45° F. Calculate the mean rate of temperature change, per second, using Celsius or Kelvin scales (they use the same units); Hint: find DT.

13. Referring to Table 16 - 1 on page 545, which would be more accurate for an all-season, outdoor use: a tape measure made of steel or one made of aluminum?

19. It is desired to slip an aluminum ring over a steel bar as shown in Figure 16-14 on page 567. At T = 10.00° C, the inside diameter of the ring is d = 4.000 cm. The bar maintains a constant temperature of T = 10.0° C.
a. In order for the ring to slip over the bar, should the ring be heated, or cooled? Explain.
b. Find the temperature, T, of the ring at which it fits over the over the bar.
c. Who is the Lord of the Rings?

25. An aluminum saucepan with a diameter of d = 25.0 cm and a height of h = 6.0 cm, is filled “to the brim” with H2O. The initial temperature of the pan and water is T = 19° C. But then the pan with the water in it is placed on a range where it is heated to T = 88° C.
a. Will the water overflow from the pan, or will it decrease? Explain.
b. Calculate the volume of water that either overflows or decreases.
c. Go make some popcorn. Turn off the stove.

29. Consider the apparatus the Joule used in his experiments on the mechanical equivalent of heat, shown in Figure 16-8 on page 550. Suppose both blocks have a mass of m1 = m2 = 0.95 kg and that they fall through a distance of h = 0.48 meter.
a. Find the expected rise in temperature, DT, of the water, given that 6200 Joules are needed for every increase of DT = 1.0° C. Give your answer in degrees Celsius (°C).
b. Would the temperature rise in degrees Fahrenheit (°F) be greater than, or less than, the result in part (a)? Explain.
c. Find the rise in temperature, +DT, in Fahrenheit degrees (°F).

31. Two objects are made of the same material, but have different temperatures, T1 and T2. Object 1 has a mass of “m” while object two has a mass of “2m.” If the objects are brought into thermal contact,
a. Is the temperature change of object 1, DT1, greater, lesser, or equal to the temperature change of object 2, DT2?
b. Choose the best explanation from among the following three choices:
I The larger object gives up more heat and therefore its temperature change is greatest
II The heat given up by one object is taken up by the other. Since the objects have the same heat capacity the temperature changes are the same
III One object loses heat of magnitude Q the other gains heat of magnitude Q. With the same magnitude of heat involved the smaller object has the greater temperature change.

39. If 2200 Joules of heat are added to a 190-gram object, its temperature increases by +DT = 12° C.
a. Find the heat capacity of the object.
b. Find the specific heat of the object.

END

LESSON 9

2010-08-19T19:10:00.000-07:00

PHYS 1100 LESSON 9 FOR
THURSDAY, AUGUST 19, 2010

I Introduction

II Logistics – Running Grades

III Turn in Assignments Due: Homework Sets 1, 2, 3. Essays 1, 2, 3, etc. and Return of Papers

IV Test 3

V Mind Game 3

VI Review of Lesson 8: Inertia, Torque, Gravity
A. Inertia is kg m2.
B. Torque is N-m
C. Gravity:

1. Kepler – P2 (m + M) = [(4 p2)/G] r3.

This is Kepler’s 3rd law dealing with the period of a planet, P, with a mass, m, traveling around the Sun, with a mass, M, at a distance between the Sun and the planet equaling “r”. However, it applies to moons, asteroids, stars, satellites, blah blah.

Period is in seconds, mass in kilograms, distance in meters (mks). The Universal Gravitational Constant is G = 6.67 x 10-11 N m2 / kg2.

2. Newton – F = GmM/r2.

This part of Newton’s 2nd Law, which deals with the force, F, of gravity.

Little ‘m’ is the mass of a smaller object, such as human; big “M” is the mass of the larger object, such as planet Earth; r is the distance between the center of the first object (like a human’s navel) and the center of the second object (like the distance from the human to Earth’s core, i.e., the radius of Earth.

As a force, instead of a period, this law applies both to moving objects, like planets, moons, stars; and to object that are not moving, like people standing on Earth, or between two bowling balls in a bowling alley.

The Universal Gravitational Constant is G = 6.67 x 10-11 N m2 / kg2.

VII Lesson 9: Waves and Fluids
A. Waves
1. Length
2. Frequency
3. Period
4. Amplitude
5. Speed
6. Types
7. Other

B. Fluid Dynamics
1. Definition of “fluid”
2. And, defn of “dynamics”
3. Density
4. Pressure
5. Bernoulli’s Law
6. Archimede’s Principle
7. And more….

VIII Laboratory Exercise 9: Kepler’s Laws

IX HWK Assignment 4: Read Chapters 15-16 and do Problems 1, 3, 9, 13, 19, 25, 29, 31, 39, 53, 63, 65, and 67 on pp 530-534 and do Problems 1, 3, 9, 13, 19, 25, 29, 31, 39, 53, 63, 65, and 67 on pp 566-569; Due 8/26

X Essay 4: Albert Einstein, due 8/26

LAB 9 for Thursday, August 19

2010-08-18T14:56:00.001-07:00

PhysicsLab9 Thursday, August 19, 2010 Name __________________
Dr Dave Menke, Instructor

I Title: Applying Kepler's Laws to Planetary Orbits

II Purpose: To determine the distance, orbital period, and velocity of the planets given, using Kepler's laws of Motion. To determine radio communication wait time in space.

Background information:
The German genius physicist, Herr Johannes Kepler (1571-1630), determined his three laws of motion in separate research, and they were published separately. However, one of his laws is P2 = k a3, where P is the period of revolution of the object; "k" is a constant, and the letter "a" represents the semi-major axis of the planet's elliptical orbit (kind of an average). If the orbit were a circle, then "a" would be the radius of that circle.

The constant, k, is equal to the number 1.00, but ONLY if the period, P, is in Earth Years; AND if the semi-major axis, "a," is given in astronomical units (AU). Otherwise, the constant
k = [4p2 /G]/(M + m), where "G" is the constant of universal gravitation, [G = 6.673 x 10-11 N-m2 /kg2 .] The letter "M" refers to the mass of the heavier object (Sun) and the letter "m" refers to the mass of the lighter object (planet) in kilograms. The units of period, P, would be in seconds. The letter "N" is a unit of forced called a "Newton."

One astronomical unit, AU, equals 150,000,000 kilometers (approximately). The speed of light, "c," is equal to 300,000 km/sec (approximately).

III Equipment -- Each Student Will Have:
1. pen, pencil, or quill;
2. calculator, abacus, or mainframe computer;
3. personal hard drive (brain);
4. personal printer (hand at end of arm);
5. graph paper
6. etc.

IV Procedure
1. Given the value of an astronomical unit, and both the perihelion and aphelion distances for several planets (in the DATA TABLE below), find for each planet given:
a. the semi-major axis (a) of its elliptical orbit, in AU’s;
b. the period of revolution about the Sun, P, in years;
c. the average velocity of all the planets listed, around the Sun (in this case,
assume all have circular orbits).

2. Pretend that we have traveled to Ganymede (a large moon of Jupiter). At that
instant, if you are given the distance from Earth to Ganymede, find out how long it will take for a radio signal to reach Earth. (The distance to Ganymede will be similar, but not exact, to the distance to Jupiter. Consult the Data section below).

MORE…

Physics Lab 9, page 2

3. With the same conditions, calculate the typical response time from an
astronaut on Ganymede to Houston's Johnson Space Flight Center. (Round trip wait time).

V Data & Calculations
Below is a table of the distances that the indicated planets are from the Sun, at their closest point, perihelion, and at their most distant point, aphelion. The units are in Astronomical Units, also known as A.U.'s, where 1.0 A.U. = 149,500,000 kilometers = 1.49 x 1011 meters, which is the average distance that Earth is from the Sun.

Planet Perihelion (AU) Aphelion (AU) SemiMajor, a (AU) Period, P (years)
Mercury 0.31 0.47
Venus 0.72 0.73
Earth 0.98 1.02
Mars 1.39 1.66
Jupiter 4.95 5.46

[Perihelion + Aphelion]/2 = a = semi-major axis

2pr = circumference

velocity = 2pr/P = circumference/Period

VI Results
The purpose was / was not achieved because

VII Error Analysis
A. Quantitative Error
Find the Average Percent Error. Do this by first finding the percent error for each of the five planets, i.e., find the percent error for Mercury, then for Venus, etc., and then add all five percent errors and divide by 5 to get Average Percent Error.

Percent error = [|True – Yours| / True ] x 100% =

What is the “truth” as far as periods go? Look it up on Wikipedia.org or some other source, like an astronomy book. Or, ask your instructor

B. Qualitative Error
Sources of error are enumerated as:
Personal –
Systematic –
Random --

VIII Questions
1. Find the average orbital velocity of each of the above planets, in km/sec. To do this, you must assume the orbits are circles in the first order approximation.

2. Pretend that we have traveled to Ganymede (a large moon of Jupiter). At that instant, if the distance from Earth to Ganymede is 5.38 AU, find out how long it will take for a radio signal to reach Earth.

3. With the same conditions, calculate the typical response time from an astronaut on Ganymede to Houston’s Johnson Space Flight Center. (Round trip wait time).

4. Use the Copernican equation (below) to determine the Sidereal (true) Period of revolution, P, from the, Synodic (observed) period, S, for Jupiter and for Saturn. The Synodic Period of Jupiter and Saturn are 13 months and 12.5 months respectively.

1/P = 1- (1/S)

SOLUTION SET 3 - COMPLETE

2010-08-18T14:54:00.001-07:00

PHYSICS 1100
SOLUTION SET 3
Problems and Conceptual Exercises

Chapter 10: Problems 1, 3, 9, 13, 19, 25, 29, 31, 39, 53, 63, and 65 on pp 323-327

1.Convert the following degree angles to radians:
a. 30°
b. 45°
c. 90°
d. 180°
Solution:
Since, by definition, 2 p radians = 360° , then half 360° = 180° would be p, and half that, 90°, would be p/2, and 45° is half of 90° so, 45° = p/4; and 30° is 1/3 of 90°, so 30° would be 1/3 of p/2 = p/6.
a. p/6
b. p/4
c. p/2
d. p

3. Find the angular speed (aka angular velocity, or, w) of
a. the minute hand of Big Ben in London
b. the hour hand of same.
Solution:
a. the minute hand of Big Ben, or any other clock, makes 360° or 2 p radians every hour, or, every 3,600 seconds. So, its angular velocity is w = 2 p radians / 3600 seconds = (2)(3.14)/3600 =
1.74 x 10-3 radians/sec.

9. The Crab Nebula has a pulsar (pulsating neutron star) embedded in it, which rotates every 33 ms. Find the angular velocity, w, of this pulsar in radians/sec.
Solution:
The period of rotation, P = 33 ms = 3.3 x 10-5 sec; frequency is n = 1/P = 1/(3.3 x 10-5 sec) = 3.03 x 104 Hz.; and angular velocity is w = 2 p n = (2)(3.14)(3.03 x 104) = 1.9 x 105 rad/s.

13. An object at rest begins to rotate with a constant angular acceleration, a. After a time, t, the object has an angular velocity of w. What was the angular velocity at ½ t?
Solution:
w = a t, so at half the time, w = ½ a t.

19. A ceiling fan is rotating at 0.96 revolutions per second. When turned off, it slows uniformly to a stop in 2.4 minutes.
a. how many revolutions does it do in this 2.4 minutes?
b. Find the number of revolutions the fan must make for its speed to decrease from 0.96 rev/s to 0.48 rev/sec.
Solution:
Initially it has a radial velocity of w i = (0.96)(2p ) radians in 1.0 sec = (0.96)(6.28)/(1.0 sec) = 6.0288 rad/sec. Then, 2.4 minutes (144 sec) later, the fan has stopped, and its radial velocity is w f = 0.0 rad/s. The angular acceleration here (actually, a deceleration) is a = (w f - w i)/Dt = (0.0 - 6.0288)/(144 sec) = - 4.19 x 10-2 rad/s2. It's negative, or a deceleration.

We can now use the angle equation, for any angle, Ɵ, this is true:

Θ = Θ0 + w0 t + ½ a t2 = (in radians)
a. How many revolutions (each revolution is 2 p radians) in 2.4 min (144 sec)? Assuming the initial angle is zero radians, i.e., Θ0 = 0.0 radians, then:

Θ = 0 + (6.0288)(144s) + (0.5)(- 4.19 x 10-2)(144)2 = 0 + 868.1472 - 434.4192 = 433.728 radians. Dividing this by 2 p radians will give us the number of revolutions:

433.728 radians / 2 p radians per rev = 69.06 revs.

b. We can now use the angular velocity equation, this is true:

(w f 2 - w I2 ) = 2 a Θ , or if we re-write this, we will have:

Θ = (w f 2 - w I2 )/(2 a) = [(0.48)2 – (0.96)2] /(2)(- 4.19 x 10-2) = [(0.2304) – (0.9216)] / (- 8.38 x 10-2) = (- 0.6912)/(-8.38 x 10-2) = + 0.0825 x 102
= 8.25 radians, or 1.31 revolutions.

25. When a carpenter shuts off his circular saw, the 10.0-inch (25.4 cm) diameter blade slows from 4440 rpm to 0.00 rpm in 2.50 seconds.
a. Find the angular acceleration, a, of the blade.
b. What's the distance traveled by a point on the rim of the blade during the deceleration?
c. What is the magnitude of the net displacement of a point on the rim of the blade during the deceleration?
Solution:
a. The angular acceleration is equal to the final angular velocity minus the initial angular velocity, all divided by the amount of time it takes to stop, or, a = (w f - w i)/Dt . We know that the final angular velocity, w f = 0.00 rad/s because it is stopped. But what is the initial radial velocity, w i ? Well, 1.0 revolution per minute, or, 1 rpm = 2 p radians per minute, or, 6.28 radians per minute. And that is the same as 6.28/60 seconds = 1.05 x 10-1 radian per second. Therefore, 4,440 rpm would equal (4440)(1.05 x 10-1) = 464.72 rad/s. And that is the initial angular velocity, w i.
We go back to the relationship, a = (w f - w i)/Dt, to find: a = (0.0 rad/s – 464.72 rad/sec)/(2.50 sec) = - 185.888 rad/s2, or ( - 1.86 x 102 rad/s2). It is negative, because it is slowing down, or, decelerating.

29. When standing at the top of a tall building,
a. Is your angular velocity, w, greater, less, or the same as if you were at sea level?
b. What is the best explanation?
I. The angular velocity is the same no matter one's height
II. On the top of the building, one is further from the axis of rotation
III. You spin faster closer to Earth's center.
Solution:
a. Since the top of the tall building has to make one complete circle of 360° in 24 hours, it has to have the same radial velocity as the ground.
b. #I

31. Two kids ride on the merry-go-round shown in Conceptual Check point 10-1 on page 306 in the book. Sally is 2.0 meters from the axis of rotation while Sue is 1.5 meters from same. If the merry-go-round completes one rev per 4.5 s,
a. Find the angular velocity, w, of each kid; v = w r
b. Find the linear velocity, v, of each kid
Solution:
a. Sally and Sue have the same angular velocity, where w = 2 p n, where n = 1/P, and P = 4.5 s, or, = (2)(3.14) /(4.5) = 1.4 rad/s.

39. A Ferris wheel with a radius of 9.5 m rotates at a constant rate, completing one revolution every 36 seconds. Find the direction and magnitude of a passenger's acceleration...
a. at the top
b. at the bottom
Solution:
First of all, the acceleration is radially, towards the center of the Ferris wheel at all times, but it would be greater at the top since you add the acceleration of gravity to it, and less at the bottom as you subtract the acceleration of gravity. And, of course, a = v2/r. We know r = 9.5 meters. But to find v, we invoke the relationship v = w r. If a = v2/r, then we replace v with w r and we have: a = (w r)2/r = w2 r. But w = 2 p n = 2 p / P = (2)(3.14)/(36s) = 1.74 rad/s. And thus, w2 r = (1.74)2(9.5) = 28.9 m/s2.
a. So, at the top, the acceleration is 28.9 + 9.8 = 38.7 m/s2 radially.
b. At the bottom, the acceleration is 28.9 – 9.8 = 19.1 m/s2 radially.

53. Tons of interplanetary dust and debris fall into Earth's atmosphere every day. What is the effect on Earth's Moment of Inertia?
Solution:
The moment of inertia of a homogenous sphere is I = 2/5 M R2. A ton of debris is about 900 Kg, so, let's say 3000 kg fall each day. Earth is 6 x 1024 kg, and a few tons is 3 x 103 kg. It's like 3 x 10-18%, so the effect is minimal.

63. Find the rate at which the rotational K.E. of Earth is decreasing. The “rate” of energy means energy over time, or, Joules/sec = watts. Earth's Moment of Inertia is 0.331 MÅ RÅ2, where MÅ = 6 x 1024 kg and R¿ = 6.4 x 106 m. Its period of rotation increases 2.3 x 10-3 seconds every century. (The iconic symbol for “Earth” is Å).
Solution:
Rotational K.E. is half the moment of inertia times radial velocity squared, or, K.E. = ½ I w2. So, the Earth's K.E. is the following: K.E. = ½ (0.331 MÅ RÅ2)w2 = ½ (0.331 MÅ RÅ2)(2p/P)2. Unless the Earth's mass or radius changes a great deal (see problem 53), the only thing different in this equation in 100 years from now will be the period, P.

Now the period is Pi. In 100 years, the period will be Pf = Pi + 0.0023 sec. Thus, K.E.f – K.E.i
= [(½) (0.331 MÅ RÅ2)(2p)2][(1/Pf)2 – (1/Pi)2].
Currently, the Earth rotates with a period of Pi = 86164.09053s and in 100 years from now, it will be Pf = Pi + 0.0023 sec = 86164.09053s + 0.0023 = 86164.09283s.

So, (Pf)2 = (86164.09283s)2= 7,424,250,893, and 1/(Pf)2 = 1.346937239 x 10-10.

And, (Pi)2 = (86164.09053s)2= 7,424,250,497, and 1/(Pf)2 = 1.346937119 x 10-10.

Thus, [(1/Pf)2 – (1/Pi)2] = [(1.346937239) - (1.346937119)] x 10-10 = 0.00000012 x 10-10 = 1.2 x 10-17.

And therefore, K.E.f – K.E.i = [(½) (0.331 MÅ RÅ2)(2p)2](1.2 x 10-17).

But, what is [(½) (0.331 M¿ R¿2)(2p)2]? Well, [(½) (0.331 M¿ R¿2)(2p)2] = [(0.5)(0.331)(6 x 1024)(6.4 x 106)2(2p)2] = [(0.5)(0.331)(6 x 1024)(40.96 x 1012)(39.4384)] = [(0.5)(0.331)(6)(1024)(40.96)(1012)(39.4384)] = [(0.5)(0.331)(6)(1024)(40.96)(1012)(39.4384)] = [(1604)(1036)] = 1.604 x 1033

And, thus, my dear friends, K.E.f – K.E.i = (1.604 x 1033)(1.2 x 10-17) = 1.925 x 1016Joules.

But we are looking for rate of loss, so we need to find Power, P = DK.E. / 1 century = (1.925 x 1016Joules)/(100 years). And how many seconds are in 100 years? There are about 86164.09053 s in a day, and 365.2422 days in a year, so, in 1 year = (86164.09053s/d)(365.2422 days d/y) = 31,470,761.99 seconds in one year, and 3,147,076,199 in 100 years.

Finally, the rate, or power is P = (1.925 x 1016Joules)/(3.147076199 x 109 sec) = 0.612908589 x 107 watts. = 6.129 x 106 watts, or 6,129 KiloWatts.

65. Consider the physical situation shown in “Conceptual Checkpoint 10-5” on page 318 in the book. Suppose this time a “sphere” is released from rest on the frictionless surface. When the ball comes to rest on the no slip surface, is its height greater, less or equal to the height from which it was released?
Solution:
Less, because it loses energy as it rolls up the friction surface.

Chapter 11: Problems 1, 3, 9, 13, 19, 25, 29, 31, 39, 53, 63, 65 and 77 on pp 366-371

1. To tighten a spark plug, it's recommended that a torque of 15 N m be applied. If a mechanic tightens the spark plug with a wrench that is 25 cm long, what is the force needed?
Solution:
We know that torque, t, is equal to the force applied, F, multiplied by the length of the lever, or, length of the moment arm, or, length of the handle, or whatever you want to call it, r. the relationship looks like this:
t = r F

Here, we are asked to find the force, F. If we re-write the relationship above, we have:

F = t/r. Do we know the torque, t? Yes, it is given as 15 N m. Do we know the length of the moment arm, r? Yes, it is given as 25 cm = 0.25 m. So all we have to do is “plug and chug” to get the answer, and we don't have to search for clues to find anything else:

F = t/r = (15)/(0.25) = 60 N.
3. A bowling trophy of mass 1.61 kg is held at arm's length, a distance of 0.65 m from the shoulder joint. What torque does the trophy exert about the shoulder if the arm is
a. horizontal
b. 22.5° below the horizontal?
Solution:
In reality, torque is the length of the moment arm, r, multiplied by the force, F, then multiplied by the sine of the angle between them, Θ. Or, in other ways of putting it,

t = r F sin Θ. If it is a right angle, Θ = 90°, then the sin 90° = 1.0, and we don't worry about it.
a. t = r F = (0.65 m)(1.61)(9.8) = 10.2557 N-m, or, 10.3 N-m.
b. Since the angle is 22.5°, or (90° – 22.5°) from the axis, then the torque is t = r F sin Θ = 10.3 sin 67.5° = 9.52 N.

9. Suppose a torque rotates your body about one of three different axes of rotation: case A, an axis through your spine; case B, an axis through your hips; and case C, an axis through your ankles. Rank these three axes of rotation in increasing order of the angular acceleration produced by the torque. Indicate ties where appropriate.
Solution:
The least torque will be through the spine, the greatest through the ankles.

13. A person holds a ladder horizontally at its center. Treating the ladder as a uniform rod of length 3.15 meter and mass of 8.42 kg, find the torque the person must exert on the ladder to give it an angular acceleration of 0.302 radians/sec2.
Solution:
One of our relationships is t = I a, and to find torque, we need the moment of inertia, I, and the angular acceleration, a. Do we have them? Yes! The moment of inertia around a uniform rod of length 3.15 meter and mass of 8.42 kg, is I = 1/12 m L2, or, I = (0)(8.42)(3.15)2. = (0.083)(8.42)(9.9225) = 6.96 kg m2.

Now we can find the torque: t = I a = (6.96)(0.302) = 2.1 N m.

19. A fish takes the bait and pulls on the line with a force of 2.2 N. The fishing reel, which rotates without friction, is a cylinder of radius 0.055 m and mass of 0.99 kg.
a. What is the angular acceleration of the fishing reel?
b. How much line does the fish pull from the reel in ¼ sec.?
Solution:
a. We are looking for acceleration and can use: t = I a = r F, so a = r F/I. Since I = ½ MR2 for a cylinder, then = 2F/Mr = 4.4/(0.55) = 8 rad/sec2.
b. First we use Θ = Θ 0 + w0 t + ½ a t2 = (in radians), but initial angle and initial angular velocity are both 0.0. So, we rewrite it to be: Θ = ½ a t2 = (0.50)(8)(0.25)2 = 0.25 radian. Since distance traveled is s = r , then s = (0.055m)(0.25) = 0.01375 or about 0.014m.

25. Refer to the person holding the baseball in problem 5 on page 366. Suppose the biceps exert just enough upward force to help keep the system in static equilibrium.
a. is the force exerted by the biceps more, less or same as the combined weight of the forearm, hand and baseball? Explain.
b. Determine the force exerted by the biceps.
Solution:
a. No, but the torque is equal
b. 1000 N >Fbiceps > 1.0 N
lo
29. A remote control gadget of mass 0.0122 kg is 23.0 cm long and rests on a table, as shown in figure 11-28 on page 368. Part of the remote control unit, of length of L, is hanging over the table’s edge. In order to operate the power button on this remote control device requires a force of 0.365 N. How far can the remote control doohickey extend beyond the edge of the table and still not tip over when one presses its power button? Assume the mass of the remote it distributed uniformly, i.e., the density is homogenous and that the power button is 1.41 cm from the overhanging end of the remote.
Solution: We need to find the point between the left end of the remote, which is hanging over the edge at is at x0 = 0.0 cm, and the point to the right of that, at the edge of the table, x = L, which will be the point of axis (or, fulcrum, or center of mass) through which it rotates. The right end of the remote will be at x = 23.0 cm.

In order for this to be in equilibrium, all torques must add up to zero, i.e.,

St = 0. And remember torque is t = r F.

the torque to the left of that point is going to be t1 = ( ½ L) F1 + (L – 1.41)(F2) ccw, such that F1 = the force of gravity on that portion of the remote that is to the left of the fulcrum point. F2 = 0.365 N, the force needed to operate the button. We need to calculate the force of gravity, F1. By the way, ccw = counter clockwise. The distance that gravity acts through has to be halfway between x = 0.0 and x = L, so it's ½ L.

F1 = m1 g where the mass here, m1, is the portion of the remote to the left of the fulcrum. But how much is m1? The entire remote has a mass of M = 0.0122 kg, and is 23.0 cm long, of which L cm is to the left of the fulcrum. Thus, m1 = (L/23)0.0122 kg; and m2 = [(23 – L)/ (23)](0.0122 kg).

Therefore, F1 = m1 g = [(L/23)(0.0122)] g. and its portion of the torque will be ( ½ L) F1 = ( ½ L)([(L/23)(0.0122)] g). ccw.

The torque from the other force will be (L – 1.41)(F2) = (L – 1.41)(0.365 N).ccw. =

Thus, the net torque on the left will be

t1 = ( ½ L)([(L/23)(0.0122)] g) + (L – 1.41)(F2) = (L – 1.41)(0.365 N). ccw.

The torque from the remote to the right of the fulcrum will be t2 = (23.0 – L) F3, such that

F3 = m2 g. = ([( ½ )(23 – L)/ (23)](0.0122 kg) g. =

Thus, t2 = (23.0 – L)[([( ½ )(23 – L)/ (23)](0.0122 kg) g]. cw,of course. =

And for equilibrium, the torques would be equal (but opposite):

t1 = t2 , or,

( ½ L)([(L/23)(0.0122)] g) + (L – 1.41)(0.365 N) = (23.0 – L)[([( ½ )(23 – L)/ (23)](0.0122 kg) g],

Which, as one can obviously see, is quite easy to work out. Yeah right.

(L)([(L/23)(0.0122)] g) + 2(L – 1.41)(0.365 N) = (23.0 – L)[(23 – L)/ (23)](0.0122 kg) g],

(L)(L/23)(0.0122)(g) + 2L(0.365) - (1.41)(0.365 N) = [(23 – L)2/ (23)](0.0122 kg) g],

(L2)(0.0122)(g) + 2L(0.365)(23) - (1.41)(23)(0.365 N) = [(23 – L)2/ (0.0122 kg)(g)],

(L2)(0.0122)2(g) + 2L(0.365)(23)(0.0122 kg) - (0.0122 kg)(1.41)(23)(0.365 N) = [(23 – L)2 ](g)],

(L2)(0.0122)2(9.8) + 2L(0.365)(23)(0.0122) - (0.0122)(1.41)(23)(0.365 N) = [(23 – L)2 ](9.8)],

0.00146 (L2) + 0.205(L) – 0.144 = [232 – 46L + L2 ](9.8)]

0.00146 (L2) + 0.205(L) – 0.144 = (529 – 451L + 9.8L2 ); or

(9.8 – 0.00146)(L2) – 450.8 L + 528.86 = 0. For the sake of sanity, we re-write it as:

9.8(L2) – 451 L + 529 = 0; and divide by 9.8, again for simplicity:

(L2) – 46 L + 54 = 0.

quadratic:

L = [(46) +/- √(2116 – 216)] / 2 = [(46) +/- √(1900)] / 2 = [(46) +/- 43.6] / 2; gives two answers:

a. 2.4/2 = 1.2 cm, or
b. 89.6 / 2 = 44.8 cm.

Clearly, it has to be 1.2 cm. So the Remote can go just slightly less than that, or,

L < 1.2 cm.

There's gotta be a simpler way.

31.Repeat Example 11-4 (on page 342 in the book); this time with a uniform diving board that weighs 225 N. (In that Example, it says: A 5.0-meter diving board of negligible mass is supported by two pillars. One pillar is at the left end of the diving board, as shown in the picture here on page 342. The other is 1.50 meters away. Find the forces exerted by the pillars when a 90.0-kg diver stands at the far [right] end of the board).
Solution: First, in the Example 11-4 on page 342 in the book, F2 = 2940 N, and F1 = – 2060 N. Now that has changed with the addition of a board that has a weight of + 225 N. This is divided up as follows: DF2 + DF1 = 225 N where = +380 N and = - 150 N due to the geometry of the set up. That means there are now new values of F2 = +3320 N and F1 = -2210 N.

39. A uniform crate with a mass of 16.2 kilograms rests on the floor with a coefficient of friction of m = 0.571. The crate is a uniform cube with sides 1.21 meters in length.
a. what horizontal force applied to the top of the crate will start the crate to tip?
b. If the horizontal force is applied halfway to the top of the crate, it will begin to slip, before it tips. Explain.
Solution: The force of friction is: Ff = mN = m m g = (0.571)(16.2)(9.8) = 90.65 N.
a. So, if a force of F > 90.65 N is applied at the top, it will tip over.
b. Pushing at halfway results in a net torque of zero (0.0).

53. Calculate the angular momentum of the Earth around its own axis due to its daily motion. Assume a sphere with uniform density.

Solution: First, a sphere with uniform density has a moment of inertia of I = (2/5) M r2.
We know M = 6 x 1024 kg; we know r = 6.4 x 106 m; we know w = 2 p radians in 24 hours, or
w = (2 p) / (86,400 sec) = 7.2685 x 10-5 radians/sec. Will want to use this relationship: l = I w.

Then if l = I w = [(2/5) M r2 ] w then l = [(2/5)(6 x 1024 kg)(6.4 x 106 m)2 ] (7.2685 x 10-5 radians/sec) =

[(0.4)(6 x 1024 kg)(4.1 x 1013 m)] (7.2685 x 10-5 radians/sec) = [(0.4)(6)(4.1)(7.2685)][(1024)(1013)(10-5)]

= [7.15 x 101] [1032] = 7.15 x 1033 kg m2 /sec.

63. A puck (like, air puck; or hockey puck, etc.) on a horizontal frictionless surface is attached to a string that passes through a hole in the surface as shown in Figure 11-34 on page 370. As the puck rotates revolves around the hole (the author must be an idiot; the puck is revolving, not rotating, so I changed it), the string is pulled downward by a magical elf, bringing the puck closer to the hole.
a. During the process, does the puck’s linear speed increase, decrease, or remain constant?
b. During the process, does the puck’s angular speed increase, decrease, or remain constant?
c. During the process, does the puck’s angular momentum increase, decrease, or remain constant?
Solution:
a. increase
b. increase
c. decrease.

65.As an ice skater begins a spin, his angular speed is 3.17 rad/s. After pulling in his arms, his angular speed increases to 5.46 rad/s. Find the ratio of the skater’s final moment of inertia to his initial moment of inertia.
Solution: We know that l = I w and that angular momentum must be conserved, i.e., l i = l f.

So, l i = I i w i and also that l f. = I f w f . Plus, l i = l f.

So the ratio of the skater’s final moment of inertia to his initial moment of inertia will be:

(I f ) / (I i) = (w i ) / (w f ) = (3.17 rad/s) / (5.46 rad/s) = 0.58 = 5.8 x 10-1 .

77. How much work must be done to accelerate a baton from rest to an angular speed of 7.4 rad/s about its center? Consider the baton to be a uniform rod of length 0.53 m and a mass of 0.44 kg.
Solution: We know that K.E. = ½ I w2.
The initial KEi = 0.0 Joules, as nothing is moving. The final KEf = ½ I w f2. So the work required to do what would be DKE = ½ I w f2 since initial radial velocity is zero (at rest).

So, what is the moment of inertial for this object? From the chart, this is like a slender rod of length L, being rotated about its center. The moment of inertia for such an object is I = (1/12) M L2. =

I = (0.085)(0.44 kg)(0.53 m)2 = (0.083)(0.44 kg)(0.281 m2) = 0.0103 kg m2.

And w f = (7.4 rad/s)2 = 54.76 rad/s2. So, to find the final K.E., which is the work required:

Work = KEf = ½ I w f2. = (0.5)(0.0103 kg m2)(54.76 rad/s2) = 0.282 Joules = 2.82 x 10-1 Joule.

Chapter 12: Problems 1, 3, 9, 13, 19, 25, 29, 31, 39, 53 and 63 on pp 409-412

1. System A has masses of m & m candies. Separated by a distance, r. System B has masses of m and “2m” separated by a distance “2r.” System C has masses of 2m and 3m separated by a distance of 2r. And, finally, System D has masses of 4m and 5m separated by a distance of 3r. Rank these systems in order of increasing gravitational force. Indicate “ties” where appropriate.
Solution: System A has a Force divided by Gravitation constant of FA/G. In the same fashion, Systems B, C, and D will be FB/G, FC/G, and FD/G. The values of each of these will give us the answer, so let's start: Remember, the force of gravity is: F = G mM/r2.

System A: FA/G = mm/r2; = (m/r)2 ;

System B: FB/G = (m)(2m)/(2r)2; = (2 m2)/(4r2) = ½ (m/r)2; thus System B is half System A;

System C: FC/G = (2m)(3m)/(2r)2; = (5 m2)/(4r2) = (5/4)(m/r)2; System C is larger than A or B

System D: FD/G = (4m)(5m)/(3r)2; = (20 m2) /(9r2) = (20/9)(m/r)2; System D is larger than C;

So, D, C, A, B.

3.A 6.1-kg bowling ball and a 7.2-kg bowling sphere rest on a rack 0.75 m apart.
a. What is the force of gravity on each of the spheres by the other one?
b. At what distance is the force of gravity between the spheres equal to 2.0 x 10-9 N?
Solution: The force of gravity is: F = G mM/r2. = [(6.67 x 10-11)(6.1)(7.2)] / (0.75)2 =
(292.95)(10-11)/(0.5625) = (2.9295)(10-9)/(0.5625) = [(2.9295)/(0.5625)] (10-9) = 5.208 x 10-9 N ~
5.2 x 10-9 N.

9.When the Earth, Moon, and Sun are aligned in such a way as to make a right triangle, with the Moon at the right angle junction as shown in Figure 12-22 on page 409 in the book, the Moon is in a phase called “Third Quarter.” The arrangement in Figure 12-22 is as if from above Earth's North Pole. If you look carefully, you may see Santa Claus. Find the magnitude and direction of the net force exerted on the Moon. Give the direction relative to the line connecting the Moon and the Sun.
Solution: The force of gravity on the Moon from the Sun is: Fs = G m M / R2. Where the mass of the Moon is 'm.' The Sun's mass is M. The distance between the Sun and the Moon is R. The Sun's force is the + x direction.
The force of gravity on the Moon from the Earth is: Fm = G m M / r2. Where the mass of the Moon is 'm.' The Earth's mass is M. The distance between the Earth and the Moon is r. The Earth's force is the - y direction (the same as 270°
.
We thus know that the net force on the Moon will be in the 4th Quadrant. But in order to get the net force we must find the magnitude and the exact angle. We also must know the masses and distances, which are: M = 2 x 1027 kg for the Sun; M = 6 x 1024 kg for the Earth; and m = 7.22 x 1022 kg for the Moon. The distance from the Sun to the Moon (on average) is R = 1.5 x 1011 meters; and from Earth to the Moon (on average) is r = 3.85 x 108 meters. Remember that G = 6.67 x 10-11.

So, now we have to plug and chug to find each Force:

Fs = [(6.67 x 10-11)(7.22 x 1022)(2 x 1027)] / (1.5 x 1011)2 =

(6.67)(7.22)(2)(10-11)(1022)(1027) / (1.5)2 (1011)2 = (96.31)(1038) / (2.25)(1022) =

(42.8)(1038/1022) = 42.8 x 1016= 4.3 x 1017 N.. And now for Earth's force:

Fe = [(6.67 x 10-11)(7.22 x 1022)(6 x 1024)] / (3.85 x 108)2 =

(6.67)(7.22)(6)(10-11)(1022)(1024) / (3.85)2 (108)2 = (288.94)(1035) / (14.82)(1016) =

(14.82)(1035/1016) = 14.82 x 1019= 1.48 x 1020 N. Now let's find the Net Force:

Fnet = Fs + Fe= (4.3 x 1017 N @ 0°) + (1.48 x 1020 N @ 270°); now let's pause and think. A right triangle with one side Fs and another side of Fe will have hypotenuse of Fnet. A similar triangle will exist if we divide both forces by 1017 to get one side of 4.3 and the other side of 1.48 x 103 = 1480. So, the length squared of the hypotenuse of that similar triangle will be l 2 = (4.3)2 + (1480)2 = (18.49) + (2,190,400) = 2,190,418.49; so the square root of that is the hypotenuse, or l = 1480.006247, or, 1.48 x 103 .

Multiply that back by 1017 we get the net force to be Fnet = 1.48 x 1020 N. This gives us the conclusion that the force from the Sun is negligible compared to Earth's. And if we were to compute the vanishingly small angle with trigonometry, it would probably be 270° (or directly to the -y direction).

13. Suppose that three astronomical objects, 1, 2, & 3, are observed to lie in a straight line, and that the distance from 1 to 3 is D. Given that object 1 has four times the mass of object 3 and seven times the mass of object 2, find the distance between objects 1 and 2 for which the net force on object 2 is zero (0.0 N).
Solution: Sounds like fun.

1 2 3
* * *
7m m 7/4 m

|----x----|------(D – x) --|

|<-------------D---------->|

F12 = G(7m)(m)/(x)2

F23 = - G(7/4 m)(m)/(D – x)2

(this force starts negative coming from another direction

And F12 + (- F23 ) = 0.0 N. Or,

Fnet = G(7m)(m)/(x)2 - G(7/4 m)(m)/(D – x)2 = 0

(7 m2 ) / (x)2 - (7m2 /4) / (D – x)2 = 0

(7 ) / (x)2 - (7 /4) / (D – x)2 = 0

(7 ) / (x)2 = (7 /4) / (D – x)2

(7)(D – x)2 = (7 /4)(x)2

(D – x)2 = (1 /4)(x)2

4(D – x)2 = (x)2

4(D2 – 2xD + x2) = (x)2

(4D2 – 8xD + 4x2) = (x)2

3x2 – 8xD + 4D2 = 0

Using the quadratic equation, then ax2 + bx + c = 0; and then

x = [- b +/- √(b2 – 4ac)] / 2a, where a = 3, b = - 8D, c = 4D2.

So, x = [+ 8D +/- √(64D2 – 48D2)] / 10, = ( 8D +/- √(16D2) / 10 = ( 8D +/- 4D) / 10.

the two mathematical solutions are x = {1.2 D, 0.4D} Since in this framework x has to be shorter then D, then the answer is x = 0.4 D.

19. At a certain distance from the center of Earth, a 4.6-kg object has a weight of 2.2 N.
a. Find the distance.
b. If the object is released at this location and allowed to fall toward the Earth, what is its initial acceleration?
c. If the object is now moved twice as far from Earth by what factor does its weight change? Explain.
d. by what factor does its initial acceleration change? Explain.
Solution: The force of gravity is: F = G mM/r2. We have everything but the distance, r for the first part, so, re-write:
(a) r2 = G mM/F, so taking the square root gives us the distance: r = √(G mM/F) =
√[(6.67 x 10-11)(4.6)(6 x 1024)/(2.2)] = √[(184 x 1013)/(2.2)] = √[(83.678 x 1013)] =
√[(8.3678 x 1014)] = √[(8.3678)(1014)] = 2.892 x 107 m ~ 2.9 x 107 m from the Earth's center. How far above the surface would be another question.
(b) At 6.4 x 106 meters (surface) the acceleration is – 9.8 m/s2. At 2.9 x 107 m, it is (29/6.4) times (4.53) farther, and since it's (gravity) an inverse square law, the new acceleration would be
(-9.8)/(4.53)2 = - 0.477 m/s2..
(c)It's an inverse square law, so 2.0 times further would be ( ½ )2 = ¼ of what it was.
(d) It's ( ¼ )(- 0.477 m/s2) = - 0.119 m/s2

25. A satellite orbits the Earth in a circular orbit of radius r. At some point its rocket engine his fired in such a way that its speed increases rapidly by a small amount. As a result,
a. does the apogee distance increase, decrease, or stay the same?
b. does the perigee distance increase, decrease, or stay the same?
Solution: Since v = w r, then radius increases, and it moves further out, but eventually comes back. The apogee gets longer, the perigee gets shorter.

29. Find the orbital speed of a satellite in a geosynchronous circular orbit 3.58 x 107 m above Earth's surface.
Solution: Using Kepler's 3rd law, P2 (m + M) = (4p2 / G) r3. The third law has the term (m + M) which is the mass of the very light satellite, m, and Earth, M. So we can ignore little 'm'. Now the formula becomes:P2 = [(4p2 / GM) r3.]

The speed or velocity will be the circumference, c, of its orbit divided by the period, or v = c/P. For convenience, let's find v2 first, or, v2 = (c/P)2 and (c/P)2 = (2 p r)2 / [(4p2 / GM) r3.] =
(2 p r)2 / [(2pr)2 / GM) r.] = GM / (r) = [(6.67 x 10-11)(6 x 1024)] / [(4.22 x 107)] =
(40 x 1013) / [(4.22 x 107)] = 9.5 x 106 . and this is v2, so v is the square root of that number, or

v = √(9.5 x 106 ) = 3.1 x 103 m/s.

31.Phobos, a moon of Mars, orbits at a distance of 9378 km from the center of Mars. What's its orbital period?
Solution: Using Kepler's 3rd law, P2 (m + M) = (4p2 / G) r3. The third law has the term (m + M) which is the mass of the very light satellite, m, and Mars, M. So we can ignore little 'm'. Now the formula becomes:P2 = [(4p2 / GM) r3.] = [(40) / (6.67 x 10-11)(7 x 1023)] (9.378 x 106 m)3.] =
[(40)(9.378 x 106 m)3] / [(6.67 x 10-11)(7 x 1023)] = {[(40)(8.25 x 1020)] / (47 x 1012)} =

{[(330 x 1020)] / (47 x 1012)} = 7.0 x 108. And this is P2, so the period is the square root of this. P = √(7.0 x 108.) = 2.65 x 104 sec (this is about 7.3 hours).

39. Find the speed of the binary stars Centauri A and Centauri B. They are separated by a distance of 3.45 x 1012 m and they have an orbital period of 2.52 x 109 seconds (that's about 80 years). They have the same mass. (Binary stars are near each other and orbit each other).
Solution: Using Kepler's 3rd law, P2 (M + M) = (4p2 / G) r3. The third law has the term (M + M) which is the mass of each of the binary stars. So we can now the formula becomes:P2 = [(4p2 / G2M) r3.]

The speed or velocity will be the circumference, c, of its orbit divided by the period, or v = c/P =
(2 p r / P) = [(2)(3.14)(3.45 x 1012)] / (2.52 x 109)] = [(21.666)(1012)] / [(2.52)(109)] = 8.6 x 103 m/s.

53. Find the Escape Velocity, vesc, for the planet Mercury. Do the same for the planet Venus. (How fast must they go to escape the Sun's gravity?)
Solution: Let's use the formula for escape velocity: vesc = √(2 G M/ r)..
So, for Mercury, M = 3.3 x 1023 kg and r = 2.4 x 103 m.

Thus, vesc = √[(2)(6.67 x 10-11)(3.3 x 1023) / (2.4 x 103)].= √[(1.834 x 1010) ]..= 1.35 x 105 m/s.

For Venus, M = 4.8 x 1024 kg; and r = 6 x 103 m.; so

Thus, vesc = √[(2)(6.67 x 10-11)(4.8 x 1024) / (6 x 103)].= √[(10.67 x 1010) ]..= 3.27 x 105 m/s.

For the Sun, M = 2 x 1030 kg; and r = 1.44 x 109 m.; so

Thus, vesc = √[(2)(6.67 x 10-11)(2 x 1030) / (1.44 x 109)].= √[(18.53 x 1010) ]..= 4.3 x 105 m/s.

63. A “dumbbell” has a weight, m, on either end of a rod of length 2a. The center of the dumbbell is at a distance, r, from Earth's center. The dumbbell is aligned radially (pointing towards Earth's center and away from it, i.e., vertical. If r >>a, show that the difference in the gravitation force exerted on the two masses by Earth is approximately equal to 4GmMa/r3. [The difference in the gravitational pull between the two weights creates a stress tension in the bar known as the tidal force. For r >>a, use 1/(r - a)2 – 1/(r + a) 2 – 4a/r3. )
Solution: This is an interesting problem, but let's skip it.

Chapter 13: Problems 1, 3, 9, 13, 19, 25, 29, 31, 39, 53, 63, 67, 69, and 77 on pp 445-447.

1. A small cart on a 5.0-m air-track moves at a speed of 0.85 m/s. (This is like the air hockey table that has no friction while two players keep hitting a round puck back and forth; however, an air track is one-dimensional, and instead of a puck, there is a small “cart.”). Bumpers at either end of the track cause the cart to reverse direction and maintain the same speed. Find the period and frequency of this motion.
Solution: First, some unknown source caused this cart to move at a constant speed. There is no acceleration. Second, there is no friction. Third, the collisions or impacts at the bumpers are 100% elastic so no energy is lost to heat, mechanical deformation, sound, or whatnot. In other words, this is fantasy land. But let's go with the flow. How much time does it take for the cart to travel from one end to the other end, and back (one cycle)? Since v = x/t, and we are looking for “t,” then t = x/v. The length, x = 5.0 meters, so the entire cycle is “2x” since it takes time to travel one length, and the same to travel back. No matter, t = (5.0 meters) / (0.85 m/s) = 5.88 seconds. The entire period would be double that, or P = 11.76 seconds. The frequency is the reciprocal of the period, so, n = (11.76 s)-1 = 0.085 cycles /sec = 8.5 x 10-2 Hz.

3. While fishing for a catfish, a fisherman suddenly notices that the bobber (a floating device on the fishing line) attached to his line is bobbing up and down with a frequency of 2.6 Hz. Find the period of the bobber's motion.
Solution: We know that the period is the reciprocal of the frequency. So, if n = 2.6 Hz, then P = (2.6)-1 = 0.3846 second = 3.8 x 10-1 s.

9. A mass moves back and forth with simple harmonic motion (SHM). The amplitude is A (this is how high the top of the crest of the wave is above the average). The period is P (this is how long it takes for one full cycle of the wave to go from one crest to another crest).
(a) How long does it take for the mass to move through a distance of x = 2A ?
(b) How long does I take for the mass to move through a total distance of 3A ?
Solution: Like the swinging pendulum that starts at maximum distance (45), the period of one complete cycle would be A + A + A + A = 4 A.
a. So to travel 2 A, it would be ½ P.
b. By the same reasoning, this would be (¾) P.

13. A mass on a spring oscillates with SHM with amplitude, A about the equilibrium point where x = 0. Its maximum speed is vmax and its maximum acceleration is amax.
(a). Find the speed of the mass at x = 0.
(b) Find the acceleration of the mass at x = A.
Solution: Like in the swinging pendulum, the acceleration will be greatest at the maximum angle of 45,
a. so amax is at x = A.
b. The bob reaches maximum speed (vmax) at the “bottom” of the swing, or at x = 0.

19. An object moves with SHM with a period T, and amplitude, A. During one complete cycle, for what length of time is the position of the object greater than A/2?
Solution: p = T for x = 4 A, so for (1/8) of 4 A = A/2, it will be 1/8 T.

25. A 30.0-gram gold finch (that's a bird, not the gold fish).lands on a slender tree branch where it oscillates up and down with SHM with A = 0.0335 m and period of T = 1.65 seconds.
(a) Find the maximum acceleration of the finch, amax. Divide this by Earth's acceleration, g = 9.8 m/s2.
(b) Find maximum speed of the gold finch, vmax.
(c) At the time the goldfinch experiences its maximum acceleration, amax, is the speed, v, maximum or minimum?
Solution: The average acceleration is defined as a = Dv / Dt, or the change of velocity over time. And the average velocity is v = Dx / Dt. So it travels x = 4 A in T, or, 4(0.0335 m) per 1.65 seconds. Thus,

v = 4(0.0335 m) / (1.65 seconds) = 4(0.0335 m) / (1.65 seconds) = (0.134 m) / (1.65 s); or,

v = 0.081 m/s. The maximum speed in the cycle is twice the average, or vmax = 0.162 m/s. Save this answer to use for (b).
a. amax = (vmax - 0.0) / ( ¼ )(1.65 s) = (0.162 m/s - 0.0 m/s) / ( ¼ )(1.65 s) =

(0.162 m/s - 0.0 m/s) / (0.4125 s) = (0.162 m/s) / (0.4125 s) = 0.393 m/s2, or 3.93 x 10-1 m/s2.

b. We found this already above: vmax = 0.162 m/s.
c. Minimum

29. The pistons in an internal combustion engine undergo a motion that approximates SHM. If A = 3.4 cm and the frequency is w = 1700 rpm, then find
(a) the maximum acceleration of the pistons, .amax ; and
(b) their maximum speed, vmax.
Solution: We know that v = w r, and if the maximum “r” is 3.4 cm, then that's easy. And we are given w = 1700 rpm = 28.3 rev/sec. But in this problem, the author uses the wrong symbol for frequency; because n = 28.3 rev/s, and w = 2 p n = 2 p (28.3) = 180 rad/s. And vmax = w rmax =

(180)(3.4 x 10- 2 m) = 6.05 m/s.

a. Since a = Dv / Dt = (6.05 – 0) / ( ¼ ) (P); however, P = (n)-1, or (P)-1 = n. So we can rewrite this as:

amax = (6.05 – 0)n / ( ¼ ) = (6.05 )(28.3) / ( ¼ ) = 4 (6.05 )(28.3) = 685 m/s2, or

amax = 6.85 x 102 m/s2.

b. We found this: vmax = w rmax = (180)(3.4 x 10- 2 m) = 6.05 m/s.

31. A bronco rider is on a mechanical bucking horse that oscillates up and down with SHM, with a period of T = 0.74 seconds with slowly increasing amplitude, A. At a certain amplitude, Ac, the rider must hang on to prevent being tossed off the horse.
(a) Explain how you'd calculate the amplitude, Ac.
(b) Carry out your method and find the amplitude, Ac.
Solution: We have been doing this for several problems, so
a. The same way we did the previous problems, looking for amax.
b. blah blah blah. S.O.S.

39. When a 0.5-kg mass is attached to a vertical spring, the spring stretches by y = - 15.0 cm. How much mass, in kilograms, must be attached to the vertical spring to stretch it so that it has a period of oscillation, T = 0.75 seconds?
Solution: Well, this is funner. We invoke Robert Hooke: F = - k y = m g. From this we can get the spring constant, k:

- k y = m g
k = - m g / y = - (0.5kg)(- 9.8m/s2) / (0.15 m) = 32.67 N/m. The spring constant is always positive (+) so if you get a negative, take the | ABSOLUTE VALUE | .

We also know that there is a relationship between the spring constant and the frequency:

P = 2 p √(m/k), so we re-write this to have “m” all alone:

P2 = (2 p)2 [(m) / (k)]

[(P) / (2 p)]2 = [(m) / (k)] so,

m = k [(P) / (2 p)]2 = (32.67 N/m)[(0.75) / (6.28)]2 = (32.67 N/m)(0.1194)2 =

(32.67 N/m)(0.0143) = 0.4659 kg.

Or, m = 4.7 x 10-1 kg.

Chapter 14: Problems 1, 3, 9, 13, 19, 25, 29, 31, 39, 53, 63, 67, 69, and 77 on pp 491-495

1. A wave has crests and troughs. From the top of one crest to the bottom of the next trough is 13 cm. The length from crest to trough along the x axis is 28 cm.
a. Find the amplitude, A.
b. Find the wavelength, l.
Solution:
a. Since the top of one crest to the bottom of the next trough is 13 cm, and since the distance from the center (zero) line to the apex (crest, top, etc.) is the Amplitude, and since the center line is halfway from the apex to the trough, we multiply the 13 cm by ½, or, ½ (13 cm) = A = 6.5 cm.
b. Since the length from crest to trough along the x-axis is 28 cm, and since the definition of wavelength the distance from crest to crest, and the distance from crest to trough is only half a wavelength, then we have to double the distance given to find the wavelength, l = 2 x 28 = 56 cm.

3. The speed of surface waves in water decreases as the water becomes more shallow. Imagine water waves crossing the surface of an otherwise calm lake at a speed of v1 = 2.0 m/s, with a wavelength of l1 = 1.5 meters. When these same waves approach shore, the speed decreases to v2 = 1.6 m/s, while the frequency remains the same. Find the wavelength, l2, in the shallow waters.
Solution: We have learned that by multiplying the wavelength, l, by the frequency, n, we get that “famous” equation, resulting in the velocity: l n = v. To find the non-changing frequency, we can rewrite the equation as n = v1/l1.= (2.0)/(1.5) = 1.33 Hz. Remember, n 1 = n 2 = n = 1.33 Hz. So, using the same equation that l n = v., to find, , we just divide both sides by thfrquency, n. And the get this: l 2 = v2 /n.= (1.6 m/s)/(1.33) = 1.2 meters.

9. Suppose that you wanted to double the wave speed, from v to 2v, on a tight string. The force that is keeping it tight is called “tension” but it is still a force and has units of Newtons. How much would you have to increase the tension so the velocity would double?
Solution: We remember from the laboratory exercise, “Tin Can Phone with Soundwaves on a String,” that the velocity is: v = (T/m)½ , where T is the tension or force, and m is the linear density. So, if we wanted the velocity to go from v to 2v, then we would have to quadruple the tension (i.e., four times more, because: 2V = (4T/d)½

13. Waves on a particular string travel at v1 = 16 m/s. To double that speed to v2 = 32 m/s, how much would you have to increase the tension?
Solution:
The clever student should quickly realize that this is the same identical problem as Number 9, so the answer is, again, . quadruple the tension.

19. Write an expression for a harmonic wave, such that amplitude, A = 0.16 m, wavelength l = 2.1 m, and period of P = 1.8 seconds.
Solution: On page in the book, it says that the expression for a harmonic wave is:
y = A cos(2px/l – 2pt/P), and thus, if you plug in the data given, your answer becomes:
y(x,t) = 0.16 Cos[(2px/2.1) – (2pt/1.8)]

25. Four waves are described by the following equations (all of them are expressions for harmonic waves), in which all distances are in centimeters and all time is in seconds:

yA = 10 Cos(3x – 4t)
yB = 10 Cos(5x + 4t)
yC = 20 Cos(-10x + 60t)
yD = 20 Cos(-4x – 20t)

a. which wave(s) travel in the +x direction?
b. which wave(s) travel in the - x direction?
c. which wave has the highest frequency?
d. which wave has the longest wavelength?
e. which wave has the fastest speed?
Solution: We immediately realize that all four of these relationships are “expressions for a harmonic wave .” But let's do this first:
a. (+x direction when a neg x is in the equation)
b. (-x direction when a pos x is in the equation)

For the next part, we plug and chug to find the truth, and the truth shall set you free.
The expression for yA would be:

yA = 10 cos(2px/l – 2pt/P), but that has to equal the exact equation that we got for yA above: yA = 10 Cos(3x – 4t). But if yA = yA, then 10 cos(2px/l – 2pt/P) = 10 Cos(3x – 4t), or (2px/l – 2pt/P) = (3x – 4t). In the first one, the coefficient of the variable “x” is 2p/l, while in the second one, the coefficient is “3”. Thus, 2p/l = 3, or lA = 2p/3= 2.27.

Repeating these steps for yB, yC, and yD, we get the following wavelengths:

l B = 1.26,
l C = 0.628,
l D = 1.57,

Thus, the shortest wavelength, and the highest frequency, would have to be from the formula:
yC = 20 Cos(-10x + 60t).

d. yd

The fastest one would have the shortest period, and using the same method as to get the wavelength, we find:

e. for ya, use 2p/T = 4, so the period, T, is 2p/4 = 1.57 s
for yb, use 2p/T = 4, so the period, T, is 2p/4 = 1.57 s (same as ya)
for yc, use 2p/T = 4, so the period, T, is 2p/60 = 0.105 s
for yd, use 2p/T = 20, so the period, T, is 2p/20 = 0.314 s.

Shortest perioid: yC = 20 Cos(-10x + 60t).

29. A soundwave in air has a frequency of n1 = 425 Hz.
a. Find its wavelength
b. If its frequency is increased, does the wavelength increase, decrease, or remain unchanged?
c. Calculate its wavelength if the frequency is n2 = 475 Hz.
Solution: Once again, we use the “famous” equation, l n = v., and to find wavelength, we divide both sides by the frequency. To get l = v/n.

a. l = v/n. = (342 m/s)/(425 Hz) = 0.805 meters. (but where did we get the speed, v? This is a sound wave. In air. The speed of sound in air is always 342 m/s.
b.l n = vsound
c. l = 342/475 = 0.72 meters.

31. A man throws a rock down to the bottom of a well. From the instant that it leaves his hand until the moment he hears the rock hit bottom, a period of 1.20 seconds passes. The depth of the well is 8.85 meters. Find the initial speed, vi, of the rock (the instant that it leaves the man's hand).

Solution: We are looking for vi, the initial speed of the rock (the instant that it leaves the man's throwing hand). So, what do we know? We know that the final speed of the rock, vf = 0.0 m/s (it is stopped when it hits the well's bottom). Maybe we can use the formula:

vf2 - vi2 = 2 a y, where since vf = 0.0, we can re-write it as:

vi2 = 2 a y. We know the number “2.” We know the distance down, y = 8.85 meters, but do we know the acceleration, a?

What is “a” ? a = g + a’. And g is Earth's gravitational pull, while a' is the extra acceleration added by the man's hand. Since I don't know where this is going, I am going to put this path to solving it on “hold” and try something else.

We know from a previous lab that the average speed as it falls is related to the initial speed, i.e., vave = ½ vi

We need to find the average velocity. We know that speed is distance over time, or, v = y/t. Let's call:

t1 = the time it takes for the object to reach the bottom after being thrown
t2 = the time for the object’s sound to reach the top after making noise at the bottom.

We do know that total time from the time the rock was released until the sound reached the man's ears is given: t1 + t2 = 1.2 s.

We know that the speed of sound here is: vs = y/t2 ; so, t2 = y/vs = 8.85 m/342 m/s = 0.026 s. If t2 = 0.26 seconds, then t1 can be found by subtraction.

t1 + 0.026 = 1.2 s
t1 = 1.2 s - 0.026 = 1.174 s

Now we can find the average speed, as the rock was falling/thrown down: distance/time, or y/t1, or vave = 8.85/1.174 = 7.54 m/s

and, if vave = ½ vi, then 2vave = vi..= (2)(7.54) = 15.08 m/s. Rounding, we get:

vi = 15 m/s.

39. Twenty identical violins are being played by 20 identical violinists at the same identical volume, for an aggregate sound level of 82.5 dB.
a. Find the dB level of one of the violins.
b. Find the dB level of twice as many identical violins (40).
Solution: According to the table provided in the book (or on the test), one sees that a dB level of 80 has an energy flux of 10-4 Watts/m2.
a. If 20 violins has an energy flux of 10-4 Watts/m2, then one violin would have 1/20th of that, or, 10-4 /20 = 100 x 10-6 / 20 = 5 x 10-6. Rechecking the table shows that an energy flux of that amount would be the equivalent to about 66 dB.
b. If 20 violins has an energy flux of 10-4 Watts/m2, then 40 violins would have twice that, or, 2 x 10-4 Watts/m2, and checking the chart, that energy flux would give that a reading of about 85 dB.

END

LESSON 8

2010-08-17T14:24:00.000-07:00

PHYS 1100 LESSON 8 FOR
TUESDAY, AUGUST 17, 2010

I Introduction

II Logistics – Running Grades

III Turn in Assignments Due and Return of Papers

IV Review of Lesson 7: Energy, Momentum, Torque, et al
Energy: interchangeable with mass, per

Einstein’s E = mc2.

41H1 = 2He4 + mc2 + 2b+.

F d = using energy = Joules. Work is useful energy. Efficiency, e = W/E, e<1

Natural gas: CH4 + 2O2 = CO2 + 2H2O + Energy

Gasoline: 2C8H18 + 25O2 =
16CO2 + 18H2O + Energy

Momentum, p = m v goes in a straight line

Angular momentum, l = m v r

V Lesson 8: Inertia, Torque, Gravity

A. Inertia is kg m2.
B. Torque is N-m
C. Gravity:

1. Kepler – P2 (m + M) = [(4 p2)/G] r3.

This is Kepler’s 3rd law dealing with the period of a planet, P, with a mass, m, traveling around the Sun, with a mass, M, at a distance between the Sun and the planet equaling “r”. However, it applies to moons, asteroids, stars, satellites, blah blah.

Period is in seconds, mass in kilograms, distance in meters (mks). The Universal Gravitational Constant is G = 6.67 x 10-11 N m2 / kg2.

2. Newton – F = GmM/r2.

This part of Newton’s 2nd Law, which deals with the force, F, of gravity.

Little ‘m’ is the mass of a smaller object, such as human; big “M” is the mass of the larger object, such as planet Earth; r is the distance between the center of the first object (like a human’s navel) and the center of the second object (like the distance from the human to Earth’s core, i.e., the radius of Earth.

As a force, instead of a period, this law applies both to moving objects, like planets, moons, stars; and to object that are not moving, like people standing on Earth, or between two bowling balls in a bowling alley.

The Universal Gravitational Constant is G = 6.67 x 10-11 N m2 / kg2.

VI Laboratory Exercise 8: Simple Harmonic Motion

VII HWK Assignment 3: 10-14 and do Problems 1, 3, 9, 13, 19, 25, 29, 31, 39, 53, 63, and 65 on pp 323-327; Problems 1, 3, 9, 13, 19, 25, 29, 31, 39, 53, 63, 65 and 77 on pp 366-371; Problems 1, 3, 9, 13, 19, 25, 29, 31, 39, 53 and 63 on pp 409-412; Problems 1, 3, 9, 13, 19, 25, 29, 31, 39, 53, 63, and 65 on pp 445-448; and Problems 1, 3, 9, 13, 19, 25, 29, 31, 39, 53, 63, 67, 69, and 77 on pp 491-495; due 8/19

VIII Essay 3: James Prescott Joule, due 8/19

LAB 8

2010-08-17T14:23:00.000-07:00

PhysicsLab8 Tuesday, August 17, 2010 Name __________________
Dr Dave Menke, Instructor

I Title: Simple Harmonic Motion

II Purpose: Study Simple Harmonic Motion using a plumb bob pendulum.

III Equipment
1. String, yarn, or cord ≥ 1.0 meter long (or as close as possible)
2. Weight - to make a plumb bob pendulum (washers?)
3. Stop Watch, face watch, digital watch, clock, or other chronometer
4. Meter stick or metric ruler
5. Protractor

IV Procedure (Some of this is similar to the Gravity lab)
1. Obtain, or make, a length of string or cord that is very close to 1.00 meter long. Slightly longer is better than slightly shorter.
2. Attach a weight to one end of the string to create a plumb bob, that we will call Bob.
3. Attach the other end of the string to some stationary object (door hinge, ceiling, weighted ring stand, etc.). Do NOT use a primate because it is not stable or other mammal to hold Bob because it is not stable.
4. Measure the length of Bob exactly (to the closest millimeter) after you have set it up. This will be from the point of connection on top to the middle of the weight. Record this length (we will call the length “r”) as accurately as possible.
5. Put the Data in the Table below; one column is for the number of the trials; another for the time (in seconds) for each cycle. And a third column for amplitude in the x-direction.
6. Have one of the lab partners pull the pendulum back, to an angle of θ = 45° (try to be exact; use protractor) as seen in the diagram a
7. Measure the x-component of the Bob's motion. If the string pendulum is exactly 1.0 meter long, and if the angle is exactly 45°, then the x-component will be (1.0 m) Sin 45° = 0.707 m = 70.7 cm. Thus, when you measure the x-component, it will be very close to 70 cm. This will be your original (and maximum) amplitude.
8. Simultaneously, release Bob and depress the stop watch button to start the time “running.” It is best to have the same homosapien release Bob and operate the stopwatch (the same brain controls both hands).
9. Allow Bob to swing freely as long it can. Every time that it returns to its starting point, note and record the time, and the distance. For example, at t = 0.00 s, the x-component will be (about) 70 cm. The next time that it comes back, about 1.8 seconds later, the x-component will be less, maybe 65 cm; next time, maybe 60 cm; and so forth. If you find it very difficult to do both the time and the x-component distance, you may substitute that data that you gathered for Lab #2 about Gravity. Keep the pendulum swinging until it stops, or, nearly stops. If you need to create another table to extend the data, please do so.
10. When done with gathering the data, plot the data points on a graph, with time in seconds, t, along the horizontal (left-right) and amplitude (length of x) along the vertical. Connect the dots as smoothly as you can.
11. Determine the period of oscillation using the graph of data.

V Data & Calculations:

Trial Time (sec) Amplitude (cm)

VI Results:
The purpose if this laboratory (Example of Simple Harmonic Motion) was / was not achieved due to:

VII Error Analysis:
A. Quantitative Error: NA

B. Qualitative Error:
1. Personal -
2. Random -
3. Systematic –

VIII Questions:
1. What is the period of oscillation?
2. What is the maximum amplitude?
3. What is the average periodic decrease in amplitude for each cycle?
4. List 4 items in your life and your world that oscillate.

LESSON 7

2010-08-16T11:43:00.000-07:00

PHYS 1100 LESSON 7 FOR
MONDAY, AUGUST 16, 2010

I Introduction

II Logistics – Running Grades

III Turn in Assignments Due and Return of Papers

IV Review of Lesson 6: Energy, Work, Momentum, Angular Momentum, Impulse

Energy: interchangeable with mass, per

Einstein’s E = mc2.

41H1 = 2He4 + mc2 + 2b+.

F d = using energy = Joules. Work is useful energy. Efficiency, e = W/E, e<1

Natural gas: CH4 + 2O2 = CO2 + 2H2O + Energy

Gasoline: 2C8H18 + 25O2 =
16CO2 + 18H2O + Energy

Momentum, p = m v goes in a straight line
Angular momentum, l = m v r

V Lesson 7: Torque, Inertia, Impulse

Impulse = I = F Dt. Baseball, golf, billiards, tennis. N-s units of Impulse

F = m a

I = (ma) Dt = m(a Dt) = m (Dv/Dt)(Dt) = m(Dv)( Dt/Dt) = mDv = p

Example: What force does a baseball hitter use when he hits a 0.15 kg ball, traveling at 160 km/hour, and the contact time, Dt = 0.12 sec?

m v = F Dt

So, WTF?

(m)(v)/(dt) = F = (0.15)(44.5)/(0.12) = 55.625 N

160 km/hr = (5/18)(160 km/h) m/s 44.5 m/s

55.625 N = 56 N ~ 5.6 E1 or x 10^1 or 101 N.

Torque has the name units as Energy; torque, t, = r F; N-m

Inertia = characteristic of matter that resists change units are kg m2.

VI Laboratory Exercise 7: PE KE

VII HWK Assignment 3: 10-14 and do Problems 1, 3, 9, 13, 19, 25, 29, 31, 39, 53, 63, and 65 on pp 323-327; Problems 1, 3, 9, 13, 19, 25, 29, 31, 39, 53, 63, 65 and 77 on pp 366-371; Problems 1, 3, 9, 13, 19, 25, 29, 31, 39, 53 and 63 on pp 409-412; Problems 1, 3, 9, 13, 19, 25, 29, 31, 39, 53, 63, and 65 on pp 445-448; and Problems 1, 3, 9, 13, 19, 25, 29, 31, 39, 53, 63, 67, 69, and 77 on pp 491-495; due 8/19

VIII Essay 3: James Prescott Joule, due 8/19

LAB 7

2010-08-16T11:42:00.000-07:00

PhysicsLab7 Monday August 16, 2010 Name __________________
Dr Dave Menke, Instructor

I Title: Potential vs. Kinetic Energy

II Purpose: To investigate the difference between potential energy and kinetic energy by using “elastic” bouncing spheres.

III Equipment:
* One “elastic” bouncing sphere of 8.5 gram mass or 32.4 grams
* Metric ruler
* Pen, calculator, lab book, etc.

IV Procedure
1. Raise the elastic sphere to h1 = 2.0 meters above the lab floor.
2. Calculate the potential energy of the elastic sphere, PE = mgh1, by placing the first elastic sphere on the ground, and then lifting it up two meters. Record. This will be in JOULES.
3. Calculate the amount of time that it takes for the elastic sphere to fall 2.0 meters. Remember the “free fall” relationship: h1 = ½ g t2. You already know h = 2.0 meters. And g = 9.8 m/s2. Record.
4. Calculate the final velocity for the elastic sphere, vf, using the relationship that vf2 – vi2 = 2 g h1. Since vi = 0, vf = √(2 g h1) = √[(2)(9.8)(2.0)]. Record.
5. Calculate the Kinetic Energy of the elastic sphere just before impact. KE = ½ m vf2. Record.
6. Compare the KE in #5 with the PE in #2. Which is larger, or are they equal? Explain why.
7. Place the elastic sphere at h1 = 2.0 meters above the floor. Release the elastic sphere. Measure how high the elastic sphere rebounds after it hits the floor, h2. Record.
8. Repeat 4 more times, for a total of 5 times. Take the average of the rebound, h2. Record.
9. Using DPE = m g (Dh), find out how much energy was “lost” by the elastic sphere when it bounced up. Remember, Dh = h1 – h2.
10. List places that the energy “went.”

V Data & Calculations
1. Mass of elastic sphere: _________grams; convert to kg____________

2. Initial PE = mgh1 = ____________ joules

3. Time for sphere to fall 2.0 meters: _________ seconds.

4. The final velocity, vf : ____________ m/s.

5. Kinetic energy: ___________ Joules

6. The difference between PE and KE, in joules: ________J.

7. The average change in height (the average of the 5 trials): _________ meters.

8. The final PE after the rebound, m g h2 : ________

9. Subtract the final PE in #8 from the initial PE in #2. DPE = __________ joules.

VI Results
Explain if there were a difference between PE and KE, and why. Also, explain where the “lost energy” went.

VII Error Analysis
A. Quantitative Error –
The true laboratory value of final velocity, vf(true) = 6.261 m/s; the lab value for the mean velocity is vave = 3.131 m/s.

% Error = [ |v(true) – v (yours)| / v(true) ] x 100%

B. Qualitative Error: Sources of Error
Personal
Systematic
Random

VIII Questions
Spheres made of rubber cause collisions that are somewhat elastic. a. name objects where the collisions would be almost perfectly elastic; b. name objects where the collisions would be almost entirely inelastic.
In light of these concepts, why do you think that some football players are “light” and others are very heavy and muscular?

LESSON 6

2010-08-16T11:41:00.000-07:00

PHYS 1100 LESSON 6 FOR
THURSDAY, AUGUST 12, 2010

I Introduction

II Logistics

III Turn in Assignments Due: Homework 1, 2, Essay 1, 2, etc. and Return of Papers

IV Test 2

V Mind Game 2

VI Review of Lesson 5: Hooke's Law, Energy, Work, Momentum

F = - k x
F = - k y = m g

Energy: heat, light, mechanical, acoustical, etc.

Kinetic, Potential, ….

Work / Energy < 1

F x d = energy, units 1.0 N-m == Joule

4.186 calorie = 1 joule

Momentum = energy of motion

p = m v

Dp = 0 in a closed system

E = F x d (in Joules)
PE = m g h (in Joules)
KE = ½ m v2 (in Joules)
p = m v (in kg m/s)
Ff = m m g (friction)

VII Lesson 6: Momentum, Impulse, Conservation, Angular Momentum

p = m v

S p = constant in a closed system, or,

Dp = 0 in a closed system

F Dt = Impulse, but same units as momentum

l = m v r = p r this is angular momentum

t = r x F this is torque

VIII Laboratory Exercise 6: Hooke's Law

IX Homework Assignment 3: Read Chapters 10-14 and do Problems 1, 3, 9, 13, 19, 25, 29, 31, 39, 53, 63, and 65 on pp 323-327; Problems 1, 3, 9, 13, 19, 25, 29, 31, 39, 53, 63, 65 and 77 on pp 366-371; Problems 1, 3, 9, 13, 19, 25, 29, 31, 39, 53 and 63 on pp 409-412; Problems 1, 3, 9, 13, 19, 25, 29, 31, 39, 53, 63, and 65 on pp 445-448; and Problems 1, 3, 9, 13, 19, 25, 29, 31, 39, 53, 63, 67, 69, and 77 on pp 491-495; due 8/19

X Essay 3: James Prescott Joule, due 8/19

LAB 6

2010-08-16T11:38:00.000-07:00

Physics Lab 6 Thursday, August 12, 2010 Name __________________

I Title: Acceleration of Gravity

II Purpose: To determine the acceleration of gravity using simple equipment.

III Equipment
1. String, yarn, or cord ≥ 1.0 meter long (or as close as possible)
2. Weight - to make a plumb bob pendulum (washers?)
3. Stop Watch, face watch, digital watch, clock, or other chronometer
4. Meter stick or metric ruler
5. Weighted Ring Stand Optional

IV Procedure
Obtain, or make, a length of string or cord that is very close to 1.00 meter long. Slightly longer is better than slightly shorter.
Attach a weight to one end of the string to create a plumb bob, that we will call Bob.
Have that same person, or another, attach the other end of the string to some stationary object (door hinge, ceiling, weighted ring stand, etc.). Do NOT use a primate because it is not stable or other mammal to hold Bob because it is not stable.

Measure the length of Bob exactly (to the closest millimeter) after you have set it up. This will be from the point of connection on top to the middle of the weight. Record this length (we will call the length “y”) as accurately as possible.
Use the Data Table below; one column is for the number of the trials; the other for the time (in seconds) for each cycle.

6. Have one of the lab partners pull the pendulum back, to about an angle of θ = 45° (but no further) as seen in the diagram to the right
Simultaneously, release Bob and depress the stop watch button to start the time “running.” It is best to have the same homosapien release Bob and operate the stopwatch (the same brain controls both hands).
Allow Bob to swing out and come back to where it was released. Stop the watch. That is one cycle. Record the time in the table. Reset the watch and get ready to repeat.
Have the same CroMagnon Repeat steps #6 - #8, nine more times, and place your data in the table. You should have a total of ten trials.
Find the average period of oscillation of the ten trials. This means, add up all the numbers in the second column, then divide by 10. Record.
Use the data to find the acceleration of Earth’s gravity, g, by re-writing Galileo’s Period-Length equation so that the acceleration, g, is all alone on the left side.

-MORE-

Physics Lab 6 Thursday, August 12, 2010 page 2 Name _____________

P = 2 p [(y / g)] ½

P is the period in seconds, y is the length in meters, and g is what we want to find. It will be in meters per square second. This equation reads “Period equals two times pi times the square root of (y/g).”

Let’s continue to re-write this relationship, in order for us to get, g all alone:

[P / 2 p ] = [(y / g)] ½

[P / 2 p ]2 = (y / g)

g = y [2 p / P] 2

Remember that “y” is the length of the string (about 1.0 meter, but make sure it’s exactly measured), and “P” is the average period of time that you found by combining the 10 periods above. And, of course, p = 3.14.

V Data & Calculations:

1. Exact length of string, in meters: _______________

Table of Data

Trial # Time (seconds) Trial # Time (seconds)

1 6
2 7
3 8
4 9
5 10

Average Period of Oscillation of these ten trials (seconds)

P = __________________

YOUR acceleration of gravity, in m/s2:

g = y [2 p / P] 2 = ______________________
-MORE-

Physics Lab 6 Thursday, August 12, 2010 page 3 Name _____________

The true value of acceleration is g0 = - 9.8 m/s2.

VI Results:
The purpose of determining the acceleration of gravity using the equipment and procedures above was or was not achieved due to: (explain in detail)

VII Error Analysis:

A. Qualitative Error:
Personal: (what did you or your partner do to screw up?)
Systematic: (what external factors happened that you could not control, e.g., broken equipment, doing the lab in a hurricane, etc.)
Random: There is always random error, unless one does multiple trials. Since we did 10 trials and took an average, there is NO random error in this lab.

B. Quantitative Error:
Find the quantitative error for this experiment: (% error). Find this by using the error analysis formula:

|[True Answer – Your Answer]| / [True Answer] x 100% = ________%

Remember, the true value of acceleration is g0 = - 9.8 m/s2.

VIII Questions
Information: The acceleration of gravity, g, for any planet, is equal to: g = GM / R2 where G is the universal constant of Gravity = 6.67 x 10-11 Nm2/kg2; M is the mass of any planet given, and R is the radius of any planet given. You are dividing GM by the square of R = R2.

1. The mass of Mars is M = 6.4 x 1023 kg; the radius of Mars is R = 3.4 x 106 meters. Find the acceleration of gravity of Mars, g♂.

2. The mass of Jupiter is M = 1.9 x 1027 kg ; the radius of Jupiter is R = 7.13 x 107 m . Find the acceleration of gravity of Jupiter, g♃.

END

LESSON 5

2010-08-10T18:37:00.000-07:00

PHYS 1100 LESSON 5 FOR
TUESDAY, AUGUST 10, 2010

I Introduction

II Logistics

III Return of tests, labs, papers; collection of overdue assignments

IV Running Grades

V Review of Lesson 4: Newton's Laws of Motion
A. Objects at rest, stay at rest; objects in motion stay in motion; UNLESS acted upon by an Fext.
B. F = m a à ties into the Universal Law of Gravity
C. Every force has an equal and opposite force, or, F1 = - F2.
D. Units of force are “Newtons”, kg m/s2. = N

F = ma
v = x/t
VI Lesson 5: Forces

A. The force of friction, Ff = mN = + mmg, which means the force of friction equals a number (m) that is called the “coefficient of friction” multiplied by the equal and opposite force to gravity, N, the Normal force; or N = mg, where m is the “mass” and g is the acceleration of gravity, “9.8 m/s2”. When I put in the “plus sign, +,” that means the Normal force is positive, whereas gravity is negative (down). “Normal” here means “perpendicular to the plane of the table,” or, up.

B. Hooke's Law, Energy, Work, Momentum
1. Hooke’s Law deals with springs (like in pens, in cars as shock absorbers, in toys like the Slinky...)

F = - k x, which means that the force, which is negative, pushes (or pulls) in the opposite direction to the external force. If I stretch a spring, it will recoil back to its original shape when I let go; if I compress a spring, it will push itself out to its original shape once I stop compressing it. The “k” is called the “spring constant.” A large k means a very strong spring, like in a shock absorber. A very small k means a very weak spring, like a Slinky. “x” means how far the spring is compressed in, or, stretched out.

In the Hooke's Law lab, we will use F = - k y instead, since we will be stretching a string downward in the negative y-direction. The force that we will apply will be a weight which will have the force of = m g. So, in the lab,
- k y = m g. We will measure “y” with a ruler, we will know the mass, and we know gravity.

2. Forms of Energy include, but are not restricted to:
· Heat
· Light
· Mechanical
· Acoustical, etc.

3. Kinetic energy (K.E.) is in motion = ½ mv2, the mass must have a speed or velocity. If it is not moving, NO K.E.
Potential energy (P.E.) is stored energy. For storing potential energy when working against the force of gravity, P.E. = mgh = (mass)(acceleration)(height above starting point).

An Example: take a 5-kg cat and carry him up a hill about 100 meters. You will have stored this much P.E. in the cat: mgh = (5kg)(- 9.8 m/s2)(100 m) ~ 5000 Joules.

If you then release (drop) the cat over the edge of a cliff that is 100 meters above where you started, 5000 Joules will be released and turn into K.E. as the cat is falling, until he impacts with the ground. The instant just before impact all the 5000 Joules have changed from P.E. to K.E. After impact, the energy then is transferred into heat (thermal energy), sound (acoustical energy), destruction (mechanical energy), and so forth.

4. The only good thing about energy is to use it to do work for us. However, nothing is 100% efficient, so the amount of work done divided by the available energy to do that work is a ratio: Work / Energy < 1, which means, less than 100%.

If I were to use a force, F, and push a mass, m, a distance, d, then the amount of work that I would do on that object would be the force times the distance moved, or F x d = energy, units 1.0 N-m == Joule

5. In chemistry, calories are used as the main concept of energy, and 4.186 calorie = 1 joule.
However, 1,000 calories, or, 1 kcal is what is called a “food calorie” or, a Calorie with a capital “C”. 1.0 Calorie = kcal.

The stuff below was NOT covered, but will be:

6. Momentum = energy of motion

p = m v

Dp = 0 in a closed system

VII Laboratory Exercise 5: Potential and Kinetic Energy

VIII HWK Assignment 2: 5-9 and do Problems 1, 3, 9, 13, 19, 25, 29, 31, and 39 on pp 140-143; Problems 1, 3, 9, 13, 19, 25, 29, 31, 39, 53, 63, and 65 on pp 177-184; Problems 1, 3, 9, 13, 19, 25, 29, 31, 39, and 53 on pp 210-213; and Problems 1, 3, 9, 13, 19, 25, 29, 31, 39, and 53 on pp 244-248; Problems 1, 3, 9, 13, 19, 25, 29, 31, 39, and 53 on pp 289-292.due 8/12

IX Essay 2: Sir Isaac Newton, due 8/12

LAB 5

2010-08-10T18:36:00.001-07:00

Physics Lab 5 Tuesday, August 10, 2010 Name __________________

I Title: Acceleration of Gravity

II Purpose: To determine the acceleration of gravity using simple equipment.

III Equipment
1. String, yarn, or cord ≥ 1.0 meter long (or as close as possible)
2. Weight - to make a plumb bob pendulum (washers?)
3. Stop Watch, face watch, digital watch, clock, or other chronometer
4. Meter stick or metric ruler
5. Weighted Ring Stand Optional

IV Procedure
1. Obtain, or make, a length of string or cord that is very close to 1.00 meter long. Slightly longer is better than slightly shorter.
2. Attach a weight to one end of the string to create a plumb bob, that we will call Bob.
3. Have that same person, or another, attach the other end of the string to some stationary object (door hinge, ceiling, weighted ring stand, etc.). Do NOT use a primate because it is not stable or other mammal to hold Bob because it is not stable.
4. Measure the length of Bob exactly (to the closest millimeter) after you have set it up. This will be from the point of connection on top to the middle of the weight. Record this length (we will call the length “y”) as accurately as possible.
5. Use the Data Table below; one column is for the number of the trials; the other for the time (in seconds) for each cycle.

6. Have one of the lab partners pull the pendulum back, to about an angle of θ = 45° (but no further) as seen in the diagram to the right
6. Simultaneously, release Bob and depress the stop watch button to start the time “running.” It is best to have the same homosapien release Bob and operate the stopwatch (the same brain controls both hands).
7. Allow Bob to swing out and come back to where it was released. Stop the watch. That is one cycle. Record the time in the table. Reset the watch and get ready to repeat.
8. Have the same CroMagnon Repeat steps #6 - #8, nine more times, and place your data in the table. You should have a total of ten trials.
9. Find the average period of oscillation of the ten trials. This means, add up all the numbers in the second column, then divide by 10. Record.
10. Use the data to find the acceleration of Earth’s gravity, g, by re-writing Galileo’s Period-Length equation so that the acceleration, g, is all alone on the left side.

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Physics Lab 5 Tuesday, August 1, 2010 page 2 Name _____________

P = 2 p [(y / g)] ½

P is the period in seconds, y is the length in meters, and g is what we want to find. It will be in meters per square second. This equation reads “Period equals two times pi times the square root of (y/g).”

Let’s continue to re-write this relationship, in order for us to get, g all alone:

[P / 2 p ] = [(y / g)] ½

[P / 2 p ]2 = (y / g)

g = y [2 p / P] 2

Remember that “y” is the length of the string (about 1.0 meter, but make sure it’s exactly measured), and “P” is the average period of time that you found by combining the 10 periods above. And, of course, p = 3.14.

V Data & Calculations:

1. Exact length of string, in meters: _______________

Table of Data

Trial # Time (seconds) Trial # Time (seconds)

1 6
2 7
3 8
4 9
5 10

Average Period of Oscillation of these ten trials (seconds)

P = __________________

YOUR acceleration of gravity, in m/s2:

g = y [2 p / P] 2 = ______________________
-MORE-

Physics Lab 5 Tuesday, August 10, 2010 page 3 Name _____________

The true value of acceleration is g0 = - 9.8 m/s2.

VI Results:
The purpose of determining the acceleration of gravity using the equipment and procedures above was or was not achieved due to: (explain in detail)

VII Error Analysis:

A. Qualitative Error:
Personal: (what did you or your partner do to screw up?)
Systematic: (what external factors happened that you could not control, e.g., broken equipment, doing the lab in a hurricane, etc.)
Random: There is always random error, unless one does multiple trials. Since we did 10 trials and took an average, there is NO random error in this lab.

B. Quantitative Error:
Find the quantitative error for this experiment: (% error). Find this by using the error analysis formula:

|[True Answer – Your Answer]| / [True Answer] x 100% = ________%

Remember, the true value of acceleration is g0 = - 9.8 m/s2.

VIII Questions
Information: The acceleration of gravity, g, for any planet, is equal to: g = GM / R2 where G is the universal constant of Gravity = 6.67 x 10-11 Nm2/kg2; M is the mass of any planet given, and R is the radius of any planet given. You are dividing GM by the square of R = R2.

1. The mass of Mars is M = 6.4 x 1023 kg; the radius of Mars is R = 3.4 x 106 meters. Find the acceleration of gravity of Mars, g♂.

2. The mass of Jupiter is M = 1.9 x 1027 kg ; the radius of Jupiter is R = 7.13 x 107 m . Find the acceleration of gravity of Jupiter, g♃.

END

Solution Set 2

2010-08-09T20:55:00.001-07:00

Chapter 5: Problems 1, 3, 9, 13, 19, 25, 29, 31, and 39 on pp 140-143

1. An object of mass, m, is initially at rest. After a force of magnitude, F, acts on it for a time, t, the object has a speed of “v.” If the mass of the object is doubled, and the force is quadrupled, How long does it take for the object to accelerate from rest to a speed of “v” now?
Solution: At t=0, v= 0 for the mass. At some other time, t, v=? A force is applied, F. We know that m v = F t, so v = F t/m = (F/m) t = a t.

3. In a grocery store, you push a 12.3-kg shopping cart with a force of 10.1 Newtons. If the cart starts at rest, how far does the cart move in 2.50 sec?
Solution: In a grocery store, you push a 12.3-kg shopping cart with a force of 10.1 Newtons. If the cart starts at rest, how far does the cart move in 2.50 sec? m, F, t; x?
F = m a, then a = F/m = 10.1/12.3 = 0.82 m/s2.
V = Dx/Dt = x/t
Vi = 0
Vf = ?
Dv/Dt = a; Vf = a t = (0.82 m/s2)(2.50 s) = 2.05 m/s.= x/t
2.5(2.05) = x = 5.125 m = 5.13 m

9. In baseball a pitcher can accelerate a 0.15-kg ball from rest to 98 mi//h in a distance of 1.7 meters.
a. Find the average force exerted on the ball during the pitch.
b. If the mass of the ball is increased is the force required of the pitcher increased, decreased, or unchanged? Explain.
Solution: m, v, x; F?
98 mi/h ~ 160 km/h = 160,000 m/3600s = 44.4 m/s
F = m a = (0.15 kg)(1162 m/s2) = 174 N.
a = v/t = 98/t; t = v/a = x/v
v = x/t; t = x/v; v2 = a x
a = v2/x = (44.4)2 / 1.7 m = 1975/1.7 = 1162 m/s2.

13. A drag racer crosses the finish line doing 202 mph and promptly deploys a drag chute.
a. What force must the drag chute exert on the 891-kg car to slow it to 45.0 mph over a distance of 185 meters.
b. Describe how you came to that solution.
Solution: vi, m, vf, x, F? Vi = 202 mph ~ 320 km/h = 320,000 m/h = 320,000/3600 m/s = 88.9 m/s, and 45 mph ~ 72 km/h = 72,000 m/h = 72,000 m/3600 sec = 20 m/s.
We know that F = m a, and we have m, but must find “a.” Using vf2 – vi2 = 2 a x, we see that a = (vf2 – vi2)/2x = [(20)2 – (88.9)2]/(2)(185) = [400-7901]/(370) = - 20.27 m/s2. Thus, F = (891)(20.27 m/s2) = - 18064 N = - 1.8 x 10^3 N.

19. A 71-kg parent and a 19-kg child meet at the center of an ice rink. They pace their hands together and push.
a. Is the force experienced by the child more than, less than, or equal to the force experienced by the parent?
b. Is the acceleration experienced by the child more than, less than, or equal to the force experienced by the parent?
c. If the acceleration of the child is 2.6 m/s2, what is the parent's acceleration?
Solution:

a. same
b. more
c. mp ap = mc ac ; mc ac / mp = ap;
(mc/mp) ac = (19/71) 2.6 m/s2 =
0.2676 (2.6 m/s2) = 0.7 m/s2.

25. A farm tractor pulls a 3700-kg trailer up an 18° incline with a steady speed of 3.2 m/s. What force does the tractor exert on the trailer (ignore friction)?
Solution: Without the tractor, the trailer would accelerate down at a = g sin (18°) = 9.8 (0.309) = 3.03 m/s2, so that means the tractor must be pulling up with an equal and opposite force, F = m a = (3700 kg)(3.03 m/s2) = 11,211 N. = 1.1 x 10^4 N. The net force is zero, as the velocity (speed) is constant.

29. A hockey puck is acted upon by one or more forces as shown in Figure 5-24 on page 142. Rank the four cases, A, B, C, and D, in order of the magnitude of the puck's acceleration, starting with the smallest. Indicate ties where appropriate.
Solution: Later

31. Before practicing his routine on the rings, a 67-kg gymnast stands motionless with one hand grasping each ring and his feet touching the ground. Both arms slope upward at an angle of 24° above the horizontal.
a. if the force exerted by the rings on each arm has a magnitude of 290 N and is directed along the length of the arm, what is the magnitude of the force exerted by the floor on his feet?
b. If the angle that his arms make with the horizontal is great than 24°, and everything else remain the same, is the force exerted by the floor on his feet greater, less, or the same as the value found in (a)? Explain.
Solution: Each arm has a y-component of (290 N)(Cos 24) = 264.93 N, so two arms has ~530 N. The 67-kg gymnast would normally have m g = (67)(9.8) = 657 N, so the difference is (657 - 530) = 127 N.

39. At the bow of ship on a stormy sea a crewman conducts an experiment by standing on a bathroom scale. In calm water, the scale reads 182 lb. During he storm the crewman finds a maximum reading of 225 lb and a minimum reading of 138 lb. Find (a) the maximum upward acceleration and (b) the maximum downward acceleration experienced by the crewman.
Solution:

Chapter 6: Problems 1, 3, 9, 13, 19, 25, 29, 31, 39, 53, 63, and 65 on pp 177-184

1. You push two identical bricks across a tabletop with constant speed, v, as shown figure 6-16 on page 178 in the book. The the first case, you place the bricks end to end. In the second case you stack the bricks one on top of the other.
a. Is the force of kinetic friction in the first case greater, less, or equal to the force of Kinetic friction in case 2?
b. Choose the best explanation
I. The normal force in case #2 is larger, and hence the bricks press down more firmly against the table.
II. The normal force is the same on both cases, and friction is independent of surface area.
III. The first case has more surface area in contact with the table top and this leads to more friction.
Solution:

3. A baseball player slides into 3rd base with an initial speed of 4.0 m/s. If the coefficient of friction between the player and the found is 0.46, how far does the player slide before coming to rest?
Solution:
9. A tie of uniform width is laid out ion a table with a fraction of its length hanging over the edge. Initially, the tie is at rest.
a. if the friction hanging from the table is increased, the tie eventually slides to the ground. Explain.
b. what is the coefficient of static friction between the tie and table if the tie begins to slide when 1/4th of its length hangs over the edge?
Solution:
13. A 97-kg sprinter wishes to accelerate from rest to a speed of 13 m/s in a distance of 22 m.
a. what coefficient of static friction is required between the sprinter's shoes and the track?
b. Explain the strategy used to get this answer.
Solution:
19. A certain spring has a force constant, k.
a. if this spring is cut in half does the resulting half spring have a force constant that is greater than, less than, or equal to k?
b. If two of the original full length springs are connected end to end, does the resulting double spring have a force constant that is greater than, less than, or equal to k?
Solution:
25. If the 52-N backpack in figure 6-19 on page 179 in the book begins to slide when the spring stretches by 2.50 cm, what is the coefficient of static friction between the backpack and the table? Let k = 150 N/m.
Solution:
29. Your friend's 13.6-g graduation tassel hangs on a string from his rear-view mirror.
a. When is accelerates from a stop, the tassel deflects backward toward the rear of the car. Explain.
b. if the tassel hangs at an angle of 6.44° relative to the vertical, what is the acceleration of the car?
Solution:
31. A picture hangs on a wall suspended by two strings, as shown in figure 6-21 on page 180 in the book. The tension in String 1 (on the left) is 1.7 Newtons.
a. Is the tension in String 2 greater, lesser, or the same as in String 1? Explain.
b. Calculate the tension in String 2 to verify
c. Find the weight of the picture in Newtons
Solution:
39. A 0.15 kg ball is placed in a shallow wedge with an open angle of 120° as shown in figure 6-27 on page 181 in the book. For each contact point between the wedge and the ball, determine the force exerted on the ball. Assume no friction.
Solution:
53. A car is driven with a constant speed around a circular track. Answer each of these following question with a yes or no.
a. Is the car's velocity constant?
b. Is the car's speed constant?
c. Is the acceleration constant?
d. Is the acceleration direction constant?
Solution:
63. At what speed must you drive over the hump in the road as seen in figure 6-35 on page 183 in the book if your passengers are going to experience the phenomenon of weightlessness? The radius of curvature of the hump is 35 meters.
Solution:
65. If you weigh yourself on a “bathroom scale” at the equator, is it higher, lower, or the same as it would be a the North Pole? Explain.
Solution:
Chapter 7: Problems 1, 3, 9, 13, 19, 25, 29, 31, 39, and 53 on pp 210-213

1. The International Space Station (ISS) orbits Earth in a circular orbit about 375 km above the surface. Over one complete orbit, is the work done by Earth on the ISS positive, negative, or zero? Explain.
Solution:
3. A plumb Bob pendulum is seen in figure 7-14 on page 211 in the book.
a. Is the work done on Bob by Earth's gravity positive, negative, or zero? Explain.
Solution:
9. A towing rope, parallel to the water pulls a water skier directly behind the boat with constant velocity for a distance of 65 meters before the skier falls. The rope's tension is 120 N.
a. Is the work done on the skier positive, negative, or zero? Explain.
Solution:
13. To clean a floor, a custodian pushes on a mop handle with a force of 50.0 N.
a. If the mop handle is at an angle of 55° above the horizontal, how much work is required to push the mop a distance of 0.5 meter?
b. If the angle is increased to 65°, does the work done increase, decrease, or stay the same? Explain.
Solution:
19. How much work is needed for a 73-kg runner to accelerate from rest to 7.7 m/s?
Solution:
25. A pine cone of 0.14 kg mass falls 16 meters to the ground landing at 13 m/s.
a. How much work was done on the pinecone by air resistance?
b. What was the average force of air resistance on the pinecone?
Solution:
29. A car of 1100 kg coasts on a horizontal road at 19 m/s. After crossing an un-paved sand stretch 32 meters long its speed decreases to 12 m/s.
a. If the sandy portion had been only 16 meters long, would the car speed have decreased by 3.5 m/s, more, or less? Explain.
b. Calculate the change of speed.
Solution:
31. A block of mass, m, and speed, v, collides with a spring, compressing it a distance of Dx. What is the compression of the spring if the force constant of the spring is increase by 4 times?
39. It takes 180 Joules of work to compress a certain spring 0.15 meter.
a. What is the force constant of the spring?
b. To compress it another 0.15 meter, will it require 180 Joules, more, or less? Explain.
Solution:
53. A certain car can accelerate from rest to a speed, v, in “t” seconds. If the power output of the car remains constant,
a. How long does it take for the car to accelerate from “v” to “2v”?
b. How fast is the car moving at a time of “2t”?
Solution:
Chapter 8: Problems 1, 3, 9, 13, 19, 25, 29, 31, 39, and 53 on pp 244-248

1. The work done by a conservative force is indicated in figure 8-14 on page 244 in the book, is for a variety of different paths connected to the point A and B. What is the work done by this force on path 1 and on path 2?
Solution:
3. Calculate the work done by friction as a 3.7-kg box is slid along a floor from point A to point B as in figure 8-16 on page 244 in the book. Do this for all three paths: 1, 2, and 3. Assume that the coefficient of kinetic friction between the box and the floor is 0.26.
Solution:
9. As an Acapulco cliff diver drops to the water from a height of 46 meters, his gravitational potential energy decreases by 25,000 Joules. Find his weight in Newtons.
13. A vertical spring stores energy (0.962 J) as spring potential energy. When a 3.5-kg mass is suspended from it.
a. by what multiplicative factor does the spring potential energy change if the mass attached to the spring is doubled?
b. Verify it.
Solution:
19. Suppose the situation descrige in Conceptual Checkpoint 8-2 (page 232 in the book) is repeated on the fictional planet Epsilon, where the acceleration due to its gravity is less than it is on Earth.
a. Would the height of a hill on Epsilon that causes a reduction in speed from 4.0 m/s to 0.0 m/s be greater, lesser, or equal to that on Earth. Explain.
b. Re-consider this Epsilonian hill. If the initial speed at the bottom of the hill is 5.0 m/s will the final speed at the top of the hill be greater, less, or equal to 3.0- m/s? Explain.
Solution:
25. At a water park, a swimmer uses a water slide to enter the main swimming pool. If the swimmer starts at the top of the slide with an initial velocity of 0.840 m/s, find the swimmer's speed at the bottom of the slide. Assume that the height of the slide is 2.31 meters and that the slide has no friction.
Solution:
29. An apple of mass 0.21 kg falls of an apple tree and lands on the ground, 4.0 meters below. Determine the apple's kinetic energy (KE), gravitational potential energy (PE), and total mechanical energy of the system (E) when the apple's height (y) is
a. 4.0 m
b. 3.0 m
c. 2.0 m
d. 1.0 m
e. 0.0 m
Ignore air friction. Also the ground is at y = 0.0 meters
Solution:
31. A rock of mass 0.26 kg is thrown “straight up” from a cliff that is 32 meters above the ground level (where y = 0.0 m). The rock rises, stops, and then is in free fall all the way to the bottom of the cliff, where its final speed, just before impact, is 29.0 m/s. Ignore air friction. For the sake of this problem, assume that the rock is thrown up from slightly beyond the cliff's edge (or else it would just hit the top of the cliff).
a. Find the initial speed of the rock (upward)
b. The maximum height of the rock (y) relative to the base of the cliff.
Solution:
39. When the Space Shuttle re-enters the Earth's atmosphere, its protective tiles get really hot.,
a. Is the mechanical energy of the Shuttle-Earth system when the Shuttle lands, greater, less, or equal to when it is in orbit?
b. Choose the best explanation from the three choices below.
I Dropping out of orbit increases the mechanical energy of the Shuttle
II Gravity is a conservative force
III A portion of the mechanical energy has been converted to heat energy.
Solution:
53. A block of mass 1.80 kg slides along a rough, horizontal surface. The block hits a spring with a speed of 2.00 m/s and compresses it a distance of 11.0 cm before coming to rest. If the coefficient of kinetic friction between the block and the surface is mk = 0.560, what is the force constant of the spring (k)?
Solution: skip this one

Chapter 9: Problems 1, 3, 9, 13, 19, 25, 29, 31, 39, and 53 on pp 289-292

1. Exercise 9-1 on page 255 states: “An 1180-kg car drives along a city street at 30.0 miles per hour = 13.4 m/s.
(a) “What is the magnitude of the car's momentum?
(b) “A major league pitcher can give a 0.142-kg baseball a speed of 101 miles per hour (45.1 m/s). Find the magnitude of the baseball's momentum.”

The answer to (a) is pcar = 15,800 kg-m/s, where “p” stands for momentum, i.e.,
p = m v.

In this first question, it asks us, “What speed must the baseball go if its momentum were to equal that of the car?” In other words, how fast must the ball go if it, too, has a momentum of p = 15,800 kg-m/s? They want the answer in miles per hour, probably because you are still converting back and forth and they don't want you to forget. (“They” refers to the authors).

pcar = 15,800 kg-m/s and pball = (0.142 kg)(vball), so now we need to find vball.

vball = (15,800 kg-m/s) / (0.142 kg) = 111,268 m/s = 111 km/sec = 69.54 miles/sec = 250,352 mi/hour = 2.5 x 105 mi/hr.

3. A 26.2-kg dog is running northward at 2.7 m/s (it's a vector!), while a 5.30-kg cat is running eastward at 3.04 m/s (another vector!). Their 74.0-kg owner has the same momentum s the two pets taken together. Find the direction and magnitude of the owner's velocity.

Solution: Dog is 26.2 kg, 2.7 m/s @90°, cat 5.3 kg, 3.04 m/s @ 0°. Homosapien, 74 kg.
p1 = m1v1 = (26.2)(2.7) = 70.7 kgm/s @ 90°
p2 = m2v2 = (5.3)(3.04) = 16.1 kgm/s @ 0°.

pt2 = (70.7)2 + (16.1)2 = 4998.5 + 259.6 = 5258

pt = (5258)½ = 72.5 m/s;

16.1 = 72.5 cos a

cos a = (16.1)/(72.5) = 0.222

a = cos-1 (0.222) = 77°

p = 72.5 kgm/s @ 77°.

v = p/m = (72.5 kgm/s)/(74.0 kg) = 0.98 m/s = 9.8 x 10-1 m/s @ 77°.

9. A net force of F = 200 N acts on a 100-kg boulder and a force of the same magnitude acts on a 100-g pebble.
(a) is the change of the boulder's momentum in one second greater than, less than, or equal to the change of the pebble's momentum over the same time period?
(b) Choose the best explanation from among the 3 following:
I. The large mass of the boulder gives it the greater momentum
II. The force causes a much greater speed in the 100-g pebble, resulting in more momentum.
III. Equal force means equal change in momentum for a given time.
Solution: Momentum, p = m v, but it is equivalent to F Dt.
(a) for the boulder, Dp = F Dt = (200 N)(1.0 second) = 200 N-s = 200 kg-m/s;
for the pebble, Dp = F Dt = (200 N)(1.0 second) = 200 N-s = 200 kg-m/s;
Thus, the same
(b) III

13. Find the magnitude of the impulse delivered to a soccer ball when a player kicks it with a force of 1,250.00 N. Assume that the player's food is in contact with the ball for 5.95 x 10-3 seconds.
Solution: Impulse is force multiplied by the small incremental change in time, I = F Dt, so in this case, I = (1250.00 N)(5.95 x 10-3 s) = 74337.5 10-3 = 74.3375 = 7.43 x 101 N-s.

19. A 0.14-kg baseball moves toward home plate with a velocity (a vector) of v = (- 36 m/s)x. After striking the bat, the ball moves vertically upward with a velocity of v = (+18 m/s)ŷ
(a) Find the direction and magnitude of the impulse delivered to the ball by the bat. Assume contact between them lasted a time of Dt = 1.5 milliseconds.
(b) How would your answer to part (a) change if the mass of the ball were doubled?
(c) How would your answer to part (a) change if the mass of the bat were doubled?
Solution: Remember that I = F Dt = D(mv) = mDv. (The mass does not change).
(a) I = (0.14 kg)(- 36 – 18) = (0.14 kg)(– 54) = - 7.5 N-s.
(b) It would double.
(c) no change.

25. A 92-kg astronaut and a 1200-kg satellite are at rest relative to a Space Shuttle near by. The astronaut pushes on the satellite, give it a speed of 0.14 m/s directly away from the Shuttle. Then, 7.5 seconds later, the astronaut comes into contact with the Shuttle. What was the initial distance from the Shuttle to the Astronaut?
Solution: In this case, there is no momentum in the system, so after the astronaut pushes the satellite, the net momentum is still zero. So, m1v1 + m2v2 = 0. But m1 = 92 kg, v1 = ?; m2 = 1200 kg, v2 = - 0.14 m/s. So, let's set up a ratio type thing:

m1v1 + m2v2 = 0

m1v1 = - m2v2

v1 = - [(m2) / (m1)] v2 = - [(1200 kg) / (92 kg)] (- 0.14 m/s) = + 1.83 m/s. If the astronaut traveled 1.83 m/s for 7.5 seconds, then he traveled a distance of x = (1.83)(7.5)= 13.7 m, or 1.37 x 101 meters.

29. In Example 9-6 on page 271, a 950-kg car moving at v = 16 m/s east crashes into a 1300-kg minivan going v = 21 m/s north; they stick together and the wreckage moves off towards the general direction of northeast.

Now, let's change some stuff …. suppose the car in Example 9-6 has an initial speed of 20.0 m/s (not 16.0 m/s) and that the direction of the wreckage after the collision is 40.0° above the x-axis (counter-clockwise from the + x-axis). Find the initial speed of the minivan and the final speed of the combined wreckage.
Solution: In a “closed” system, which this is, we don't have any outside sources of speed, momentum, force, or acceleration. So, the momentum of both vehicles before the impact must equal the momentum of both vehicles (stuck together) after the collision.

pi = pc + pv = mcvc + mvvv = pf = (mc + mv)vf .And we are looking for vv.

mcvc = (950 kg)(20 m/s) = 19,000 kg-m/sec @ 0°. This is the initial momentum for the car.

mvvv = (1300 kg)(21 m/s) = 27,300 kg-m/sec @ 90°. This is the initial momentum for the van.

As the initial momenta are at right angles, then the net original momentum would be the vector addition of pc + pv = 19,000 @ 0° + 27,300 @ 90°. In this case, using Pythagoras, a = 19,000 and b = 27,300, so we need c, which we get from the Pythagorean relationship: a2 + b2 = c2, or, (1.9 x 104)2 + (2.73 x 104)2 = (c)2 ; = (3.61 x 108) + (7.45 x 108) = 11.06 x 108. Therefore, c = √(11.06 x 108) = 3.33 x 104 kg-m/s. But what is the angle, θ?
Tanθ = (27300)/(19000) = (27.3)/(19) = 1.437; so the angle is: θ = tan-1 (1.437) = 55.17°.
So, in the end, the final momentum of the wreckage is 3.33 x 104 kg-m/s @ 55.17°

However, what if the angle is 40.0°? Then that means that Tanθ = tan 40.0° = 0.839. So, pv/pc = 0.839; and pc is still 19,000 kg-m/s @ 0°. Thus, pv/pc = (pv)/(19,000) = 0.839. Or, pv = (19000)(0.839) = 15,943 kg-m/s. Since the mass of the van is 1300 kg, then the initial velocity of the mini van is: pv = mvvv = (1300 kg)(vv) = 15,943 kg-m/s, or, vv = (15,943 kg-m/s) / (1300 kg) = 12.26 m/s.

31. Exercise 9-2 on page 268 states: “A 1200-kg car moving at 2.5 m/s is struck in the rear by a 2600-kg truck moving at 6.2 m/s. If the vehicles stick together after the collision, what's their speed immediately after the collision?”

The answer is vf = 5.0 m/s.

In this question, it asks us,(a) is the final KE of the car+truck greater than, less than, or equal to the sum of the initial KE's of the car and the truck separately? In other words, is KEf > KE1 + KE2, is KEf < KE1 + KE2, or is KEf = KE1 + KE2. Explain.
(b) Verify your answer to part (a) by calculating the initial and final KE's of the system.
Solution: Okay, the relationship for kinetic energy is: KE = ½ m v2.
(a) The KE of the car, or KE1 = ½ m1 v12 = (0.5)(1200)(2.5)2 = 3,750 Joules.
The KE of the truck, or KE2 = ½ m2 v22 = (0.5)(2600)(6.2)2 = 49,972 Joules.
So, KE1 + KE2 = 3750 Joules + 49,972 Joules = 53,722 Joules. But KEf = ½ (m1 + m2)vf2 = (0.5)(3800 kg)(5.0 m/s)2 = 47,500 Joules. Since 53,722 Joules > 47,500 Joules, then KEf < KE1 + KE2 or KE1 + KE2 > KEf.
(b) we did this already in part (a)

39. A charging bull elephant with a mass of M = 5240 kg comes directly towards you at ve = 4.55 m/s. In a panic, you toss a rubber ball of mass, m = 0.150 kg , at the elephant, with a speed of vb = 7.81 m/s.
(a) When the ball bounces back towards you, what is its speed?
(b) How do you account for the fact that the ball's KE has increased?
Solution:
(a) 4.55 + 7.81 = 12.36 m/s.
(b) The hefalump gave it more energy.

53. Three uniform meter sticks, each of mass, m, are placed on the floor as follows: stick 1 lies along the y-axis from (0,0) to (0,1); stick 2 lies along the x-axis from (0,0) to (1,0). Stick 3 lies along the x-axis from (1,0) to (2,0).
(a) Find the location of the center of mass of the meter sticks.
(b) How would the location of the center of mass be affected if the mass of the meter sticks were doubled?
Solution: This is the same as if we had a wooden stick of 1 meter whose center of mass is at 0.5 meters in the y-direction, and a wooden stick of 2 meters, whose center of mass is at 1.0 meters in the x-direction.
(a) So the center of mass is some where between (0, 0.5) and (1.0, 0). And in fact, it's at
(1.0, 0.5).
(b) no change

END

LAB EXERCISE 4

2010-08-08T06:16:00.001-07:00

PhysicsLab4, Monday, August 9, 2010 NAME __________________
Dr Dave Menke, Instructor
I Title: Centripetal Acceleration

II Purpose: To study centripetal acceleration and have fun
Theory: Planets, like Earth, travel around the Sun similar to how a weight on a string travels in a circular path if you swing it around. For the Sun and the planets, there is no “string,” but the force is Gravity. The planets are like weights. Each planet has a velocity or speed and an acceleration. You will notice that the force in this lab is a central force, so that the acceleration is a central one, i.e., ac.

III Equipment
- White String
- Metal weight (washer, nut, whatever)
- Wooden metric ruler
- stopwatch
- Scissors

IV Procedure
1. Select a weight
2. Obtain approximately a 1.0-meter length of string
3. Attach the weight to one end of the string, just like before
4. Suspend the (string + weight) by holding the top of the string tightly at one end. This is your friend, “plumb bob.” He’s related to Sponge Bob, Bob the Builder, and 97.5 Bob FM. He’s the same guy as last time,
5. Measure exactly the length, l, of Bob (from your fingers to the middle of the weight) in meters.
6. Leave the classroom and to find an open area reasonably clear of muggles*.
7. Have one lab partner to practice - carefully - swinging Bob in a circle until he/she/it has achieved a relative constant velocity. Don’t hit anyone. Some students swing it overhead, like a lasso. It is not likely that you will hit anyone who is walking on the ceiling.
8. Have another lab partner practice using the stop watch.
9. When ready, have the swinging partner (SP) begin swinging Bob in circles at a constant rate. When ready, have the stopwatcher lab partner (SWLP) click the stop watch and count 10 cycles, then have the SWLP stop the watch. Record. The SP can keep swinging or not. Personal preference.
10. Repeat this three times to get an average amount of time for each 10-cycle period. Record. Now stop the SP if he/she/it hasn’t already.

*muggle (1) common, ordinary, ignorant person; (2) someone with NO magical powers – from the Harry Potter series of books; (3) a marijuana “joint” – from the 1920’s New Orleans

More…

Lab 3, page 2, August 9 Name _____

11. Return to the classroom, and encourage the SP and SWLP to join you. Put away your toys, and write up your report.

12. Divide your average cycle time by 10 to get the period, P, of one cycle. Record.

13. Find the circumference, c, of the orbital path. Do this by multiplying Bob’s length that you found in #5, l, by the number 2 pi or 2p = 2(3.14). Record.

14. Calculate the average linear velocity, v, of the mass. Do this by dividing the circumference that you found in #13 by the period (time) that you found in #12. Record.

15. Calculate the mean centripetal acceleration, ac, of the mass. Do this by squaring the velocity, v2 (multiply it by itself) that you found in #14 and dividing that by the length of the string that you find in #5, l. Record.

V Data & Calculations (This is where you put your data)
1. Bob’s length, l, in meters: _____________________
2. Trials and Times

TRIAL NUMBER of 10 Cycles TIME IN SECONDS of each 10 Cycles
1
2
3
AVE

3. Period of one cycle, P (divide the average of 10 cycles by 10) ______ s

4. The circumference of the orbital path, 2 p l = ______________ m

5. The average linear velocity of the mass, v = _______________m/s

6. The mean centripetal acceleration of the mass, ac = __________m/s2

VI Results
“The purpose of the lab was to go Bob-Bob-Bobbin’ along.” No, for “reals” it was to study orbital revolutions and have fun, and it (was, was not) [circle one] achieved because …

More…
Lab 3, page 3, August 9 Name _____

VII Error Analysis
A. Quantitative Error – NA

B. Qualitative Error:
1. Personal
2. Systematic
3. Random
VIII Questions
1. Find the circumference of Earth’s orbit around Sun (in meters) if Bob’s length, l, (the radius of Earth’s orbit) is 150,000,000 km, just like you did in Procedure #13 above.
2. Find the period of the Earth’s orbit (in seconds). Do this by multiplying the number of seconds in a day, 86,400, by the number of days in a year, 365.
3. Find the linear velocity of Earth (in m/s). Do this by dividing what you found in Question #1 with what you found in Question #2.
4. Find the centripetal acceleration of Earth around the Sun (in m/s2). Do this by squaring the velocity that you found in Question #3 and then dividing it with the radius of Earth’s orbit, 150,000,000 km.
5. There is no number 5.

HOMEWORK SET 2

2010-08-08T06:03:00.000-07:00

Chapter 5: Problems 1, 3, 9, 13, 19, 25, 29, 31, and 39 on pp 140-143

1. An object of mass, m, is initially at rest. After a force of magnitude, F, acts on it for a time, t, the object has a speed of “v.” If the mass of the object is doubled, and the force is quadrupled, how long does it take for the object to accelerate from rest to a speed of “v” now?
3. In a grocery store, you push a 12.3-kg shopping cart with a force of 10.1 Newtons. If the cart starts at rest, how far does the cart move in 2.50 sec? m, F, t; x?
9. In baseball a pitcher can accelerate a 0.15-kg ball from rest to 98 mi/h in a distance of 1.7 meters. m, v, x; F?
a. Find the average force exerted on the ball during the pitch.
b. If the mass of the ball is increased is the force required of the pitcher increased, decreased, or unchanged? Explain.
13. A drag racer crosses the finish line doing 202 mph and promptly deploys a drag chute.
a. What force must the drag chute exert on the 891-kg car to slow it to 45.0 mph over a distance of 185 meters?
b. Describe how you came to that solution.
19. A 71-kg parent and a 19-kg child meet at the center of an ice rink. They place their hands together and push.
a. Is the force experienced by the child more than, less than, or equal to the force experienced by the parent?
b. Is the acceleration experienced by the child more than, less than, or equal to the force experienced by the parent?
c. If the acceleration of the child is 2.6 m/s2, what is the parent's acceleration?
25. A farm tractor pulls a 3700-kg trailer up an 18° incline with a steady speed of 3.2 m/s. What force does the tractor exert on the trailer (ignore friction).
29. A hockey puck is acted upon by one or more forces as shown in Figure 5-24 on page 142. Rank the four cases, A, B, C, and D, in order of the magnitude of the puck's acceleration, staring with the smallest. Indicate ties where appropriate.
31. Before practicing his routine on the rings, a 67-kg gymnast stands motionless with one hand grasping each ring and his feet touching the ground. Both arms slope upward at an angle of 24° above the horizontal.
a. if the force exerted by the rings on each arm has a magnitude of 290 N and is directed along the length of the arm, what is the magnitude of the force exerted by the floor on his feet?
b. If the angle that his arms make with the horizontal is great than 24°, and everything else remain the same, is the force exerted by the floor on his feet greater, less, or the same as the value found in (a)? Explain.
39. At the bow of ship on a stormy sea a crewman conducts an experiment by standing on a bathroom scale. In calm water, the scale reads 182 lb. During the storm the crewman finds a maximum reading of 225 lb and a minimum reading of 138 lb. Find (a) the maximum upward acceleration and (b) the maximum downward acceleration experienced by the crewman.

Chapter 6: Problems 1, 3, 9, 13, 19, 25, 29, 31, 39, 53, 63, and 65 on pp 177-184

1. You push two identical bricks across a tabletop with constant speed, v, as shown figure 6-16 on page 178 in the book. In the first case, you place the bricks end to end. In the second case you stack the bricks one on top of the other.
a. Is the force of kinetic friction in the first case greater, less, or equal to the force of Kinetic friction in case 2?
b. Choose the best explanation
I. The normal force in case #2 is larger, and hence the bricks press down more firmly against the table.
II. The normal force is the same on both cases, and friction is independent of surface area.
III. The first case has more surface area in contact with the table top and this leads to more friction.
3. A baseball player slides into 3rd base with an initial speed of 4.0 m/s. If the coefficient of friction between the player and the ground is 0.46, how far does the player slide before coming to rest?
9. A tie of uniform width is laid out on a table with a fraction of its length hanging over the edge. Initially, the tie is at rest.
a. if the friction hanging from the table is increased, the tie eventually slides to the ground. Explain.
b. what is the coefficient of static friction between the tie and table if the tie begins to slide when 1/4th of its length hangs over the edge?
13. A 97-kg sprinter wishes to accelerate from rest to a speed of 13 m/s in a distance of 22 m.
a. what coefficient of static friction is required between the sprinter's shoes and the track?
b. Explain the strategy used to get this answer.
19. A certain spring has a force constant, k.
a. if this spring is cut in half does the resulting half spring have a force constant that is greater than, less than, or equal to k?
b. If two of the original full length springs are connected end to end, does the resulting double spring have a force constant that is greater than, less than, or equal to k?
25. If the 52-N backpack in figure 6-19 on page 179 in the book begins to slide when the spring stretches by 2.50 cm, what is the coefficient of static friction between the backpack and the table? Let k = 150 N/m.
29. Your friend's 13.6-g graduation tassel hangs on a string from his rear-view mirror.
a. When is accelerates from a stop, the tassel deflects backward toward the rear of the car. Explain.
b. if the tassel hangs at an angle of 6.44° relative to the vertical, what is the acceleration of the car?
31. A picture hangs on a wall suspended by two strings, as shown in figure 6-21 on page 180 in the book. The tension in String 1 (on the left) is 1.7 Newtons.
a. Is the tension in String 2 greater, lesser, or the same as in String 1? Explain.
b. Calculate the tension in String 2 to verify
c. Find the weight of the picture in Newtons

39. A 0.15 kg ball is placed in a shallow wedge with an open angle of 120° as shown in figure 6-27 on page 181 in the book. For each contact point between the wedge and the ball, determine the force exerted on the ball. Assume no friction.

53. A car is driven with a constant speed around a circular track. Answer each of these following questions with a yes or no.
a. Is the car's velocity constant?
b. Is the car's speed constant?
c. Is the acceleration constant?
d. Is the acceleration direction constant?
63. At what speed must you drive over the hump in the road as seen in figure 6-35 on page 183 in the book if your passengers are going to experience the phenomenon of weightlessness? The radius of curvature of the hump is 35 meters.

65. If you weigh yourself on a “bathroom scale” at the equator, is it higher, lower, or the same as it would be at the North Pole? Explain.

Chapter 7: Problems 1, 3, 9, 13, 19, 25, 29, 31, 39, and 53 on pp 210-213

1. The International Space Station (ISS) orbits Earth in a circular orbit about 375 km above the surface. Over one complete orbit, is the work done by Earth on the ISS positive, negative, or zero? Explain.
3. A plumb Bob pendulum is seen in figure 7-14 on page 211 in the book.
a. Is the work done on Bob by Earth's gravity positive, negative, or zero? Explain.
9. A towing rope, parallel to the water pulls a water skier directly behind the boat with constant velocity for a distance of 65 meters before the skier falls. The rope's tension is 120 N.
a. Is the work done on the skier positive, negative, or zero? Explain.
13. To clean a floor, a custodian pushes on a mop handle with a force of 50.0 N.
a. If the mop handle is at an angle of 55° above the horizontal, how much work is required to push the mop a distance of 0.5 meter?
b. If the angle is increased to 65°, does the work done increase, decrease, or stay the same? Explain.
19. How much work is needed for a 73-kg runner to accelerate from rest to 7.7 m/s?
25. A pine cone of 0.14 kg mass falls 16 meters to the ground landing at 13 m/s.
a. How much work was done on the pinecone by air resistance?
b. What was the average force of air resistance on the pinecone?
29. A car of 1100 kg coasts on a horizontal road at 19 m/s. After crossing an un-paved sand stretch 32 meters long its speed decreases to 12 m/s.
a. If the sandy portion had been only 16 meters long, would the car speed have decreased by 3.5 m/s, more, or less? Explain.
b. Calculate the change of speed.
31. A block of mass, m, and speed, v, collides with a spring, compressing it a distance of Dx. What is the compression of the spring if the force constant of the spring is increase by 4 times?
39. It takes 180 Joules of work to compress a certain spring 0.15 meter.
a. What is the force constant of the spring?
b. To compress it another 0.15 meter, will it require 180 Joules, more, or less? Explain.
53. A certain car can accelerate from rest to a speed, v, in “t” seconds. If the power output of the car remains constant,
a. How long does it take for the car to accelerate from “v” to “2v”?
b. How fast is the car moving at a time of “2t”?

Chapter 8: Problems 1, 3, 9, 13, 19, 25, 29, 31, 39, and 53 on pp 244-248

1. The work done by a conservative force is indicated in figure 8-14 on page 244 in the book, is for a variety of different paths connected to the point A and B. What is the work done by this force on path 1 and on path 2?
3. Calculate the work done by friction as a 3.7-kg box is slid along a floor from point A to point B as in figure 8-16 on page 244 in the book. Do this for all three paths: 1, 2, and 3. Assume that the coefficient of kinetic friction between the box and the floor is 0.26.
9. As an Acapulco cliff diver drops to the water from a height of 46 meters, his gravitational potential energy decreases by 25,000 Joules. Find his weight in Newtons.
13. A vertical spring stores energy (0.962 J) as spring potential energy. When a 3.5-kg mass is suspended from it.
a. by what multiplicative factor does the spring potential energy change if the mass attached to the spring is doubled?
b. Verify it.
19. Suppose the situation describe in Conceptual Checkpoint 8-2 (page 232 in the book) is repeated on the fictional planet Epsilon, where the acceleration due to its gravity is less than it is on Earth.
a. Would the height of a hill on Epsilon that causes a reduction in speed from 4.0 m/s to 0.0 m/s be greater, lesser, or equal to that on Earth? Explain.
b. Re-consider this Epsilonian hill. If the initial speed at the bottom of the hill is 5.0 m/s will the final speed at the top of the hill be greater, less, or equal to 3.0- m/s? Explain.
25. At a water park, a swimmer uses a water slide to enter the main swimming pool. If the swimmer starts at the top of the slide with an initial velocity of 0.840 m/s, find the swimmer's speed at the bottom of the slide. Assume that the height of the slide is 2.31 meters and that the slide has no friction.
29. An apple of mass 0.21 kg falls of an apple tree and lands on the ground, 4.0 meters below. Determine the apple's kinetic energy (KE), gravitational potential energy (PE), and total mechanical energy of the system (E) when the apple's height (y) is
a. 4.0 m
b. 3.0 m
c. 2.0 m
d. 1.0 m
e. 0.0 m
Ignore air friction. Also the ground is at y = 0.0 meters
31. A rock of mass 0.26 kg is thrown “straight up” from a cliff that is 32 meters above the ground level (where y = 0.0 m). The rock rises, stops, and then is in free fall all the way to the bottom of the cliff, where its final speed, just before impact, is 29.0 m/s. Ignore air friction. For the sake of this problem, assume that the rock is thrown up from slightly beyond the cliff's edge (or else it would just hit the top of the cliff).
a. Find the initial speed of the rock (upward)
b. The maximum height of the rock (y) relative to the base of the cliff.
39. When the Space Shuttle re-enters the Earth's atmosphere, its protective tiles get really hot.
a. Is the mechanical energy of the Shuttle-Earth system when the Shuttle lands, greater, less, or equal to when it is in orbit?
b. Choose the best explanation from the three choices below.
I Dropping out of orbit increases the mechanical energy of the Shuttle
II Gravity is a conservative force
III A portion of the mechanical energy has been converted to heat energy.
53. A block of mass 1.80 kg slides along a rough, horizontal surface. The block hits a spring with a speed of 2.00 m/s and compresses it a distance of 11.0 cm before coming to rest. If the coefficient of kinetic friction between the block and the surface is mk = 0.560, what is the force constant of the spring (k)?

Chapter 9: Problems 1, 3, 9, 13, 19, 25, 29, 31, 39, and 53 on pp 289-292

1. Exercise 9-1 on page 255 states: “An 1180-kg car drives along a city street at 30.0 miles per hour = 13.4 m/s.
(a) “What is the magnitude of the car's momentum?
(b) “A major league pitcher can give a 0.142-kg baseball a speed of 101 miles per hour (45.1 m/s). Find the magnitude of the baseball's momentum.”

The answer to (a) is pcar = 15,800 kg-m/s, where “p” stands for momentum, i.e., p = m v.

In this first question, it asks us, “What speed must the baseball go if its momentum were to equal that of the car?” In other words, how fast must the ball go if it, too, has a momentum of p = 15,800 kg-m/s? They want the answer in miles per hour, probably because you are still converting back and forth and they don't want you to forget. (“They” refers to the authors).

3. A 26.2-kg dog is running northward at 2.7 m/s (it's a vector!), while a 5.30-kg cat is running eastward at 3.04 m/s (another vector!). Their 74.0-kg owner has the same momentum s the two pets taken together. Find the direction and magnitude of the owner's velocity.

9. A net force of F = 200 N acts on a 100-kg boulder and a force of the same magnitude acts on a 100-g pebble.
(a) is the change of the boulder's momentum in one second greater than, less than, or equal to the change of the pebble's momentum over the same time period?
(b) Choose the best explanation from among the 3 following:
I. The large mass of the boulder gives it the greater momentum
II. The force causes a much greater speed in the 100-g pebble, resulting in more momentum.
III. Equal force means equal change in momentum for a given time.

13. Find the magnitude of the impulse delivered to a soccer ball when a player kicks it with a force of 1,250.00 N. Assume that the player's food is in contact with the ball for 5.95 x 10-3 seconds.

19. A 0.14-kg baseball moves toward home plate with a velocity (a vector) of v = (- 36 m/s)x. After striking the bat, the ball moves vertically upward with a velocity of v = (+18 m/s)ŷ
(a) Find the direction and magnitude of the impulse delivered to the ball by the bat. Assume contact between them lasted a time of Dt = 1.5 milliseconds.
(b) How would your answer to part (a) change if the mass of the ball were doubled?
(c) How would your answer to part (a) change if the mass of the bat were doubled?

25. A 92-kg astronaut and a 1200-kg satellite are at rest relative to a Space Shuttle near by. The astronaut pushes on the satellite, give it a speed of 0.14 m/s directly away from the Shuttle. Then, 7.5 seconds later, the astronaut comes into contact with the Shuttle. What was the initial distance from the Shuttle to the Astronaut?

29. In Example 9-6 on page 271, a 950-kg car moving at v = 16 m/s east crashes into a 1300-kg minivan going v = 21 m/s north; they stick together and the wreckage moves off towards the general direction of northeast. Now, let's change some stuff …. suppose the car in Example 9-6 has an initial speed of 20.0 m/s (not 16.0 m/s) and that the direction of the wreckage after the collision is 40.0° above the x-axis (counter-clockwise from the + x-axis). Find the initial speed of the minivan and the final speed of the combined wreckage.

31. Exercise 9-2 on page 268 states: “A 1200-kg car moving at 2.5 m/s is struck in the rear by a 2600-kg truck moving at 6.2 m/s. If the vehicles stick together after the collision, what's their speed immediately after the collision?”

The answer is vf = 5.0 m/s.

In this question, it asks us,(a) is the final KE of the car+truck greater than, less than, or equal to the sum of the initial KE's of the car and the truck separately? In other words, is KEf > KE1 + KE2, is KEf < KE1 + KE2, or is KEf = KE1 + KE2. Explain.
(b) Verify your answer to part (a) by calculating the initial and final KE's of the system.

39. A charging bull elephant with a mass of M = 5240 kg comes directly towards you at ve = 4.55 m/s. In a panic, you toss a rubber ball of mass, m = 0.150 kg , at the elephant, with a speed of vb = 7.81 m/s.
(a) When the ball bounces back towards you, what is its speed?
(b) How do you account for the fact that the ball's KE has increased?

53. Three uniform meter sticks, each of mass, m, are placed on the floor as follows: stick 1 lies along the y-axis from (0,0) to (0,1); stick 2 lies along the x-axis from (0,0) to (1,0). Stick 3 lies along the x-axis from (1,0) to (2,0).
(a) Find the location of the center of mass of the meter sticks.
(b) How would the location of the center of mass be affected if the mass of the meter sticks were doubled?

END

LESSON 3

2010-08-05T09:56:00.000-07:00

PHYS 1100 LESSON 3 FOR
THURSDAY, AUGUST 05, 2010

I Introduction

II Logistics: Seating, Syllabus, etc.

III Turn in Assignments Due: Homework 1, Essay 1, etc. and Return of Papers

IV Test 1

V Mind Game 1

VI Review of Lesson 2: Vector Physics/Motion

Vectors
Virtual arrow: magnitude (size) and direction

heads, tails

adding vectors ≠ adding algebraically

adding vectors

align the head of one vector with the tail of another; never put 2 tails together, or 2 heads

together: →→ is

okay; NOT →←

and NOT ←→

→↑ is okay; but ↑→ is not okay; and →↑ is not okay
A vector usually has an arrow (→) above it: , or a “hat” or “carrot” (^) above it:

Pythagoras (576 BC – 495 BC)

Velocity, acceleration, force, momentum, or any number of other concepts can be represented as vectors
Components: see above; the x-component of vector A+B is A; the y-component is B.
If a vector is not directly along the x-axis or along the y-axis, it can be broken down into its x- and y- components
Acceleration vector, along an inclined plane: a = g sin θ, where θ is the angle shown:

Projectile Motion

Range, height, angle (above)

Lesson 3: Circular motion
v = 2pr/P Э 2pr = c (circumference); r = radius; P = period, in seconds, to make one trip around the circle; and P = 1/n, Э n = the frequency in cycles per second (Hz).

v2/r = 4p2r/P2, but since P = (1/n), then P2 = (1/ n)2, or (1/P2) = n2

So, v2/r = 4p2rn2 which can be written as v2/r = (2pn)2 r

And, in circular motion, a = v2/r =
(2pn)2 r, “centripetal acceleration”
However, 2pn = w in rad/sec, thus a = w2r

If part of a circle, say, s, is the arc, AB, then we can say that for small “s” that r sin θ = s, and if it’s even smaller, then r θ = s

because for small θ, sin θ = θr where the angle, θ, is measured in radians, not degrees. 360° = 2p radians, so 1 radian = 57.3°.

Since v = dist/time, then v = s/t = r (θ/t) but is another way of writing (θ/t) = w, so
v = wr and v2/r = w2r = a, “acceleration”

V Lesson 3: Newton's Laws of Motion
Objects at rest, stay at rest; objects in motion stay in motion; UNLESS acted upon by an Fext.
F = m a -à Universal Law of Gravity
Every force has an equal and opposite force, or, F1 = - F2.
Units of force are “Newtons”, kg m/s2. = N

F = ma
V = x/t

V Laboratory Exercise 3: Circular Motion

Centrifugal force is a reaction to a real force. Centripetal

VI HWK Assignment 2: 5-9 and do Problems 1, 3, 9, 13, 19, 25, 29, 31, and 39 on pp 140-143; Problems 1, 3, 9, 13, 19, 25, 29, 31, 39, 53, 63, and 65 on pp 177-184; Problems 1, 3, 9, 13, 19, 25, 29, 31, 39, and 53 on pp 210-213; and Problems 1, 3, 9, 13, 19, 25, 29, 31, 39, and 53 on pp 244-248; Problems 1, 3, 9, 13, 19, 25, 29, 31, 39, and 53 on pp 289-292.due 8/12

VII Essay 2: Sir Isaac Newton, due 8/12

LAB 3 FOR THURSDAY AUGUST 5

2010-08-03T15:31:00.000-07:00

PhysicsLab3 Thursday, August 5, 2010 Name __________________
Dr Dave Menke, Instructor
I. Title: Linear Motion

II. Purpose: To observe objects moving at a constant speed. Graph the relationships; interpret the graphs

III. Equipment
A rolling object (on wheels or a ball)
Brick or block
Graph paper, pencil, ruler
Masking Tape
Metric ruler
Timing Chronometer (Stop Watch or similar)

IV. Procedure
1. Find a clear, flat surface a few meters long.
2. Using masking tape, mark a starting point, known as “The Starting Point.”
3. Place rolling object on starting point.
4. Have the lab partner practice pushing the object with a consistent force to get the same initial speed each time.
5. After practicing to get a consistent speed, push the object, and timing device simultaneously (same person).
6. At the 2.0-second point, shout “2-second point!” while the second lab partner notes the displacement of the object and marks it with tape. Do this 4 more times. The ball should pass the same point every time, or very close to it. Mark this point, or its average, and label it the 0.00-meter point. (This 0.0 meter point is NOT “The Starting Point.”)
7. Now you are ready. Give the stopwatch to the second lab partner. The first lab partner will then place the object at the starting point, and push it to go. When it crosses the 0.00-meter point, the second lab partner will start the timing device. After 10 seconds, the 2nd lab partner will shout “10-second point!” while a third lab partner notes the displacement of the object and marks it with tape. Have a lab partner note the distance traveled from 0.00-meters. Repeat this 8 more times: one for 9 sec, 8, 7, 6, 5, 4, 3, and 2 seconds. Record the distances v. times in a table. Measure the exact lengths with the meter stick - from the 0.00 point.
9. After putting all the data in the table, graph the nine points.

V. Data & Calculations
1. Distance that the ball travels in 2.0 seconds (on average): ______________
2. The distance traveled for each second, from 10 seconds all the way to 2 seconds:
Make a table and attach

Time (seconds) Distance (m)
10 24
9 23

v1 = (x2 – x1) / (1.0 s)
v2 = (x3 – x2) / (1.0 s)
etc.
vave = Svi / n
3.Make graph of distance traveled, in meters, vs. time (in seconds).
Make graph and attach.

VI. Results
The purpose of this lab was to observe objects moving at a constant speed. Explain how well this was achieved…..

VII. Error Analysis
A. Personal
B. Systematic
C. Random

VIII. Questions
1. Did the ball speed up, slow down, or stay the same speed as it traveled?
Explain or support.
2. What is the shape of the graph you made?
3. How far did the ball travel during each 1.0-second interval?
4. Predict the position of the ball after 12.0 seconds, if you had actually done it. Support
your prediction.

LESSON 2

2010-08-03T15:27:00.002-07:00

PHYS 1100 LESSON 2 FOR
TUESDAY, AUGUST 03, 2010

I Introduction

II Logistics: Seating, Syllabus, etc.

Here are two additional resources that may help in solving physics problems (solutions are here)

www.cramster.com

www.wolframalpha.com

III Review of Lesson 1: Measurements, Ch 1
Units – metric (mks; also known as SIU)
Meters for length
Length / Distance / Displacement, “x”
Kilograms for mass
Mass; 1kg = 2.2 lbs at sea level
Seconds for time
Time; second; 3600 sec = 1 hour
Coordinates: 3D: x, y, z
Speed: v = Dx/Dt; D = “change of” or
(x2 – x1) / (t2 – t1)
Velocity = speed with a vector
Acceleration = Dv / Dt or (v2 – v1) / (t2 – t1) ; often a vector
Scientific Notation:
5,617 = 5.617 x 103 or 5.617 x 10^3 or 5.617 E3
Significant Figures: 5,617 x 27 = 151,659 ? NO! 150,000 or 1.5 x 105
Estimating (ball parking; educated guessing): 5,617 x 321 = (5.6 x 3) x 105 = 16.8 x 105 = 1.68 x 106 for a guess. Exact is 1,803,057 = 1.80 x 106
Trigonometry: Create a right triangle inside a “unit” circle of radius r = 1. Then the x-component will be r cosine(θ) and the y-component will be r sine(θ) . However, since r = 1, we realize that the x-component will almost always be cosθ while the y-component will almost always be sinθ. Plus, sinθ / cosθ = y/x = tangent of the angle, or, tanθ.

IV Lesson 2: Vector Physics/Motion, Ch 2

Vectors
Virtual arrow: magnitude (size) and direction

heads, tails

adding vectors ≠ adding algebraically

adding vectors

align the head of one vector with the tail of another; never put 2 tails together, or 2 heads

together: →→ is

okay; NOT →←

and NOT ←→

→↑ is okay; but ↑→ is not okay; and →↑ is not okay
A vector usually has an arrow (→) above it: , or a “hat” or “carrot” (^) above it:

Pythagoras (576 BC – 495 BC)

KISMIE
Velocity, acceleration, force, momentum, or any number of other concepts can be represented as vectors
Components: see above; the x-component of vector A+B is A; the y-component is B.
If a vector is not directly along the x-axis or along the y-axis, it can be broken down into its x- and y- components
Acceleration vector, along an inclined plane: a = g sin θ, where θ is the angle shown:

Projectile Motion

Range, height, angle (above)

Circular motion
v = 2pr/P Э 2pr = c (circumference); r = radius; P = period, in seconds, to make one trip around the circle; and P = 1/n, Э n = the frequency in cycles per second (Hz).

v2/r = 4p2r/P2, but since P = (1/n), then P2 = (1/ n)2, or (1/P2) = n2

So, v2/r = 4p2rn2 which can be written as v2/r = (2pn)2 r

And, in circular motion, a = v2/r =
(2pn)2 r, “centripetal acceleration”
However, 2pn = w in rad/sec, thus a = w2r

If part of a circle, say, s, is the arc, AB, then we can say that for small “s” that r sin θ = s, and if it’s even smaller, then r θ = s

because for small θ, sin θ° = θr where the angle, θ, is measured in radians, not degrees. 360° = 2p radians, so 1 radian = 57.3°.

Since v = dist/time, then v = s/t = r θ/t but is another way of writing w, so v = wr and v2/r = w2r = a, “acceleration”

V Laboratory Exercise 2: Density of Water

VI Conclusion
Finish Homework 1 Problems until end of period: HWK Assignment 1: Read Chapters* 1-4 and do Problems 1, 3, 9, 13, 19, 25, 29, 31, and 39 on pp 15-16; Problems 1, 3, 9, 13, 19, 25, 29, 31, 39, 53, 63, 65 and 71 on pp 48-54; Problems 1, 3, 9, 13, 19, 25, 29, 31, 39, and 53 on pp 76-80; and Problems 1, 3, 9, 13, 19, 25, 29, 31, 39, and 53 on pp 103-107. (due 5/6)
Prepare for Test 1 (review and problems)

HOMEWORK SOLUTION SET 1

2010-08-03T15:27:00.001-07:00

Chapter 1: Problems 1, 3, 9, 13, 19, 25, 29, 31, 39 on pp 15-16.

The movie Spiderman brought in $114,000,000 during its opening weekend. Express this amount in (a) gigadollars, (b) teradollars
Solution: 114,000,000 = 1.14 x 10^8, so
1.0 gigadollar = 1.0 x 10^9, so (1.14 x 10^8)/(1.0 x 10^9) = 0.114 gigadollar.
(remember that 10^9 means “ten to the 9th power)”
1.0 teradollar = 1.0 x 10^12, so (1.14 x 10^8)/(1.0 x 10^12) = 1.14 x 10^(-4) teradollar.

The speed of light is about 0.3 Giga-m/s. Express that in meters/sec.
Solution: 0.3 Giga-meters is 0.3 x 10^9 meters = 3 x 10^8 meters, so 0.3 Giga-m/s = 3 x 10^8 meters/sec.

9. If acceleration is expressed as, a = 2xtp, then find out what the number “p” is (the exponent of t). Here, x is distance, t is time.
Solution: a = 2xtp = (2)(m)(s-2),
Because
(a) x is in units of meters (m)
(b)t is in units of seconds (s)
(c)p is the exponent (a number), and has to be “-2”,

Why? Because (s-2) ≡ 1/s2 (in this case, the symbol “≡” means “defined as” or “is the same as”

13. The irrational number p = 3.14159265358979…..
a. Round this to three significant figures
b. Round it to five sig figs
c. Do it again to 7 s.f.
Solution:
a. Round this to three significant figures p = 3.14 ;
b. Round it to five sig figs, p = 3.1416 ;
c. Do it again to 7 s.f., p = 3.141593 .

19. Peacock Mantis Shrimp can crush the shell of a snail by slamming it at 23 m/s. Find how fast this is in (a) feet per second; (b) miles/hour.
Solution: 1.0 m = 39.37 inches = 3 feet 3.37 inches = 3.281 feet
23 m = 23(3.281) = 75.459 ft, so, 23 m/s = 75.459 ft /s. Or, sig. figs, 75 ft/s.
23 m/s = 75 ft/s = (3600 s)(75 ft/s) = 271,653 ft/hour =(271,653 ft/h) /(5280 ft/mi) = 51.449 mi/h, or, 51 mph.

25. The largest blue whale observed was 108 feet long. Find that in meters.
Solution: 1.0 m = 3.281 ft, so (108 ft)/(3.281 ft/m) = 32.91679366 m = 32.9 m.

29. Woody the Woodpecker can accelerate its beak to 98 m/s2. Express that in feet per square second, ft/s2.
Solution: Dr Dave already knows that 9.8 m/s2 = 32 ft/s2. 320 ft/s2 ? KISMIE?
Or 1.0 m = 3.281 feet, so 98 m = (98)(3.281) = 321.538 ft/s2 = 320 ft/s2.
3.280833333

31. a. A Mutchkin is a Scottish unit of liquid measure, about 0.42 liter. This is not a munchkin, which is a fictional small resident of the Land of Oz.) How many mutchkins will fill a cube that measures 1.0 foot on a side? (Convert this to metric first).
b. A noggin is 0.28 mutchkin. Give the conversion factor between noggins and gallons. (Convert gallons to metric first).
Solution: 1.0 ft3 = 0.028 m3. [This is because 1.0 ft = 12 inches = (12)(2.54) = 30.48 cm = 0.3048 m; so 1.0 ft3 = (0.3048 m)3 = 0.028344726 m3.]
1.0 mutch = 420 cm3 = (0.0748887 m)3. = 0.00042 m3. So, (0.028344726)/(0.00042) = 67.4875074 = 67.
1.0 noggin = 0.28 mutchkin = (0.28)(0.42) = 0.1176 liters; but 3.78 liters = 1.0 gal, so 0.1176 liters = (0.1176 liters)/(3.78 liters/gal) = 0.031 gallons = 1.0 noggin, or, 32.14 noggins per gallon = 32.

39. Antonio just won a $12 million pay out from the local state lottery.
a. If he took all $12 million in quarters, how much mass is that?
b. If he took it all in $1 bills, how heavy, in mass, is that?
Solution: A quarter is 5.67 grams*; a dollar bill is 1.0 g (Wiki answers).
*per http://www.webs1.uidaho.edu/dl2/on_target/mass_of_quarter.htm
a. $12 million = 48 million quarters = (48 x 10^6)(5.67 x 10^-3 kg) = 272.17 x 10^3 = 2.7217 x 10^5 = 2.7E5 kg = 5.98774E5 = 6.0 E5 lbs.

Chapter 2: Problems 1, 3, 9, 13, 19, 25, 29, 31, 39, and maybe 53, 63, 65, and 71 on pp 48-54.
Take a look at the diagram in the book. Imagine leaving your house and walking east to the Library. When you are finished, you turn around and walk west to the local Park. The distance from your house to the park is 0.75 miles and the Library is another 0.6 miles east of that.
How far did you walk, i.e., total distance from your house to the Library to the Park?
What was your displacement, i.e., net distance from your house to the Park?
Solution: Distance is total amount of length traveled, while displacement is the difference between your starting point and your ending point, so…
From the house to the library = 0.75 + 0.60 = 1.35, plus, from the library to the park = 0.6, for a total of 1.35 + 0.6 = 1.95 mi.
You started at your house and ended at the park, which is 0.75 mi.

3. The golfer, seen in the book, stands 10 meters to the west of the hole, and sinks the ball in two putts. His first putt misses the hole and travels 2.5 meters further east of the hole. On his second putt, the ball travels 2.5 meters west and falls in.
a. How far did the ball travel overall?
b. What was the total displacement of the ball?
Solution: The ball travels past the hole and has to come back, retracing its steps, so to speak
The ball traveled 10 + 2.5 + 2.5 = 15 meters.
The ball started at the golfer and ended up 10 meters away.

9. The Olympic record for the 200 meter dash was 19.75 seconds in 1988. How fast is that in meters per second? Miles per hour?
Solution: Some runner traveled 200 meters in 19.75 seconds. 200 meters is 0.2 kilometers. 1.0 kilometer = 0.6214 mile*.
*http://en.wikipedia.org/wiki/Kilometer
a. (200 m)/(19.75 s) = 10.12658228 m/s, or 10.1 m/s.
b. 10.12658228 m/s = (10.12658228 m/sec)(3600 sec/hour) = 36,455.69621 m/hour = 36.45569621 km/hr, and since 1.0 kilometer = 0.6214 mile, then 36.45569621 km = 22.65356962 mi, so, the answer is 22.65356962mi/hour = 22.7 mi/hr.

13. Radio waves are light waves and thus travel at the speed of light, about 186,000 miles per second. How much time would it take for a radio wave to travel from Earth to the Moon and back? (The Moon is, on average, about 240,000 miles from Earth).
Solution: The distance from Earth to the Moon, and back, is about 480,000 miles. So, the light travels 480,000 miles at 186,000 mi/hr; thus, divide 480,000 by 186,000: (480,000)/(186,000) = (480)/(186) = 2.58 sec.

19. A dog named Fido runs back and forth between Jack and Jill, as seen in the book. However, Jack is walking towards Jill at 1.3 m/s while Jill is walking towards Jack at 1.3 m/s. If Fido begins to run when Jack and Jill are 10.0 meters apart, and if he travels at 3.0 m/s, how far will Fido travel when Jack and Jill crash into each other?
Solution: In order to KISMIE, forget about the dog until later. Then, pretend either Jack or Jill is standing still while the other is moving. Relative to each other, it’s the same as if Jack is moving at 2.6 m/s towards a non-moving Jill. How long would it take Jack to travel 10.0 meters at 2.6 m/s? Divide 10 by 2.6 to get approximately 4 seconds: (10.0)/(2.6) = 3.846153846 sec. Now, let’s go back to the dog. How far does Fido travel in if he travels at a constant speed of 3.0 m/s? x = (3.846153846 sec)(3.0 m/s) = 11.53846154 m. or 12 m.

25. The position of a particle as a function of time is given by this formula: x = (6 m/s)t + (- 2 m/s2)t2.
a. Plot a graph of x vs. t for the time from 0.0 s to 2.0 s.
b. Find the average velocity of the particle from t = 0.0 sec to 1.0 sec.
c. Find the average speed during that same time period.
Solution: In this problem since no direction is given, speed = velocity
See graph somewhere
Find the velocity at 0.0 sec and at 1.0 sec, then add them and divide by 2.0:
v = x/t, so v = (6 m/s) + (-2 m/s2)t ; at t = 0, v = 6 m/s; at t = 1.0 sec, v = 6 m/s + (-2 m/s) = 4 m/s. Average is 6 + 4 divided by 2, or, 10/2 = 5 m/s.

29. The position of a particle as a function of time is given by this formula: x = (2 m/s)t + (- 3 m/s3)t3.
a. Plot a graph of x vs. t for the time from 0.0 s to 1.0 s.
b. Find the average velocity of the particle from t = 0.35 sec to 0.45 sec.
c. Find the average velocity of the particle from t = 0.39 sec to 0.41 sec.
d. Do you expect the instantaneous velocity at t = 0.40 s to be closer to 0.54 m/s, 0.56 m/s, or 0.58 m/s? Explain.
Solution: v = (2 m/s) + (-3m/s2)t2 so…
See the graph somewhere
v(0.35) = (2 m/s) + (-3m/s2)(0.35)2 = (2 m/s) + (-3m/s2)(0.1225) = (2 m/s) +
(-0.3675) = 1.6325; and v(0.45) = (2 m/s) + (-3m/s2)(0.45)2 = (2 m/s) +
(-3m/s2)(0.2025) = (2 m/s) + (-0.6075) = 1.3925. The average is then [½(1.6325 + 1.3925)] = ½ [3.025] = 1.5125 m/s, or 1.5 m/s.
v(0.39) = (2 m/s) + (-3m/s2)(0.39)2 = (2 m/s) + (-3m/s2)(0.1521) = (2 m/s) +
(-0.4563) = 1.5437; and v(0.41) = (2 m/s) + (-3m/s2)(0.41)2 = (2 m/s) +
(-3m/s2)(0.1681) = (2 m/s) + (-0.5043) = 1.4957. The average is then [½(1.5437 + 1.4957)] = ½ [3.0394] = 1.5197 m/s, or 1.5 m/s.
d. TBD

31. Two bows shoot identical arrows with the same launch speed. The accomplish this, the string in bow 1 must be pulled back farther when shooting its arrow than the string in bow 2.
a. Is the acceleration of the arrow shot by bow 1 greater than, less than, or equal to the acceleration of the arrow shot by bow 2?
b. Choose the best explanation from among the following:
I. The arrow in bow 2 accelerates for a great time.
II. Both arrows start from rest.
III. The arrow in bow 1 accelerates for a greater time.
Solution: Both arrows are identical, same size, shape, mass, density. So, no tricks there.
The same
None. The bows are different in tensile strength.

39. Assume that the brakes in your car create a constant deceleration of 4.2 m/s2 regardless of how fast you are driving. If you double your driving speed from 16 m/s to 32 m/s
a. does the time required to stop increase by a factor of two or a factor of four? Explain.
b. Verify your answer by calculating the stopping times for the initial speeds of 16 m/s
c. Verify your answer by calculating the stopping times for the initial speeds of 32 m/s.
Solution: Realize that acceleration equals velocity divided by time, or, a = v/t.
Since a = v/t, then t = v/a, which means that time and velocity are directly related. Thus, double the velocity and you double the time (factor of two).
Okay, I did. Thanks.
Ditto.

53. Approximate 0.1% of the bacteria in the intestine are E coli. These bacteria have been observed to move with speeds of up to 15 m/s (microns per second) and max accelerations of 166 m/s2. Suppose an E coli bacterium in your intestine starts at rest and accelerates at 156 m/s2.
a. How much time is required for the bacterium to reach a speed of 12 m/s?
b. How much distance is required for the bacterium to reach a speed of 12 m/s?
Solution: From problem 39, we know that a = v/t, or, t = v/a.
t = (12 m/s)/(156 m/s2) = 0.076923076 seconds, or = 0.077 sec.
The standard distance equation, when starting from zero (0.0 m) and with an initial velocity of zero (0.0 m/s) is: x = ½ a t2. Since a = (156 m/s2) and the time we just found to be t = 0.076923076 seconds, then x = ½ a t2 =
½ (156 m/s2)(0.076923076 s)2 = ½ (156 m/s2)(0.005917159621 s2) = 0.46153845 m. Or, just 0.46 m.

63. A model rocket rises with constant acceleration to a height of 3.2 m, at which point its speed is 26.0 m/s.
a. How much time does it take for the rocket to reach this height?
b. What was the magnitude of the rocket’s acceleration?
c. Find the height and speed of the rocket 0.10 s after launch.
Solution: The standard height equations, when starting from zero (0.0 m) and with an initial velocity that is non-zero (0.0 m/s) are:

(i) y = y0 + v0t + ½ a t2. And

(ii) v = v0 + a t

We assume the rocket starts from ground level, i.e., y0 = 0.0 meters, and that it starts from rest, i.e., v0 = 0.0 m/s. So…we can rewrite equation (i) above as:

y = y0 + v0t + ½ a t2. Or, y = 0 + 0 + ½ a t2. Which leaves only y = ½ a t2.

We know “y”, but don’t know “a” even though we know that it is a constant acceleration, so we can’t find “t” just from equation (i). So, let’s use equation (ii) to see if it helps:

v = v0 + a t, but since v0 = 0.0 m/s, that leaves us with v = a t. Again, here we know “v” but, again, don’t know “a” even though we know that it is a constant acceleration.

Let’s use both equations.
First, we found that y = ½ a t2. Which we re-write as t = √ [(2y)/a].
Second, we found that v = a t. Which we re-write as t = v/a.

Since they both equal “t”, we can set them equal to find “a” then go back to get “t.”

Or, √ [(2y)/a] = v/a. Now, square both sides to get rid of the square root:

[(2y)/a] = (v/a)2. = (v2)/(a2). Now, multiply both sides by a2 to get

2ya = v2. Now, divide both sides by “2y” to get: a = (v) 2 / (2y) = (26.0 m/s)2/[(2)(3.2 m)] = (676)/(6.4) = 105.625 m/s2.

Now we use v = a t and rewrite it as t = v/a = (26.0)/(105.625) = 0.246153846 seconds, or 0.25 sec.

We found this while doing part (a.), so a =105.625 m/s2 or 106 m/s2. Or 1.06 x 10^2 m/s2.
To find the height, we use equation (i): y = ½ a t2 or, y(0.1) = ½ (105.625)(0.1)2 = ½ (105.625)(0.01) = 0.528125 m, or 0.528 m.
To find its speed, we use equation (ii): v = a t = (105.625)(0.10s) = 10.5625 m/s or 10.6 m/s.

65. A bicyclist named Bob is finishing his repair of a flat tire when a friend named Bill rides by with a constant speed of 3.5 m/s. Two seconds (2.0 s) later Bob hops on his bike and accelerates at 2.4 m/s2 until he catches up with Bill.
a. How much time does it take for Bob to catch up with Bill?
b. How far has Bob traveled in this time?
c.What is Bob’s speed when catches up with Bill?
Solution: The standard distance equations, when starting from zero (0.0 m) and with an initial velocity that is non-zero (0.0 m/s) are:

(i) x = x0 + v0t + ½ a t2. And

(ii) v = v0 + a t

When Bob starts, his initial distance (x0) and velocity (v0) are zero. Thus, these two equations for Bob now are modified to be:

(i) x = ½ a t2. And we know “a” but don’t know “t”, so we can’t find “x” yet, plus

(ii) v = a t. Here, we also know “a” but not “t” so we can’t find “v” yet.

When Bob starts (t0 = 0.0 s), Bill is at a distance of x = v t = (3.5 m/s)(2.0 s) = 7.0 meters from Bob, and at the end, “t,” Bill is at a distance of zero (0.0 m) from Bob. This means that Bob has to travel some distance, “x” to reach Bill, who is also at the same place, “x.”

For Bill, the two equations are a little different:

(x = x0 + v0t + ½ a t2) becomes (x = 0 + v0t + 0 = v0t) because Bill’s initial distance, x0, is the same as Bob’s, or, zero (0.0 m) and Bill’s initial speed is not zero (0.0 m/s) but 3.5 m/s. Plus, since we are told that Bill has a constant velocity, there is NO acceleration, or, for Bill, acceleration is zero (0.0 m/s2). Thus, for Bill:
x = v0t = (3.5 m/s)t

(ii) v = v0 + a t. Which becomes v = v0 since a = 0.0 m/s2.

So, Bob’s “x” and Bill’s “x” are the same:

½ a t2. = v0t, or ½ a t = v0. Thus, t = (2 v0)/a = (2)(3.5)/(2.4) = 7.0/2.4 = 2.917 s or 2.9 sec.
How far has Bob traveled? We use equation (i):

x = x0 + v0t + ½ a t2. = 0.0 m + (0.0 m/s)(2.9 sec) + ½ (2.4)(2.9)2 = 0 + 0 +
½ (2.4)(8.5) = 10.2 meters.

What is Bob’s speed at that time? Use equation (ii):

v = a t = (2.4)(2.9) = 7.0 m/s.

71. Review the image of the car falling off a cliff. The driver says, “Let’s see if we can go from zero to sixty in three seconds.” Prove the driver right or wrong.
Solution: The standard height equations, when starting from zero (0.0 m) and with an initial velocity that is non-zero (0.0 m/s) are:

(i) y = y0 + v0t + ½ a t2. And

(ii) v = v0 + a t

In this question, y0 = 0.0 meters, or, the edge of the cliff; v0=0.0 m/s as it is falling down not zooming down (it is moving to the left, but that’s the x-axis and not relevant). And we assume that “zero to sixty” means from v0= 0.0 m/s (and 0.0 miles per hours) to v = 60.0 miles/hour = 96.6 km/hour = 96,600 meters/hour = 26.8 m/s. We also know that the acceleration here is ONLY gravity, or, a = - 9.8 m/s2. Also, we don’t seem to need equation (i), so we will use only equation (ii). So,

v = v0 + a t = 0 + (g)(t) = (- 9.8 /s2)(3.0 s) = - 29.4 m/s. (it is negative, as it is falling down, not up). So, the statement is false. It cannot go from 0.0 m/s to 60 miles per hour (- 26.8 m/s) in 3.0 seconds of falling. But it comes close!

Chapter 3: Problems 1, 3, 9, 13, 19, 25, 29, 31, 39, and 53 on pp 76-80.

1.Suppose that each component of a certain vector is doubled.
By what multiplicative factor does the magnitude of the vector change?
By what multiplicative factor does the direction angle of the vector change?
Solution:
Double
No change

3.Rank the vectors in figurer 3-31 on page 76 in order of increasing value of their x-component.
Solution: C, B, A, D. Although D will be in the negative direction.

9. A baseball diamond is a square with sides 90 feet (27.4 meters) in length. If the positive x-axis is home plate to first base, and if the positive y-axis points from home plate to third base , find the displacement vector of a base runner who has just hit a
a. double (safely hit the ball and made it to second base)
b. triple (safely hit the ball and made it to third base)
c. home run (safely hit the ball and made it all the way back to home plate)
Solution:
(27.4 m)√2 = 38.7 meters, or (90ft)√2 = 127ft.
27.4 meters = 90 feet
Zero (0.0 m = 0.0 feet)

13. In 1755, John Harrison completed his 4th precision chronometer (clock), known as the “H4”. When the minute hand of the H4 indicated 10 minutes past the hour, it extended 3.0 cm in the horizontal direction.
a. How long was the H4’s minute hand?
b. At 10 minutes past the hour was the extension of the minute hand in the vertical direction more than, less than, or equal to 3.0 cm? Explain.
c. Calculate the vertical extension of the minute hand at 10 minutes past the hour.
Solution: 10 minutes is 1/6 of an hour, or 1/6 of 360°, or 60°.
The x-component is 3.0 cm, we are told, and since in this case, x = h sinθ, where h = the length of the minute hand (the hypotenuse). So, h = (x)/(sinθ) = (3.0 cm)/(sin60°) = (3.0)/(0.866) = 3.46 cm.
Longer.
It will be (3.0 cm)/(cos60) = (3.0)/(0.5) = 6.0 cm.

19. Refer to figure 3-36 on page 78. A refers to a vector.
a. Is the magnitude of A + D greater than, less than, or equal to the magnitude of A + E?
b. Is the magnitude of A + E greater than, less than, or equal to the magnitude of A + F?
Solution: D is in a negative direction, while A, E, and F are all in the positive direction.
the magnitude of A + D is less than, or equal to the magnitude of A + E
the magnitude of A + E equal to the magnitude of A + F? Since E and F are equal in the y-direction.

25. Vector A points in the negative y-direction and has a magnitude of 5 units. Vector B has twice the magnitude and points in the positive x-direction. Find the direction and magnitude of
a. A + B
b. A – B
c. B – A
Solution: A = - 5 ŷ and B = + 10 x
a. A + B will create a vector that will be in Quadrant IV, with a magnitude of √(125) = 11.2 and an angle of 333°.
b. A – B will create a vector that will be in Quadrant III, with a magnitude of √(125) = 11.2 and an angle of 206.8°.
c. B – A will create a vector that will be in Quadrant I with a magnitude of √(125) = 11.2 and an angle of 26.8°.

29. Vector A has a length of 6.1 meters and points in the negative x-direction.
a. Find the x-component of – 3.7 A.
b. Find the magnitude of the vector – 3.7 A.
Solution: Since vector is in the negative x-direction, multiplying it by a negative would make a vector in the positive direction
x-component is (+3.7)(6.1) = 22.57, or 23 meters along the x-axis.
Same

31. Find the direction and magnitude of the vectors. x refers to the unit vector in the x-direction, and ŷ refers to the unit vector in the y-direction.
a. A = (5.0 m)x + (-2.0 m)ŷ
b. B = (-2.0 m)x + (5.0 m)ŷ
c. A + B.
Solution: a. A = (5.0 m)x + (-2.0 m)ŷ will create a vector that will be in Quadrant IV, with a magnitude of √(29) = 5.4 and an angle of 338°.
b. B = (-2.0 m)x + (5.0 m)ŷ will create a vector that will be in Quadrant II, with a magnitude of √(29) = 5.4 and an angle of 112°.
c. A + B. will create a vector, C = (3.0 m)x + (3.0 m)ŷ that will be in Quadrant I, with a magnitude of (3.0) √(2) = 4.2 and an angle of 45°.

39. A neighborhood cat takes 45 minutes to travel 120 meters due north (+ y-direction) and then takes 17 minutes to travel 72 meters due west (- x-direction). Find the magnitude and direction of its average velocity over its 62 minute (3720 sec) jaunt.
Solution: The feline’s foray sketches a right triangle with one side (120 meters)ŷ and the other side (– 72 meters)x. The cat ends up in Quadrant II, thus, it will be more than 90 degrees (North) but less than 180 degrees (West). So, it’s average velocity (or speed) is distance over time, or, (120 + 72)/(3720 sec) = (192 m)/(3720 sec) = 0.0516 m/s. About 5 cm/sec is not too fast. The cat’s displacement will be the hypotenuse of the right triangle made, or, letting the hypotenuse = h, then x2 + y2 = h2. Or, (72)2 + (120)2 = (5184) + (14400) = 19,584 = h2. So, h = √(19,584) = 139.9428455 meters. The cat traveled 192 meters, but its displacement was only 140 meters. And what direction? y = h cosθ here, so cos θ = y/h = (120 m)/(140 m) = 0.857142857, thus the angle, θ, equals the arc-cosine of (0.857142857), or
cos-1(0.857142857) = 31°. So the hypotenuse is 31° to the west of north, but the proper way would be to say how far it is away from the + x-axis, thus we need to add 90° to get 31° + 90° = 121°. The question did NOT ask for the vector, for if it did, then the answer would be 140 m @ 121°.

53. An airplane pilot wants to fly due north but a 65 km/hour wind blowing eastward makes him make changes.
a. In what direction should the pilot head if its relative air speed is 340 km/h?
b. Draw a vector diagram of this
c. If the pilot decreases the plane’s air speed but still wants to fly due north, should the angle found in (a) be increased or decreased?
Solution: If the pilot is going 340 km/h north and there is a west wind (wind from the west, blowing him eastward) of 65 km/h, then he is really traveling √[(340)2 + (65)2] = √[(115600) + (4225)] = √(119825) = 346 km/h at an angle of cos-1(340/346) =
cos-1(0.982658959) = 10.7° east of north (or, 346 km/hr @ 79.3°).
To compensate, he will have to travel 10.7° west of north, or, 346 km/h @ 100.7°.
See a drawing somewhere
Increased to compensate

Chapter 4: Problems 1, 3, 9, 13, 19, 25, 29, 31, 39, and 53 on pp 103-107.

Walking down the street, you toss a ball up into the sky.
If you want the ball to land in your hand when it comes back down, should you toss the ball straight up, forward, or backwards to your movement?
Choose the best explanation
I If the ball is thrown straight up you will leave it behind and it will fall in back of you
II You have to throw the ball in the direction that you are walking
III The ball moves in the forward direction with your walking speed at all times.
Solution:
If you want the ball to land in your hand when it comes back down, you should you toss the ball straight up to your movement.
Choose the best explanation - III The ball moves in the forward direction with your walking speed at all times.

3.While walking to class at a constant speed of 1.75 m/s, you are moving in a direction that is 18.0° north of east. How much time does it take to change your displacement by
a. 20.0 meters east
b. 30.0 meters north
Solution: Essentially, what is the x-component of this velocity vector, and, what is the y-component, then use the distances and speed to find time. The x-component of the speed, vx = (1.75)Cos 18.0° = (1.75)(0.95) = 1.66 m/s; and vy = (1.75) Sin 18.0° = (1.75)(0.31) = 0.54 m/s
vx = x/tx so, tx = x/vx = (20.0 meters)/(1.66 m/s) = 12 sec.
vy = y/ty so, tx = y/vy = (30.0 meters)/(0.54 m/s) = 55 sec.

9. Two kids jump/drop off a cliff overhang into a lake. Johnny drops straight down, while Jimmy runs off the cliff with an initial horizontal speed of v0. (Assume no air friction)
a. Is Jimmy’s splashdown speed larger, the same, or less than Johnny’s splashdown speed? (splashdown speed is the vertical speed that the boys hit the water).
b. Choose the best explanation
I. Both kids are in free fall and hence they will have the same splashdown speed.
II. The kids have the same vertical speed but Jimmy has more horizontal speed
III. Johnny hits faster since he goes straight down.
Solution:
a. Jimmy’s splashdown is the same as Johnny’s splashdown speed.
b. Choose the best explanation - II. The kids have the same vertical speed but Jimmy has more horizontal speed

13. An astronaut exploring the asteroid Ceres tosses a rock horizontally at 6.95 m/s. The rock falls vertically 1.4 meters and lands a horizontal distance of 8.75 meters from the astronaut. Find the acceleration of gravity on Ceres.
Solution: we will use the relationships of v = x/t, and y = ½ g t2. In this case, we are looking for gCeres. First, how much time does it take to go 8.75 meters if the rock travels at 6.95 m/s? It will be x/v = (8.75 m)/(6.95 m/s) = 1.26 sec. What is the g for this asteroid? It will be (2y)/t2 = (2)(1.4)/(1.26)2 = (2.8)/(1.58) = 1.77 m/s2.

19. There is a tradition in Denver, Colorado, in which children bring their old pumpkins from Halloween and toss them off a 9.0-meter high tower, while trying to hit a target that is 3.5 meters from the base of the tower (See figure 4-15 on page 105). Find the horizontal speed needed to hit the target.
Solution: Well, the pumpkin must travel 3.5 meters in the x-direction in the same amount of time it takes to fall 9.0 meters down = y. Denver is on Earth, so we need to use the acceleration to be 9.8 m/s2. But we won’t use a negative, as we are not looking for a vector, and we can’t find a time that is “imaginary” if we took the square root of a negative number. So, we need to find the time, t, and divide x by it to get vx. How long does it take for an object to fall 9.0 meters? We use y = ½ g t2 and re-write it to be t = √(2y/g) = √[(2)(9)/(9.8)] = √(18/9.8) = √(1.8367) = 1.36 sec. So, the pumpkin needs to move 3.5 meters in 1.36 seconds (in the x-direction) to hit the spot, so the horizontal speed is vx = (3.5)/(1.36) = 2.58 m/s.

25. A ball rolls off a table at Sea Level on Earth, and falls 0.75 meters to the floor, landing with a speed of 4.0 m/s. Ignore air friction.
a. What is the acceleration of the ball just before it strikes the ground?
b. What was the initial speed of the ball?
c. What initial speed must the ball have it if is to land with a speed of 5.0 m/s?
Solution: First of all, it is the feet of the table’s legs that are at sea level, so the ball is 0.75 meters above sea level, but the difference in gravitational acceleration due to that is negligible.
The acceleration of any object on Earth, g, near the surface, is 9.8 m/s2.
The initial vertical speed of the ball was zero, vy = 0.0 m/s if it merely rolls off the table. However, for a ball to land at 4.0 m/s after traveling 0.75 meters, it means that we have to use a special equation: vf2 = vi2 + 2 gy, or to find vi, we re-write it to be vi2 = vf2 - 2 gy = [(4)2 – (2)(9.8)(0.75)] = (16 – 14.7) = 1.3. So, if vi2 = 1.3, then vi = √(1.3) = 1.14 m/s. We must conclude that a “fairy” or some other mythical creature propelled it downward the moment it rolled off the table.
If it lands at 5.0 m/s instead of 4.0 m/s, we use the same equations and substitute “5” for “4”: vi2 = vf2 - 2 gy = [(5)2 – (2)(9.8)(0.75)] = (25 – 14.7) = 10.3. So, if vi2 = 10.3, then vi = √(10.3) = 3.2 m/s. Probably from another more powerful fairy.

29. In a baseball game, the second baseman picks up the baseball and tosses it to the first baseman, who catches it at the same level from which is it was thrown. The throw is made with an initial speed of 18.0 m/s at an angle 37.5° from the horizontal.
a. What is the horizontal component of the ball’s velocity just before it’s caught?
b. How long is the ball “in the air”?
Solution: Ignoring air friction, there are no other forces along the x-axis. And, of course, the ball takes as long to reach its maximum height as it does to come back to the original level.
vx is constant, so it’s always (18.0 m/s)Cos(37.5°) = (18.0)(0.793) = 14.3 m/s.
vy is, initially, (18.0)Sin(37.5°) = (18.0)(0.608) = 10.96 m/s, but as it goes up, it slows to, eventually, zero (0.0 m/s) before falling back down. We use the relationship of vf2 = vi2 + 2 gy, where vf = 0.0 m/s at the top. So, now we have the relationship 0 = vi2 + 2 gy, and we can now find “y” the distance it went up before stopping: vi2 = - 2 gy, or y = - (vi2)/(2g) = - [(10.96)2] /[(2)(-9.8)] = - (120)/(- 19.6) = + 6.13 meters. So, the ball goes up 6.13 meters, stops, and goes down 6.13 meters. Since vf = vi + gt we can find how much time it took to get up. We know that vf = 0.0 m/s. So, 0 = vi + gt, and vi = - gt, and t = -(vi/g) = - (10.96)/(-9.8) = + 1.12 sec. Obviously, it takes another 1.12 seconds to come down, or 2.4 sec.

31. A cork shoots out of a champagne bottle at an angle of 35.0° from the horizontal. If the cork travels a horizontal distance 1.30 meters in 1.25 seconds, what was its initial speed?
Solution: If it travels 1.30 meters per 1.25 sec, that means that the x-component of the velocity is vx = (1.30)/(1.25) = 1.04 m/s. Since vx = v0Cos(35.0°), then
v0 = (vx)/[ Cos(35.0°)] = (1.04 m/s)/(0.8192) = 1.27 m/s.

39. The “hang time” of a punt (the kick of the football in American Football) is 4.50 seconds. If the ball were kicked at an angle of 63.0° to the horizontal and was caught at the same level from which it was kicked, what was its initial speed?
Solution: This means it took t = 2.25 seconds to reach its maximum height, y. We are looking for v0. (It then takes another 2.25 seconds to come back down). We know that vx = v0 Cos(63.0°) and vy = v0 Sin (63.0°). So, if we find vy, then we will be able to get v0. We learned from an earlier problem that vy = - gt = - (-9.8)(2.25) = + 22.05 m/s. So, v0 = (vy)/[Sin (63.0°)] = (22.05 m/s)/(0.89) = 24.7 m/s.

53. A soccer ball is kicked with an initial speed 10.2 m/s in a direction 25.0° above the horizontal. Find the magnitude and direction of its velocity
a. 0.250 seconds after being kicked
b. 0.500 seconds after being kicked
c. Is the ball at its greatest height before or after 0.500 second? Explain.
Solution: Using all the same stuff that we already had, we can find out how much time it stays above ground. We have to assume that the ball was kicked from the ground by a midget about 1 micron in size. We learned from before that vy = v0 Sin (25.0°) = (10.2 m/s) Sin (25.0°) = (10.2 m/s)(0.4226) = 4.31 m/s. And remember, since vy = - gt, we find that t = - (vy)/(g) = - (4.31 m/s)/(- 9.8 m/s2) = + 0.44 sec.
Okay, v(0.250sec) = - (g)(0.250 sec) @ 25.0°. = - (- 9.8 m/s2)(0.250 sec) = + 2.45 m/s.
At 0.500 sec, we know that it stopped at 0.44 seconds when it reached the top. So this is like having the ball in free fall for a period of (0.500 – 0.44) sec = 0.06 seconds. And thus, v = gt = (- 9.8)(0.06 sec) = - 0.588 m/s. It’s negative, since it is going down.
We did this already. It reaches its greatest height at 0.44 seconds, or 0.06 seconds before 0.500 sec.

END

LAB EXERCISE 2

2010-08-02T15:00:00.002-07:00

Physics 1100 Lab 2 Tuesday, August 3, 2010 Name __________________
Dr Dave Menke, Instructor, Brown Mackie College

I Title: Density of Water

II Purpose: Practice in Measurement, and to find Density of Water

III Equipment: beaker (50-ml) or Pyrex measuring cup, graduated cylinder (50-ml), digital scale, water

IV Procedure:
1. Weigh (in grams) dry beaker/cup. Record.
2. Use graduated cylinder to add 25 50 ml of water to beaker or ¼ cup water
3. Weigh (in grams) the beaker/cup with the water in it. Record.
4. Find the weight (grams) of the water (subtract #1 from #3). Record.
5. Find the density (grams / ml) of the water (divide #4 by 25 ml). Record.
(If using a Pyrex measuring cup, conversions from cups to liters will be required)

V Data
1. Mass of dry beaker/cup ___________grams

2. Mass of beaker/cup + water ___________ grams

3. Mass of the water ___________grams

4. Density of water ___________ grams / ml

(The True Density of Water is 1.0 grams / ml or 1000 kg/m3)

VI Results: In this section, explain how successful you were at achieving the Purpose. [If you found (or were close) the density of water, then you were successful. If not, you were not.]

VII Error
Qualitative – sources of error
Personal – what things did you or your partner(s) do to screw things up, if anything at all?
Systematic – what was wrong with the equipment that caused the lab to go “bad”, or, what external factors (hurricane, earthquake, etc) contributed to difficulties in this lab?
Random – there are always random errors, unless you do this lab exercise 5 times and take an average
Quantitative – Find the Percent Error

[True Answer – Your Answer] / [True Answer] x 100% = ________%

VIII Questions
Which water is denser, salt water (ocean) or fresh water (lake or river)? Find the density of each.
What is denser, solid water (ice) or liquid water? Give an example.

LAB EXERCISE 1

2010-08-02T15:00:00.001-07:00

Physics Lab 1 Monday, August 2, 2010 Name __________________
Dr Dave Menke, Instructor, Brown Mackie College

I Title: Measurements

II Purpose: Practice in Measurements

III Equipment: any object that is appropriate* to measure, meter stick or metric ruler, scale

*Not too small or too large to measure in class with the equipment available: small book or other item is fine.

IV Procedure:
1. Pick an appropriate object to measure. Record.
2. Measure the length, width, and height of the object in centimeters. Record.
3. Calculate the surface area (approximately) of this 3-D object, in cm2. Record.
4. Calculate the volume of the object in cm3. Record.
5. Use the scale to measure the mass of the object in grams. Record.
6. Determine the density of the object (in g/cm3) by dividing the mass that you found in Step 5, with the volume that you found in Step 4. Record.

V Data
1. Object selected:_________________________
2. The length x width x height of object, in cm: _______ x _______ x ________
3. The surface area of the object: ___________ cm2.
4. Volume of the object: ___________cm3.
5. Mass of the object: ___________ grams
6. Density of the object: ____________ g/cm3 .

VI Results (Conclusions): In this section, explain how successful you were at achieving the Purpose. (Use back or extra sheets)

VII Error
Qualitative – sources of error
Personal – what things did you or your partner(s) do to screw things up, if anything at all?
Systematic – what was wrong with the equipment that caused the lab to go “bad”, or, what external factors (hurricane, earthquake, etc) contributed to difficulties in this lab?
Random – there are always random errors, unless you do this lab exercise 5 times and take an average
Quantitative – NA

VIII Questions (Hint: Use any source available, including Wikipedia.org)
What is a vernier caliper?
What is a micrometer caliper?

APPENDIX TO LESSON 1

2010-08-02T14:59:00.000-07:00

Appendix to Lesson 1
Chapter 1 #9
a = 2xtp = (2)(m)(s-2), s-2 = 1/s2
Because
a. x is in units of meters (m)
b. t is in units of seconds (s)
c. p is the exponent (a number), and has to be “-2”,
Why? Because (s-2) ≡ 1/s2 (in this case, the symbol “≡” means “defined as” or “is the same as”

Remember, the units of acceleration, or, ‘a’ are m/s2

Remember, that distance, ‘x’ is in units of meters (m)

Also, velocity (speed), ‘v’ is distance/time, or, x/t, and it is in units of meters per second, ‘m/s’

acceleration ‘a’ is velocity/time = v/t = (m/s)/s = [m/(s)(s)] = m/s2.

s ≡ second
1/s ≡ per second (same as s-1)
1/s2 ≡ per second squared (same as s-2)
Meters times per second squared = (x)(s-2)

Other examples:
v = 3xtr, so what is “r” = -1
distance = 4xtw, so what is w? 0

Chapter 1 #19
1.0 m = 39.37 inches = 3 feet, 3.37 inches; 3.28 feet

23 m/s = 75 ft/sec = 0.0142 mi/sec = (0.0142 mi/s)(3600 sec/hour) = 51.13 mi/hr = 5.1 x 101 mi/hr

75 feet = 75/5280 = 0.0142 mi

Chapter 1 #31
(1 foot)3 = (30.48 cm)3 = 28,317 cm3. = 28,317 ml = 28.3 L, how many 0.42 L’s (mutchkin) fit into 28.3 L?
(28.3 / 0.42) = 67

1.0 gallon = 3.79 liters

1 noggin = 0.28(0.42 L) = 0.1176 L

So, how many (0.1176’s) go into 3.79?

3.79/0.1176 = 32
32 noggins = 1 gallon

Chapter 2 #9
200 meters = 0.2 km = (0.2 km)(5/8 mi/km) = 1/8 mi = 0.125 mi

0.125 mi/19.75 sec = 0.006329 mi/sec = (0.006329 mi/sec)(3,600 sec/hour) = 22.78 mi/hour

Chapter 2, #19

The relative speed of Jack to Jill is: v = 2.6 m/s; The distance traveled is10 meters; so, how long does it take to travel 10 meters at a speed of 2.6 m/s? Like this >>> (10m)/(2.6m/s) = 3.85 s (until Jack and Jill crash)

Since velocity is v = x/t, we know that the dog’s distance traveled is: x = v t = (3 m/s)(3.85 s) = 11.55 m.

Chapter 2, #25
y = ax2 + bx + c standard quadratic equation, but for this:

y = x
a = -2
b = 6
c = 0
x = t

x = -2t2 + 6t, as if it were y = -2x2 + 6x
make table, choose any number for x, then

x(t) y(x)
0s 0
1s 4
0.5s 2.5
etc.
After this, plot these dots, connect the dots…
… and “plug n chug” to get what x is for each “t” that is selected… 0 s, 1 s, etc.

HOMEWORK SET 1

2010-08-02T14:57:00.000-07:00

Chapter 1: Problems 1, 3, 9, 13, 19, 25, 29, 31, 39 on pp 15-16.

1.The movie Spiderman brought in $114,000,000 during its opening weekend. Express this amount in (a) gigadollars, (b) teradollars

3.The speed of light is about 0.3 Giga-m/s. Express that in meters/sec.
9. If acceleration is expressed as, a = 2xtp, then find out what the number “p” is (the exponent of t). Here, x is distance, t is time.
13. The irrational number p = 3.14159265358979…..
a. Round this to three significant figures
b. Round it to five sig figs
c. Do it again to 7 s.f.
19. Peacock Mantis Shrimp can crush the shell of a snail by slamming it at 23 m/s. Find how fast this is in (a) feet per second; (b) miles/hour.
25. The largest blue whale observed was 108 feet long. Find that in meters.
29. Woody the Woodpecker can accelerate its beak to 98 m/s2. Express that in feet per square second, ft/s2.
31. a. A Mutchkin is a Scottish unit of liquid measure, about 0.42 liter. This is not a munchkin, which is a fictional small resident of the Land of Oz.) How many mutchkins will fill a cube that measures 1.0 foot on a side? (Convert this to metric first).
b. A noggin is 0.28 mutchkin. Give the conversion factor between noggins and gallons. (Convert gallons to metric first).
More....

Physics 1100, HOMEWORK SET 1, page 2 of 8

39. Antonio just won a $12 million pay out from the local state lottery.
a. If he took all $12 million in quarters, how much mass is that?
b. If he took it all in $1 bills, how heavy, in mass, is that?
Chapter 2: Problems 1, 3, 9, 13, 19, 25, 29, 31, 39, and maybe 53, 63, 65, and 71 on pp 48-54.
Take a look at the diagram below. Imagine leaving your house and walking east to the Library. When you are finished, you turn around and walk west to the local Park. The distance from your house to the park is 0.75 miles and the Library is another 0.6 miles east of that.
How far did you walk, i.e., total distanced from your house to the Library to the Park?
What was your displacement, i.e., net distance from your house to the Park?
H______E_____G
House …………………Park…………………Library
0.75 mi 0.60 mi
3. The golfer, below, stands 10 meters to the west of the hole, and sinks the ball in two putts. His first putt misses the hole and travels 2.5 meters further east of the hole. On his second putt, the ball travels 2.5 meters west and falls in.
a. How far did the ball travel overall?
b. What was the total displacement of the ball?
m________w__r
Golfer dude………………hole…….where ball stops after first putt
10 m 2.5 m
9. The Olympic record for the 200 meter dash was 19.75 seconds in 1988. How fast is that in meters per second? Miles per hour?
More....
Physics 1100, HOMEWORK SET 1, page 3 of 8

13. Radio waves are light waves and thus travel at the speed of light, about 186,000 miles per second. How much time would it take for a radio wave to travel from Earth to the Moon and back? (The Moon is, on average, about 240,000 miles from Earth).
19. A dog named Fido runs back and forth between Jack and Jill, as seen in the diagram below. However, Jack is walking towards Jill at 1.3 m/s while Jill is walking towards Jack at 1.3 m/s. If Fido begins to run when Jack and Jill are 10.0 meters apart, and if he travels at 3.0 m/s, how far will Fido travel when Jack and Jill crash into each other?
>

> <
25. The position of a particle as a function of time is given by this formula: x = (6 m/s)t + (- 2 m/s2)t2.
a. Plot a graph of x vs. t for the time from 0.0 s to 2.0 s.
b. Find the average velocity of the particle from t = 0s to 1.0 sec.
c. Find the average speed during that same time period.
29. The position of a particle as a function of time is given by this formula: x = (2 m/s)t + (- 3 m/s3)t3.
a. Plot a graph of x vs. t for the time from 0.0 s to 1.0 s.
b. Find the average velocity of the particle from t = 0.35 sec to 0.45 sec.
c. Find the average velocity of the particle from t = 0.39 sec to 0.41 sec.

d. Do you expect the instantaneous velocity at t – 0.40 s to be closer to 0.54 m/s, 0.56 m/s, or 0.58 m/s? Explain.
31. Two bows shoot identical arrows with the same launch speed. The accomplish this, the string in bow 1 must be pulled back farther when shooting its arrow than the string in bow 2.
a. Is the acceleration of the arrow shot by bow 1 greater than, less than, or equal to the acceleration of the arrow shot by bow 2?
b. Choose the best explanation from among the following:
More....
Physics 1100, HOMEWORK SET 1, page 4 of 8

I. The arrow in bow 2 accelerates for a great time.
II. Both arrows start from rest.
III. The arrow in bow 1 accelerates for a greater time.
39. Assume that the brakes in your car create a constant deceleration of 4.2 m/s2 regardless of how fast you are driving. If you double your driving speed from 16 m/s to 32 m/s
a. does the time required to stop increase by a factor of two or a factor of four? Explain.
b. Verify your answer by calculating the stopping times for the initial speeds of 16 m/s
c. Verify your answer by calculating the stopping times for the initial speeds of 32 m/s.
53. Approximate 0.1% of the bacteria in the intestine are E coli. These bacteria have been observed to move with speeds of up to 15 m/s (microns per second) and max accelerations of 166 m/s2. Suppose an E coli bacterium in your intestine starts at rest and accelerates at 156 m/s2.
a. How much time is required for the bacterium to reach a speed of 12 m/s?

b. How much distance is required for the bacterium to reach a speed of 12 m/s?
63. A model rocket rises with constant acceleration to a height of 3.2 m, at which point its speed is 26.0 m/s.
a. How much time does it take for the rocket to reach this height?
b. What was the magnitude of the rocket’s acceleration?
c. Find the height and speed of the rocket 0.10 s after launch.
65. A bicyclist named Bob is finishing his repair of a flat tire when a friend named Bill rides by with a constant speed of 3.5 m/s. Two seconds (2.0 s) bob hops on his bike and accelerates at 2.4 m/s2 until he catches up with Bill. b
a. How much time does it take for Bob to catch up with Bill?
b. How far has Bob traveled in this time?
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Physics 1100, HOMEWORK SET 1, page 5 of 8

c.What is Bob’s speed when catches up with Bill?
71. Review the image of the car falling off a cliff. The d

river says, “Let’s see if we can go from zero to sixty in three seconds.” Prove the driver right or wrong.
Chapter 3: Problems 1, 3, 9, 13, 19, 25, 29, 31, 39, and 53, on pp 76-80.
1.Suppose that each component of a certain vector is doubled.
By what multiplicative factor does the magnitude of the vector change?
By what multiplicative factor does the direction angle of the vector change?
3.Rank the vectors in figurer 3-31 on page 76 in order of increasing value of their x-component.

9. A baseball diamond is a square with sides 90 feet (27.4 meters) in length. If the positive x-axis is home plate to first base, and if the positive y-axis points from home plate to third base , find the displacement vector of a base runner who has just hit a
a. double (safely hit the ball and made it to second base)
b. triple (safely hit the ball and made it to third base)
c. home run (safely hit the ball and made it all the way back to home plate)

13. In 1755, John Harrison completed his 4th precision chronometer (clock), known as the “H4”. When the minute hand of the H4 indicated 10 minutes past the hour, it extended 3.0 cm in the horizontal direction.
a. How long was the H4’s minute hand?
b. At 10 minutes past the hour was the extension of the minute hand in the vertical direction more than, less than, or equal to 3.0 cm? Explain.
c. Calculate the vertical extension of the minute hand at 10 minutes past the hour.
More....

Physics 1100, HOMEWORK SET 1, page 6 of 8

19. Refer to figure 3-36 on page 78. A refers to a vector.
a. Is the magnitude of A + D greater than, less than, or equal to the magnitude of A + E?
b. Is the magnitude of A + E greater than, less than, or equal to the magnitude of A + F?
25. Vector A points in the negative y-direction and has a magnitude of 5 units. Vector B has twice the magnitude and points in the positive x-direction. Find the direction and magnitude of
a. A + B
b. A – B
c. B - A
29. Vector A has a length of 6.1 meters and points in the negative x-direction.
a. Find the x-component of – 3.7 A.
b. Find the magnitude of the vector – 3.7 A.
31. Find the direction and magnitude of the vectors. x refers to the unit vector.
a. A = (5.0 m)x + (-2.0 m)y
b. B = (-2.0 m)x + (5.0 m)y

c. A + B.
39. A neighborhood cat takes 45 minutes to travel 120 meters due north (+ y-direction) and then takes 17 minutes to travel 72 meters due west (- x-direction). Find the magnitude and direction of its average velocity over its 62 minute jaunt.
53. An airplane pilot wants to fly due north but a 65 km/hour wind blowing eastward makes him make changes. Q
More....
Physics 1100, HOMEWORK SET 1, page 7 of 8
a. In what direction should the pilot head if its relative air speed is 340 km/h?
b. Draw a vector diagram of this
c. If the pilot decreases the plane’s air speed but still wants to fly due north, should the angle found in (a) be increased or decreased?
Chapter 4: Problems 1, 3, 9, 13, 19, 25, 29, 31, 39, and 53, on pp 103-107.
Walking down the street, you toss a ball up into the sky.
If you want the ball to land in your hand when it comes back down, should you toss the ball straight up, forward, or backwards to your movement?
Choose the best explanation
I If the ball is thrown straight up you will leave it behind and it will fall in back of you
II You have to throw the ball in the direction that you are walking
III The ball moves in the forward direction with your walking speed at all times.
3.While walking to class at a constant 1.75 m/s2, you are moving in a direction that is 18.0° north of east. How much time does it take to change your displacement by
a. 20.0 meters east

b. 30.0 meters north
9. Two kids jump/drop off a cliff overhang into a lake. Johnny drops straight down, while Jimmy runs off the cliff with an initial horizontal speed of v0. (Assume no air friction)
a. Is Jimmy’s splashdown speed larger, the same, or less than Johnny’s splashdown speed? (splashdown speed is the vertical speed that the boys hit the water).
b. Choose the best explanation
I. Both kids are in free fall and hence they will have the same splashdown speed.
II. The kids have the same vertical speed but Jimmy has more horizontal speed
More....
Physics 1100, HOMEWORK SET 1, page 8 of 8
III. Johnny hits faster since he goes straight down.
13. An astronaut exploring the asteroid Ceres tosses a rock horizontally at 6.95 m/s. T

he rock falls vertically 1.4 meters and lands a horizontal distance of 8.75 meters from the astronaut. Find the acceleration of gravity on Ceres.
19. There is a tradition in Denver, Colorado, in which children bring their old pumpkins from Halloween and toss them off a 9.0-meter high tower, while trying to hit a target that is 3.5 meters from the base of the tower (See figure 4-15 on page 105). Find the horizontal speed needed to hit the target.
25. A ball rolls off a table at Sea Level on Earth, and falls 0.75 meters to the floor, landing with a speed of 4.0 m/s. Ignore air friction.
A. What is the acceleration of the ball just before it strikes the ground?
b. What was the initial speed of the ball?
c. What initial speed must the ball have it if is to land with a speed of 5.0 m/s?
29. In a baseball game, the second baseman picks up the baseball and tosses it to the first baseman, who catches it at the same level from which is it was thrown. The throw is made with an initial speed of 18.0 m/s at an angle 37.5° from the horizontal.
a. What is the horizontal component of the ball’s velocity just before it’s I caught?
b. How long is the ball “in the air”?

31. A cork shoots out of a champagne bottle at an angle of 35.0° from the horizontal. If the cork travels a horizontal distance 1.30 meters in 1.25 seconds, what was its initial speed?
39. The “hang time” of a punt (the kick of the football in American Football) is 4.50 seconds. If the ball were kicked at an angle of 63.0° to the horizontal and was caught at the same level from which it was kicked, what was its initial speed?

53. A soccer ball is kicked with an initial speed 10.2 m/s in a direction 25.0° above the horizontal. Find the magnitude and direction of its velocity
a. 0.250 seconds after being kicked
b. 0.500 seconds after being kicked
c. Is the ball at its greatest height before or after 0.500 second? Explain.
END
Chapter 1: Problems 1, 3, 9, 13, 19, 25, 29, 31, 39 on pp 15-16.
The movie Spiderman brought in $114,000,000 during its opening weekend. Express this amount in (a) gigadollars, (b) teradollars
The speed of light is about 0.3 Giga-m/s. Express that in meters/sec.
9. If acceleration is expressed as, a = 2xtp, then find out what the number “p” is (the exponent of t). Here, x is distance, t is time.
13. The irrational number p = 3.14159265358979…..
a. Round this to three significant figures
b. Round it to five sig figs
c. Do it again to 7 s.f.
19. Peacock Mantis Shrimp can crush the shell of a snail by slamming it at 23 m/s. Find how fast this is in (a) feet per second; (b) miles/hour.
25. The largest blue whale observed was 108 feet long. Find that in meters.
29. Woody the Woodpecker can accelerate its beak to 98 m/s2. Express that in feet per square second, ft/s2.
31. a. A Mutchkin is a Scottish unit of liquid measure, about 0.42 liter. This is not a munchkin, which is a fictional small resident of the Land of Oz.) How many mutchkins will fill a cube that measures 1.0 foot on a side? (Convert this to metric first).
b. A noggin is 0.28 mutchkin. Give the conversion factor between noggins and gallons. (Convert gallons to metric first).
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Physics 1100, HOMEWORK SET 1, page 2 of 8

39. Antonio just won a $12 million pay out from the local state lottery.
a. If he took all $12 million in quarters, how much mass is that?
b. If he took it all in $1 bills, how heavy, in mass, is that?
Chapter 2: Problems 1, 3, 9, 13, 19, 25, 29, 31, 39, and maybe 53, 63, 65, and 71 on pp 48-54.
Take a look at the diagram below. Imagine leaving your house and walking east to the Library. When you are finished, you turn around and walk west to the local Park. The distance from your house to the park is 0.75 miles and the Library is another 0.6 miles east of that.
How far did you walk, i.e., total distanced from your house to the Library to the Park?
What was your displacement, i.e., net distance from your house to the Park?
H______E_____G
House …………………Park…………………Library
0.75 mi 0.60 mi
3. The golfer, below, stands 10 meters to the west of the hole, and sinks the ball in two putts. His first putt misses the hole and travels 2.5 meters further east of the hole. On his second putt, the ball travels 2.5 meters west and falls in.
a. How far did the ball travel overall?
b. What was the total displacement of the ball?
m________w__r
Golfer dude………………hole…….where ball stops after first putt
10 m 2.5 m
9. The Olympic record for the 200 meter dash was 19.75 seconds in 1988. How fast is that in meters per second? Miles per hour?
More....
Physics 1100, HOMEWORK SET 1, page 3 of 8

13. Radio waves are light waves and thus travel at the speed of light, about 186,000 miles per second. How much time would it take for a radio wave to travel from Earth to the Moon and back? (The Moon is, on average, about 240,000 miles from Earth).
19. A dog named Fido runs back and forth between Jack and Jill, as seen in the diagram below. However, Jack is walking towards Jill at 1.3 m/s while Jill is walking towards Jack at 1.3 m/s. If Fido begins to run when Jack and Jill are 10.0 meters apart, and if he travels at 3.0 m/s, how far will Fido travel when Jack and Jill crash into each other?
>

> <
25. The position of a particle as a function of time is given by this formula: x = (6 m/s)t + (- 2 m/s2)t2.
a. Plot a graph of x vs. t for the time from 0.0 s to 2.0 s.
b. Find the average velocity of the particle from t = 0s to 1.0 sec.
c. Find the average speed during that same time period.
29. The position of a particle as a function of time is given by this formula: x = (2 m/s)t + (- 3 m/s3)t3.
a. Plot a graph of x vs. t for the time from 0.0 s to 1.0 s.
b. Find the average velocity of the particle from t = 0.35 sec to 0.45 sec.
c. Find the average velocity of the particle from t = 0.39 sec to 0.41 sec.

d. Do you expect the instantaneous velocity at t – 0.40 s to be closer to 0.54 m/s, 0.56 m/s, or 0.58 m/s? Explain.
31. Two bows shoot identical arrows with the same launch speed. The accomplish this, the string in bow 1 must be pulled back farther when shooting its arrow than the string in bow 2.
a. Is the acceleration of the arrow shot by bow 1 greater than, less than, or equal to the acceleration of the arrow shot by bow 2?
b. Choose the best explanation from among the following:
More....
Physics 1100, HOMEWORK SET 1, page 4 of 8

I. The arrow in bow 2 accelerates for a great time.
II. Both arrows start from rest.
III. The arrow in bow 1 accelerates for a greater time.
39. Assume that the brakes in your car create a constant deceleration of 4.2 m/s2 regardless of how fast you are driving. If you double your driving speed from 16 m/s to 32 m/s
a. does the time required to stop increase by a factor of two or a factor of four? Explain.
b. Verify your answer by calculating the stopping times for the initial speeds of 16 m/s
c. Verify your answer by calculating the stopping times for the initial speeds of 32 m/s.
53. Approximate 0.1% of the bacteria in the intestine are E coli. These bacteria have been observed to move with speeds of up to 15 m/s (microns per second) and max accelerations of 166 m/s2. Suppose an E coli bacterium in your intestine starts at rest and accelerates at 156 m/s2.
a. How much time is required for the bacterium to reach a speed of 12 m/s?

b. How much distance is required for the bacterium to reach a speed of 12 m/s?
63. A model rocket rises with constant acceleration to a height of 3.2 m, at which point its speed is 26.0 m/s.
a. How much time does it take for the rocket to reach this height?
b. What was the magnitude of the rocket’s acceleration?
c. Find the height and speed of the rocket 0.10 s after launch.
65. A bicyclist named Bob is finishing his repair of a flat tire when a friend named Bill rides by with a constant speed of 3.5 m/s. Two seconds (2.0 s) bob hops on his bike and accelerates at 2.4 m/s2 until he catches up with Bill. b
a. How much time does it take for Bob to catch up with Bill?
b. How far has Bob traveled in this time?
More....
Physics 1100, HOMEWORK SET 1, page 5 of 8

c.What is Bob’s speed when catches up with Bill?
71. Review the image of the car falling off a cliff. The d

river says, “Let’s see if we can go from zero to sixty in three seconds.” Prove the driver right or wrong.
Chapter 3: Problems 1, 3, 9, 13, 19, 25, 29, 31, 39, and 53, on pp 76-80.
1.Suppose that each component of a certain vector is doubled.
By what multiplicative factor does the magnitude of the vector change?
By what multiplicative factor does the direction angle of the vector change?
3.Rank the vectors in figurer 3-31 on page 76 in order of increasing value of their x-component.

9. A baseball diamond is a square with sides 90 feet (27.4 meters) in length. If the positive x-axis is home plate to first base, and if the positive y-axis points from home plate to third base , find the displacement vector of a base runner who has just hit a
a. double (safely hit the ball and made it to second base)
b. triple (safely hit the ball and made it to third base)
c. home run (safely hit the ball and made it all the way back to home plate)

13. In 1755, John Harrison completed his 4th precision chronometer (clock), known as the “H4”. When the minute hand of the H4 indicated 10 minutes past the hour, it extended 3.0 cm in the horizontal direction.
a. How long was the H4’s minute hand?
b. At 10 minutes past the hour was the extension of the minute hand in the vertical direction more than, less than, or equal to 3.0 cm? Explain.
c. Calculate the vertical extension of the minute hand at 10 minutes past the hour.
More....

Physics 1100, HOMEWORK SET 1, page 6 of 8

19. Refer to figure 3-36 on page 78. A refers to a vector.
a. Is the magnitude of A + D greater than, less than, or equal to the magnitude of A + E?
b. Is the magnitude of A + E greater than, less than, or equal to the magnitude of A + F?
25. Vector A points in the negative y-direction and has a magnitude of 5 units. Vector B has twice the magnitude and points in the positive x-direction. Find the direction and magnitude of
a. A + B
b. A – B
c. B - A
29. Vector A has a length of 6.1 meters and points in the negative x-direction.
a. Find the x-component of – 3.7 A.
b. Find the magnitude of the vector – 3.7 A.
31. Find the direction and magnitude of the vectors. x refers to the unit vector.
a. A = (5.0 m)x + (-2.0 m)y
b. B = (-2.0 m)x + (5.0 m)y

c. A + B.
39. A neighborhood cat takes 45 minutes to travel 120 meters due north (+ y-direction) and then takes 17 minutes to travel 72 meters due west (- x-direction). Find the magnitude and direction of its average velocity over its 62 minute jaunt.
53. An airplane pilot wants to fly due north but a 65 km/hour wind blowing eastward makes him make changes. Q
More....
Physics 1100, HOMEWORK SET 1, page 7 of 8
a. In what direction should the pilot head if its relative air speed is 340 km/h?
b. Draw a vector diagram of this
c. If the pilot decreases the plane’s air speed but still wants to fly due north, should the angle found in (a) be increased or decreased?
Chapter 4: Problems 1, 3, 9, 13, 19, 25, 29, 31, 39, and 53, on pp 103-107.
Walking down the street, you toss a ball up into the sky.
If you want the ball to land in your hand when it comes back down, should you toss the ball straight up, forward, or backwards to your movement?
Choose the best explanation
I If the ball is thrown straight up you will leave it behind and it will fall in back of you
II You have to throw the ball in the direction that you are walking
III The ball moves in the forward direction with your walking speed at all times.
3.While walking to class at a constant 1.75 m/s2, you are moving in a direction that is 18.0° north of east. How much time does it take to change your displacement by
a. 20.0 meters east

b. 30.0 meters north
9. Two kids jump/drop off a cliff overhang into a lake. Johnny drops straight down, while Jimmy runs off the cliff with an initial horizontal speed of v0. (Assume no air friction)
a. Is Jimmy’s splashdown speed larger, the same, or less than Johnny’s splashdown speed? (splashdown speed is the vertical speed that the boys hit the water).
b. Choose the best explanation
I. Both kids are in free fall and hence they will have the same splashdown speed.
II. The kids have the same vertical speed but Jimmy has more horizontal speed
More....
Physics 1100, HOMEWORK SET 1, page 8 of 8
III. Johnny hits faster since he goes straight down.
13. An astronaut exploring the asteroid Ceres tosses a rock horizontally at 6.95 m/s. T

he rock falls vertically 1.4 meters and lands a horizontal distance of 8.75 meters from the astronaut. Find the acceleration of gravity on Ceres.
19. There is a tradition in Denver, Colorado, in which children bring their old pumpkins from Halloween and toss them off a 9.0-meter high tower, while trying to hit a target that is 3.5 meters from the base of the tower (See figure 4-15 on page 105). Find the horizontal speed needed to hit the target.
25. A ball rolls off a table at Sea Level on Earth, and falls 0.75 meters to the floor, landing with a speed of 4.0 m/s. Ignore air friction.
A. What is the acceleration of the ball just before it strikes the ground?
b. What was the initial speed of the ball?
c. What initial speed must the ball have it if is to land with a speed of 5.0 m/s?
29. In a baseball game, the second baseman picks up the baseball and tosses it to the first baseman, who catches it at the same level from which is it was thrown. The throw is made with an initial speed of 18.0 m/s at an angle 37.5° from the horizontal.
a. What is the horizontal component of the ball’s velocity just before it’s I caught?
b. How long is the ball “in the air”?

31. A cork shoots out of a champagne bottle at an angle of 35.0° from the horizontal. If the cork travels a horizontal distance 1.30 meters in 1.25 seconds, what was its initial speed?
39. The “hang time” of a punt (the kick of the football in American Football) is 4.50 seconds. If the ball were kicked at an angle of 63.0° to the horizontal and was caught at the same level from which it was kicked, what was its initial speed?

53. A soccer ball is kicked with an initial speed 10.2 m/s in a direction 25.0° above the horizontal. Find the magnitude and direction of its velocity
a. 0.250 seconds after being kicked
b. 0.500 seconds after being kicked
c. Is the ball at its greatest height before or after 0.500 second? Explain.
END

LESSON 1

2010-08-02T14:56:00.000-07:00

PHYS 1100 LESSON 1
MONDAY, AUGUST 2, 2010

I Introduction
II Logistics
Syllabus
Seating
Questions
Brain
III Measurements – Chapters 1, 2
Units – metric (mks; also known as SIU)
Meters for length
Length / Distance / Displacement, “x”
Kilograms for mass
Mass; 1kg = 2.2 lbs at sea level
Seconds for time
Time; second; 3600 sec = 1 hour
Coordinates: 3D: x, y, z
Speed: v = Dx/Dt; D = “change of” or
(x2 – x1) / (t2 – t1)
Velocity = speed with a vector
Acceleration = Dv / Dt or (v2 – v1) / (t2 – t1) ; often a vector
Scientific Notation:
5,617 = 5.617 x 103 or 5.617 x 10^3 or 5.617 E3
Significant Figures: 5,617 x 27 = 151,659 ? NO! 150,000 or 1.5 x 105
Estimating (ball parking; educated guessing): 5,617 x 321 = (5.6 x 3) x 105 = 16.8 x 105 = 1.68 x 106 for a guess. Exact is 1,803,057 = 1.80 x 106

IV Trigonometry: Create a right triangle inside a “unit” circle of radius r = 1. Then the x-component will be r cosine(θ) and the y-component will be r sine(θ) . However, since r = 1, we realize that the x-component will almost always be cosθ while the y-component will almost always be sinθ . Plus, sinθ / cosθ = y/x = tangent of the angle, or, tanθ.

V Lab 1: Measurements (Handout)

VI Homework 1: Read Chapters* 1-4 and do Problems 1, 3, 9, 13, 19, 25, 29, 31, and 39 on pp 15-16; Problems 1, 3, 9, 13, 19, 25, 29, 31, 39, 53, 63, 65 and 71 on pp 48-54; Problems 1, 3, 9, 13, 19, 25, 29, 31, 39, and 53 on pp 76-80; and Problems 1, 3, 9, 13, 19, 25, 29, 31, 39, and 53 on pp 103-107. Turn in Thursday, 8/5

VII Essay 1: Aristotle, due 8/5

VIII Problem Solving until end of period.

Weekly Workload

2010-08-02T14:55:00.000-07:00

Weekly Workload for Physics Students

Week 1: August 2 - 6

1. Monday
a. Introduction
b. Syllabus
c. Lesson 1 – Units, chapter 1; vectors, chapter 2
d. Lab Exercise #1: Measurements
e. Homework Assignment 1: Read Chapters* 1-4 and do Problems 1, 3, 9, 13, 19, 25, 29, 31, and 39 on pp 15-16; Problems 1, 3, 9, 13, 19, 25, 29, 31, 39, 53, 63, 65 and 71 on pp 48-54; Problems 1, 3, 9, 13, 19, 25, 29, 31, 39, and 53 on pp 76-80; due 8/5
f. Essay 1: Aristotle; due 8/5

2. Tuesday
a. Introduction and Logistics
b. Lesson 2 – Motion in 1 D, 2D and 3D; chapter 3; Trigonometry & Vectors chapter 4
c. Lab Exercise #2: Density of Water
d. Homework Assignment 1: Read Chapters* 1-4 and do Problems 1, 3, 9, 13, 19, 25, 29, 31, and 39 on pp 15-16; Problems 1, 3, 9, 13, 19, 25, 29, 31, 39, 53, 63, 65 and 71 on pp 48-54; Problems 1, 3, 9, 13, 19, 25, 29, 31, 39, and 53 on pp 76-80; due 8/5
e. Essay 1: Aristotle; due 8/5

3. Wednesday – No Class; Study for Test 1

4. Thursday
a. Introduction
b. Turn in Homework 1, Essay 1
c. Test 1 on Chapters 1 - 4
d. Mind Game 1
e. Lesson 3 – Circular Motion, chapter 5; Torque, Chapter 6
f. Lab Exercise #3, Linear Motion
g. Homework Assignment 2: Read Chapters 5-9 and do Problems 1, 3, 9, 13, 19, 25, 29, 31, 39, and 53 on pp 103-107; Problems 1, 3, 9, 13, 19, 25, 29, 31, 39, 53, 63, and 65 on pp 177-184; Problems 1, 3, 9, 13, 19, 25, 29, 31, 39, and 53 on pp 210-213; and Problems 1, 3, 9, 13, 19, 25, 29, 31, 39, and 53 on pp 244-248; Problems 1, 3, 9, 13, 19, 25, 29, 31, 39, and 53 on pp 289-292; due 8/12
h. Essay 2 on Isaac Newton, due 8/12

5. Friday – No Class

Week 2: August 9 - 13

1. Monday
a. Introduction/Logistics
b. Return papers
c. Running Grades
d. Review of Lesson 3 – Circular Motion
e. Lesson 4 – Newton's Laws
PHY1100 Addendum to Course Syllabus, August 2010, Dr Dave Menke, page 4 of 6

f. Lab Exercise #4: Circular Motion
g. Homework Assignment 2: Read Chapters 5-9 and do Problems 1, 3, 9, 13, 19, 25, 29, 31, 39, and 53 on pp 103-107; Problems 1, 3, 9, 13, 19, 25, 29, 31, 39, 53, 63, and 65 on pp 177-184; Problems 1, 3, 9, 13, 19, 25, 29, 31, 39, and 53 on pp 210-213; and Problems 1, 3, 9, 13, 19, 25, 29, 31, 39, and 53 on pp 244-248; Problems 1, 3, 9, 13, 19, 25, 29, 31, 39, and 53 on pp 289-292; due 8/12
h. Essay 2 on Isaac Newton, due 8/12

2. Tuesday
a. Introduction and Logistics
b. Running Grades
c. Review of Lesson 4 – Newton's Laws
d. Lesson 5 – Hooke's Law; Energy; Momentum
c. Lab Exercise #5: Hooke’s Law
g. Homework Assignment 2: Read Chapters 5-9 and do Problems 1, 3, 9, 13, 19, 25, 29, 31, 39, and 53 on pp 103-107; Problems 1, 3, 9, 13, 19, 25, 29, 31, 39, 53, 63, and 65 on pp 177-184; Problems 1, 3, 9, 13, 19, 25, 29, 31, 39, and 53 on pp 210-213; and Problems 1, 3, 9, 13, 19, 25, 29, 31, 39, and 53 on pp 244-248; Problems 1, 3, 9, 13, 19, 25, 29, 31, 39, and 53 on pp 289-292; due 8/12
h. Essay 2 on Isaac Newton, due 8/12

3. Wednesday – No Class; Study for Test 2

4. Thursday
a. Introduction/Logistics
b. Running Grades
c. Turn in Homework 2, Essay 2
d. Test 2 on Chapters 5 - 9
d. Mind Game 2
e. Lesson 6 – Momentum, Impulse, Conservation, Angular Momentum
f. Lab Exercise #6, Gravity
g. Homework Assignment 3: Read Chapters 10-14 and do Problems 1, 3, 9, 13, 19, 25, 29, 31, 39, 53, 63, and 65 on pp 323-327; Problems 1, 3, 9, 13, 19, 25, 29, 31, 39, 53, 63, 65 and 77 on pp 366-371; Problems 1, 3, 9, 13, 19, 25, 29, 31, 39, 53 and 63 on pp 409-412; Problems 1, 3, 9, 13, 19, 25, 29, 31, 39, 53, 63, and 65 on pp 445-448; Problems 1, 3, 9, 13, 19, 25, 29, 31, 39, 53, 63, 65, 69, and 85 on pp 491-495; due 8/19
h. Essay 3 on James Prescott Joule, due 8/19

5. Friday – No Class

Week3: August 16 – 20
1. Monday
a. Introduction/Logistics
b. Return papers
c. Running Grades
d. Review of Lesson 6 - Momentum, Impulse, Conservation, Angular Momentum
e. Lesson 7 – Torque, Inertia
f. Lab Exercise #7: KE PE

PHY1100 Addendum to Course Syllabus, August 2010, Dr Dave Menke, page 5 of 6

g. Homework Assignment 3: Read Chapters 10-14 and do Problems 1, 3, 9, 13, 19, 25, 29, 31, 39, 53, 63, and 65 on pp 323-327; Problems 1, 3, 9, 13, 19, 25, 29, 31, 39, 53, 63, 65 and 77 on pp 366-371; Problems 1, 3, 9, 13, 19, 25, 29, 31, 39, 53 and 63 on pp 409-412; Problems 1, 3, 9, 13, 19, 25, 29, 31, 39, 53, 63, and 65 on pp 445-448; Problems 1, 3, 9, 13, 19, 25, 29, 31, 39, 53, 63, 65, 69, and 85 on pp 491-495; due 8/19
h. Essay 3 on James Prescott Joule, due 8/19

2. Tuesday
a. Introduction and Logistics
b. Running Grades
c. Review of Lesson 7 – Torque, Inertia
d. Lesson 8 – Inertia, Torque, and Gravity
c. Lab Exercise #8 – Simple Harmonic Motion
g. Homework Assignment 3: Read Chapters 10-14 and do Problems 1, 3, 9, 13, 19, 25, 29, 31, 39, 53, 63, and 65 on pp 323-327; Problems 1, 3, 9, 13, 19, 25, 29, 31, 39, 53, 63, 65 and 77 on pp 366-371; Problems 1, 3, 9, 13, 19, 25, 29, 31, 39, 53 and 63 on pp 409-412; Problems 1, 3, 9, 13, 19, 25, 29, 31, 39, 53, 63, and 65 on pp 445-448; Problems 1, 3, 9, 13, 19, 25, 29, 31, 39, 53, 63, 65, 69, and 85 on pp 491-495; due 8/19
h. Essay 3 on James Prescott Joule, due 8/19

3. Wednesday – No Class; Study for Test 3

4. Thursday
a. Introduction/Logistics
b. Running Grades
c. Turn in Homework 3, Essay 3
d. Test 2 on Chapters 10 - 14
e. Mind Game 3
f. Review Lesson 8 – Inertia, Torque, Gravity
g. Lesson 9 – Waves and Fluids
h. Lab Exercise #9, Kepler's Orbits
i. Homework Assignment 4: Read Chapters 15-16 and do Problems 1, 3, 9, 13, 19, 25, 29, 31, 39, 53, and 63 on pp 530-534; Problems 1, 3, 9, 13, 19, 25, 29, 31, 39, 53, 63, 65, and 67 on pp 566-569; due 8/26
j. Essay 4 on Albert Einstein, due 8/26

5. Friday – No Class

Week4: August 23 - 28
1. Monday
a. Introduction/Logistics
b. Return papers
c. Running Grades
d. Review of Lesson 9 – Waves & Fluids
e. Lesson 10 – Density, Archimedes Principle, Bernoulli's Equation
f. Lab Exercise #10: Waves on a String
g. Homework Assignment 4: Read Chapters 15-16 and do Problems 1, 3, 9, 13, 19, 25, 29, 31, 39, 53, and 63 on pp 530-534; Problems 1, 3, 9, 13, 19, 25, 29, 31, 39, 53, 63, 65, and 67 on pp 566-569; due 8/26PHY1100 Addendum to Course Syllabus, August 2010, Dr Dave Menke, page 6 of 6

h. Essay 4 on Albert Einstein, due 8/26

2. Tuesday
a. Introduction and Logistics
b. Running Grades
c. Review of Lesson 10 - Density, Archimedes Principle, Bernoulli's Equation
d. Lesson 11 – Preparation for Final Examination
c. Lab Exercise: NO MORE LABS
g. Homework Assignment 4: Read Chapters 15-16 and do Problems 1, 3, 9, 13, 19, 25, 29, 31, 39, 53, and 63 on pp 530-534; Problems 1, 3, 9, 13, 19, 25, 29, 31, 39, 53, 63, 65 and 67 on pp 566-569; due 8/26
h. Essay 4 on Albert Einstein, due 8/26

3. Wednesday – No Class; Study for Final Exam

4. Thursday
a. Introduction/Logistics
b. Running Grades
c. Turn in Homework 4, Essay 4
d. Final Exam on Chapters 15 - 16
d. Mind Game NO MIND GAME
e. Lesson 12 – NO MORE LESSONS
f. Lab Exercise: NO MORE LABS
g. Homework Assignment 5: NO MORE HOMEWORK
h. Essay 5: NO MORE ESSAYS

5. Friday – No Class; END OF TERM

Final Exam is on Thursday August 27. It will have 15 questions directly from Homework Assignment 4. *All chapters and Review Exercises / Critical Thinking are from the textbook: Physics, Volume 1, 4th Edition, By J.S. Walker

END

Syllabus Physics 1

2010-08-02T14:50:00.000-07:00

Title: Physics

Textbook: Physics, Volume 1, 4th Edition, By J.S. Walker, Chapters 1 - 16

Supplies and Equipment That Are Helpful
1. Note pad & scratch paper to take notes and scribble or laptop
2. Scientific calculator
3. Computer (or access to a computer)
4. Textbook

All of the above are sold somewhere.

Website for this course: http://drdhmenkebmc.blogspot.com/

Each Student must learn how to log on to the Course Website

Purpose: To Understand the basics of Mechanics, Heat, Sound

Grading:
Students will be graded on a percentage scale from points earned. There are 1000 points available. Students may earn points from tests, homework, in-class lab exercises, “mind challenge” games, and weekly essays. The total points that a student earns, divided by the total number of required points available, will yield a percentage, and the grades will be as such: 90% A; 80% B; 70% C; 60% D.

Tests There will be three (3) tests and one (1) final exam this session. Each test will be on the following Thursdays: May 6, May 13, and May 20. Each test is approximately one hour (1 h) long and will have 10 questions, taken directly from the homework and lecture material that we are studying that week. Each test is worth 100 points. All test questions will be physics problems. 3 x 100 = 300 points.

Final Exam There will be one (1) comprehensive Final Exam on Thursday, May 27. The Final Exam is about two hours (2 h) long and worth 500 points. All final exam questions will be problems. 1 x 500 = 500 points.

Homework: There are four (4) homework sets. All homework sets are listed in the “Weekly Work” file. Each Homework Set has a series of problems or questions at the end of the chapters in the book. Each Homework Set is worth twenty (20) points. Research has shown if you do the homework, then you do well on the tests. If you don’t do the homework, most of the time you fail the tests. To get credit for homework, it is due each Thursday before the test; a hard copy - either hand written or printed – is fine. For the Final Exam, it is due on that Thursday, May 27. Total credit: 4 x 20 = 80 points.

In-Class Lab Exercises – There will be ten (10) “In Class Lab Exercises” related to the subject at hand. Each is worth up to 5 points. The dates are listed in “Weekly Work.” Total credit: 10 x 5 = 50 points.

Challenging Mind Games Following each test (but not after the Final Exam), a challenging “mind game” will be offered, worth up to 10 points. Examples include solving class-related crossword puzzles. Total credit: 3 x 10 = 30 points.

Weekly Essays* Once a week a topic related to the course will be required. The topic is listed in the “Weekly Work.” It will also be announced in class. To earn up to 10 points, you must write an ORIGINAL essay about the topic. Length? About 300 words is fine. However, plagiarism is not acceptable -- you must not cut and paste from
electronic media sources or from an Encyclopedia on CD. Nor can you copy another student’s description. A hard copy of the essay is due each Thursday. You may also turn in your essay via email – to dmenke@brownmackie.edu – and then it is due at midnight on Friday. Total credit: 4 x 10 = 40 points.

If you email it, you MUST place in the Subject Heading “SA1” for the first week (it stands for “essay number 1”), SA2 for the 2nd week, etc. If the subject heading is wrong, or mislabeled, or missing, your essay may be deleted as if you had never done it at all.

*Credit for essays will be a function of content only. This is not a writing class, literature class, or spelling class. You may think, “I can’t write.” So what? It’s okay. Can’t spell? As long as I can understand it, your essay will be acceptable.

Students will be graded on a percentage scale from points earned. There are 1000 points available. Students may earn points from tests, homework, “mind challenge” games, and weekly essays. The total points that a student earns, divided by the total number of required points available, will yield a percentage, and the grades will be as such: 90% A; 80% B; 70% C; 60% D.

Activity Point Value Number Total Value
Weekly Tests 100 3 300
Final Exam 300 1 300
Weekly Homework Ass’ts 20 4 80
Weekly Essays 10 4 40
Mind Games 10 3 30
Lab Exercises 25 10 250
TOTAL POINTS 1000

Attendance policy: Students are required to attend each session of their classes. Missing sessions may cause students to be dropped. Exceptions may include conditions over which students have no control prevent their being present. In any case, students MUST contact the instructor as soon as they can. Prospective employers will evaluate attendance habits.

Last day to drop class without grade penalty: Within first four weeks

Last day to drop class without financial penalty: Within the first week

Course description: A review of the historical and scientific evolution and application of the principles of anatomy and of physiology

Specific objectives: At the completion of the course, the student will be able to:
· Leap Tall Buildings in a Single Bound
· Walk on Water
· Take over the World

Policy on Incomplete Work: Missed work may be made up with the approval of the instructor. The instructor will entertain request for an Incomplete grade on a case-by-case basis with no guarantee of approval, although requesting an Incomplete grade is not recommended by the instructor.

Classroom behavior: It is expected that students will comport themselves in a mature and civil manner and govern themselves accordingly. Please keep cell phones turned off, or put to vibrate, and if you must talk on the cell phone, please excuse yourself. It is suggested that food and drink (except for bottled water) not be brought into the classroom.

ADA Policy Statement: If you have a disability that requires special accommodations, you are strongly urged to contact the Student Services office at the beginning of the term so that reasonable accommodations can be made in a timely manner. You may call 520.319.3300.

Course Outline/Calendar for PHY 1100 (Tentative)*

Week # Week of Topics/Chapters
1 Aug 2 Linear, Vectors, Spatial; 1,2,3,4
2 Aug 9 Newton, Work, Energy, Momentum; 5,6,7,8,9
3 Aug 16 Rotation, Torque, Equilibrium, Gravity, Oscillation, Waves, Sound;10,11,12,13,14
4 Aug 23 Fluids; Heat;15,16
5 Aug 30 There is no week 5.

*Subject to change; students will be informed by the instructor or the College as soon as possible if there is a change. Chapters are from Physics, Volume 1, 3rd Edition, by J.S. Walker

How to Contact the Instructor:
1. On site, before, during, or after class.
2. During Office Hours: By appointment only.
3. By phone: Call 520-319-3300 and leave message; emergency, Instructor’s Cell
Phone: 664-4109
4. By E-Mail. Send any questions, comments, or suggestions, which are directly
related to the course, to dmenke @brownmackie.edu

DISCLAIMER: Aspects of this syllabus may be subject to change. In the event of a change, students will be informed as soon as possible.