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course Phy 202
More questions to be submitted.
`q001. Include a copy of your data for the domino measurements, including your estimated uncertainty:
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Paper Ruler
A- Length 7.9 cm, Width 1.7 cm, Thickness 1.6 cm
B- Length 7.7 cm, Width 1.9cm, Thickness 1.9 cm
C- Length 7.9 cm, Width 0.9 cm, Thickness 0.9 cm
D- Length 7.9 cm, Width 1.9 cm, Thickness 2 cm
E- Length 7.9 cm, Width 1.9 cm, Thickness 1.8 cm
F- Length 7.9 cm, Width 0.6 cm, Thickness 0.6 cm
Meter Stick
A- Length 4.9, Width 1.8 cm, Thickness 1.6 cm
B- Length 4.9, Width 0.9 cm, Thickness 1.9 cm
C- Length 4.9 cm, Width 0.8 cm, Thickness 0.9 cm
D- Length 4.7 cm, Width 0.9 cm Thickness 1 cm
E- Length 4.8 cm, Width 1 cm, Thickness 1 cm
F- Length 4.9 cm, Width 0.7 cm, Thickness 0.6 cm
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Your lengths all look fine.
However the widths of the dominoes don't vary much at all, so your widths can't be correct.
You've also indicated thicknesses that would imply some dominoes are three times as thick as others (e.g., 1.8 cm is three times as great as 0.6 cm).
Measured with an accurate meter stick the legnths would be around 5 cm, the widths around 2.5 cm and the thicknesses between about .8 cm and .9 cm.
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Percent Uncertainty For Length
Paper Ruler
A- 0.1/7.9*100%= 1.3%
B- 0.2/7.7*100%= 2.6%
C- 0.1/7.9*100%= 1.3%
D- 0.1/7.9*100%= 1.3%
E- 0.1/7.9*100%= 1.3%
F- 0.1/7.9*100%= 1.3%
Meter Stick
A- 0.1/4.9*100%= 2.0%
B- 0.1/4.9*100%= 2.0%
C- 0.1/4.9*100%= 2.0%
D- 0.2/4.7*100%= 4.3%
E- 0.1/4.8*100%= 2.1%
F-0.1/4.9*100%= 2.0%
Percent Uncertainty for Width
Paper Ruler
A- 0.2/1.7*100%= 11.8%
B- 0.1/1.9*100%= 1.3%
C- 0.1/0.9*100%= 11.1%
D- 0.1/1.9*100%= 5.3%
E- 0.1/1*100%= 10%
F- 0.2/0.6*100%= 33.3%
Meter Stick
A- 0.2/1.8*100%= 11.1%
B- 0.1/0.9*100%= 11.1%
C- 0.2/0.8*100%= 25%
D- 0.1/0.9*100%= 11.1%
E- 0.1/1*100= 10%
F- 0.2/0.7*100%= 28.6%
Percent Uncertainty for Thickness
Paper Ruler
A- 0.3/1.6*100%= 18.6%
B-0.1/1.9*100%= 5.3%
C- 0.1/0.9*100%= 11.1%
D- 0.1/2*100%= 5%
E- 0.2/1.8*100%= 11.1%
F- 0.3/0.6*100%= 50%
Meter Stick
A- 0.2/0.8*100%= 25%
B- 0.2/0.8*100%= 25%
C- 0.1/0.9*100%= 11.1%
D- 0.1/1*100%= 10%
E- 0.1/1*100%= 10%
F- 0.3/0.6*100%= 50%
Percent Uncertainty for volume
Paper Ruler
A- 0.1/21.5*100%= .47%
B- 0.2/27.8*100%= .72%
C- 0.2/6.4*100%= 3.1%
D- 0.2/30.0*100%= .67%
E- 0.1/27.0*100%= .37%
F- 0.3/2.8*100%= 10.7%
Meter Stick
A- 0.2/7.1*100%= 2.8%
B- 0.1/3.5*100%= 2.9%
C- 0.1/3.5*100%= 2.9%
D- 0.2/4.2*100%= 4.8%
E- 0.2/4.8*100%= 4.2%
F- 0.1/2.1*100%= 4.8%
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Based on your data, calculate the volume of each domino in cm^3, and in (cm_reduced)^3, where cm_reduced is the measurement made with your paper ruler.
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Paper Ruler
A- 7.9*1.7*1.6= 21.5 cm^3
B- 7.7*1.9*1.9= 27.8 cm^3
C- 7.9*0.9*0.9= 6.4 cm^3
D- 7.9*1.9*2.0= 30.0 cm^3
E- 7.9*1.9*1.8= 27.0 cm^3
F- 7.9*0.6*0.6= 2.8 cm^3
Meter Stick
A- 4.9*1.8*0.8= 7.1 cm^3
B- 4.9*0.9*0.8= 3.5 cm^3
C- 4.9*0.8*0.9=3.5 cm^3
D- 4.7*0.9*1= 4.2 cm^3
E- 4.8*1*1= 4.8 cm^3
F- 4.9*0.7*0.6= 2.1 cm^3
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Your uncertainty estimates are good, based on the reported lengths, thicknesses and widths. However the widths and thicknesses aren't consistent.
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`q002. Describe your experience in working with the buoyant balance.
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Balancing the piece of the lattice on the dominos and the paper clips between them was very hard when you had to counter balance the end with the paper clip.
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`q003. Report all data obtained during class, and include a description of what was measured and how:
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Bottle data
20 oz height 21
w/tubing
1.5 in down= 70 cm
1.75 in down= 65 cm
2 in down= 64 cm
2.5 in=62 cm
3 in= 59 cm
3.5 in= 52 cm
4 in= 50 cm
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I can't tell what any of the data to this point mean. Can you provide a little more explanation?
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Mark 1= 50 cm
2 “cal”= 45 cm
3 total calorie= 38 cm
4 10 calories= 32 cm
5 20 fl oz= 22 cm
Bottom label= 16 cm
Hole only
Bottom label= 5 cm
Mark 5= 7 cm
Mark 4= 8 cm
Mark 3= 9 cm
Mark 2= 10.5 cm
Mark 1= 11.5 cm
4 in= 13 cm
3 in= 15 cm
2.5 in= 16.5 cm
2= 18 cm
1.5 in= 19 cm
Measuring temperature of water
60 seconds.- 32 degrees C
120 seconds.- 31.5 degrees C
180 seconds.- 31 degrees C
240 seconds- 29 degrees C
300 seconds- 29 degrees C
360 seconds- 28.5 degrees C- before ice
w/ice
60 seconds- 10 degrees C
120 seconds- 10 degrees C
180 seconds- 10 degrees C
240 seconds- 11 degrees C
300 seconds- 11 degrees C
360 seconds- 11 degrees C
w/ 2 washer (2 small washers) - 60 degrees C
60 seconds- 12 degrees C
120 seconds- 15 degrees C
180 seconds- 15.1 degrees C
240 seconds- 15.1 degrees C
300 seconds- 15.2 degrees C
Adding water to a starifoam cup we measure the temperature of the water from room temperature in degrees C. Then adding washers to the cup, we then determined the temperature change from the heated washers.
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`q004. Once the buoyant balance is constructed, a small mass is placed on the free end (i.e., the end on which the paperclips aren’t hanging into the water). The system is observed to rotate in such a way that this end moves lower. Taking the direction of this rotation as the positive rotational direction, we conclude that the added mass results in an additional positive torque on the system.
The system oscillates for a short time and comes to rest, with the end containing the added mass a little lower than before.
Once the system is at rest, the net torque on it is zero.
Explain how we know that the torque in the positive direction is greater than before the mass was added.
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The torque is in the positive direction was greater before the mass was added because it was spinning in the opposite direction
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Is the torque in the negative direction now greater or less than before we added the mass?
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The torque in the negative direction is now less than before the mass was added.
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When the system is in equilibrium the torque in the positive direction and the torque in the negative direction are equal and opposite. They have the same magnitude. If one is greater in magnitude than before, so it the other.
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What caused the torque in the negative direction to change?
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When the mass was added the torque spun back to the opposite direction.
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Adding the mass increased the positive torque.
The negative torque also had to increase in magnitude.
What happened on the negative side to make it so?
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Suppose we add a square of paper having mass 0.6 gram to the ‘free’ side of the system, at a point 20 cm from the balancing point. How much additional torque will then be acting on the system?
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T= rf
= 20 cm* 0.6 g
= 12 g * cm
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Right idea, but 0.6 grams is a mass, not a force.
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Assuming that the paper clips are suspended from a point 20 cm from the balancing point, on the end opposite the ‘free end’, what must be the change in the torque on this side once the system has come to rest in its new position?
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T= rf
= -20 cm * 2.8 g
= -56 g* cm
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How is it that the torque on this end changes?
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The weight of the paper clips torque is reversed.
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Does the weight of the paper clips on this end change? If so does it increase or decrease, and by how much?
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The weight of the paper clips does change it, it increases by 1.4 g
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The paper clips don't change sides and they don't significantly change their distance from the fulcrum. Their weight stays the same. The torque doesn't reverse.
What force does change?
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#$&*
Does the torque due to the weight of the paper clips on this end change? If so does it increase or decrease, and by how much?
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Yes the torque does change the torque decreases by -56 g* cm.
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Does the buoyant force on the paper clips on this end change? If so does it increase or decrease, and by how much?
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The buoyant force on the paper clips on this end does not change.
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If more or less of the paper clips are under water, then the buoyant force has changed.
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Suppose we add a square of paper having mass 0.6 gram 3 cm on each edge, to the ‘free’ side of the system, at a point 20 cm from the balancing point. A sheet of this paper, 8.5 inches x 11 inches, has a mass of 5 grams. By how much does the torque in the positive direction change as a result?
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T=rf
= -20 cm* 5 g
= -100 g* cm
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If the hanging paper clips are 15 cm from the balancing point, then after the system has come to rest, by how much has the weight of the paperclips changed, and by how much has the buoyant force on the paper clips changed?
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The weight of the paper clips changed by 2.8 g. The buoyant force on the paper clips changed by 1 or 2 N.
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`q005. BB’s are bouncing back and forth, without loss of energy, between two parallel walls, one on the left and one on the right. Assume that the walls are 36 cm apart, and that gravitational influences have negligible effect. A BB shot from one wall, straight toward the other, will therefore keep bouncing back and forth between the walls forever. The speed of the BB, when it’s not in contact with the wall, will always be the same as the speed from which it was shot. (This is of course an idealization; the coefficient of restitution for an actual BB colliding with a real wall is less than 1 so energy will in the real world be lost with each collision).
A BB is shot toward the right wall from the position of the left wall. A second BB is shot at the same velocity from the same position, just as the first BB hits the right wall, from which it rebounds toward the left wall without any loss of speed.
How far from the left wall will the two BB’s be when they meet up?
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The BBs are 36 cm walls apart and halfway of 36 cm is 18 cm.
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Assuming that they narrowly miss one another and continue bouncing back and forth between the walls, how far from the left wall will they be the next time they pass one another?
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Half of 18 cm is 9 cm.
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At what distances from the left wall will they pass for the third, fourth and fifth time?
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Third- 4.5 cm
Fourth- 2.25 cm
Fifth- 1.125 cm
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As they continue bouncing back and forth, assuming their paths are just far enough apart to avoid collision, how many ‘passing points’ will there be between the two walls?
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There will be 3 passing points between the 2 walls.
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If the second BB was just a little slower than the first, how would the ‘passing point’ change as time goes on?
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The passing points will become slower.
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Suppose now that the second BB is fired at the instant the first BB is halfway to the right wall. How far from the left wall will they be when they meet?
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They will be about 9 cm from the left wall.
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If we continue firing a new BB every time the last BB reaches the halfway point, assuming that the BB’s always manage always to narrowly miss one another, at what distances from the left wall will BB’s pass one another?
At the instant of firing, at what possible distances from the left wall will all the other BB’s be located?
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The BB’s will be located 4.5 cm away from the left wall.
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At the instant the first BB passes by the second, at what possible distances from the left wall will all the other BB’s be located?
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When the 1st and the 2nd BB passes each other, the other BB’s could be 9 cm or 4.5 cm from the left wall.
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In what other ways might the firing be timed so that when two BB’s pass, every other BB is at the same distance from the left wall as at least one other BB?
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When firing one make sure that one BB is a few centimeters before the middle between the left and the right wall.
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`q006. If an ideal gas is kept at constant volume, then the percent change in its pressure is the same as the percent change in its absolute temperature. Every additional 10 cm of water supported in a thin vertical tube requires an increase in pressure equal to about a 1% of standard atmospheric pressure.
Based on this, how much additional pressure do you estimate corresponds to one unit on your ‘squeeze scale’?
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The additional pressure on the squeeze scale would be a 2 or 3.
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`q007. If the initial absolute temperature of the gas is 300 Kelvin, then if it is heated enough to raise a column of water 40 cm high, then:
By how much will its temperature have changed?
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The temperature would have to be 40 or 50 K degrees more to raise a column of water 40 cm.
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What percent of an atmosphere is required to support a 40 cm column of water?
What percent temperature increase would therefore be required?
How many degrees would this be?
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If the bottle contains 500 milliliters of air, how much energy would be required to achieve this change? Note that the volume of water in a thin tube can be regarded as negligible, so that the volume of the gas will remain unchanged. (You should know how to find approximately how many moles are contained in 500 milliliters of air at atmospheric pressure and typical ambient temperatures).
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500 mL of air has 1 N/m^2 of energy. 500 mols of air
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The air is diatomic.
How do you figure out the energy required to raise the temperature of a diatomic gas?
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How much would the temperature of the air in the bottle have to change to have the same effect as one unit on your ‘squeeze scale’?
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The temperature would have to change by 3 or 4 degrees K to have the same effect as one unit on my squeeze scale.
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