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course Phy 202
submitted 02/16/2012
120123
The Introductory Problem Set problems for your course are at
http://vhmthphy.vhcc.edu/ph2introsets/default.htm
Problem Set 5 corresponds to work we are doing with gases and fluids. You should work through that set and make notes indexing topics, concepts, solution methods, etc.. By Feb. 2 you will be expected to be familiar with the problems in this set.
Problem Set 6 corresponds to work we are doing related to waves and optics. You will be expected to familiarize yourself with this set by Feb. 8.
General College Physics students should do a quick read-through of Chapters 10, 13, 14 and 15
University Physics students should do a quick read-through of Chapters 14, 17, 18, 19 and 20
You are expected to have made brief notes, at least recognizing things in the text that you think might be relevant to the systems we are working with in class.
`q001. Include a copy of your data for the domino measurements, including your estimated uncertainty:
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Paper Ruler
Length (cm reduced) Width (cm reduced) Height (cm reduced)
A 12.2 6.1 2.0
B 12.2 6.0 2.2
C 12.3 6.0 2.3
D 12.1 6.0 2.4
E 12.2 6.2 2.3
F 12.3 6.1 1.8
Meter Stick
Length (cm) Width (cm) Height (cm)
A 5.2 2.5 0.9
B 5.2 2.5 0.9
C 5.2 2.6 1.0
D 5.1 2.6 1.0
E 5.2 2.6 1.0
F 5.2 2.6 0.8
Uncertainty Paper Ruler
Length (%) Width (%) Height (%)
A 0 0 9.1
B 0 1.6 0
C 0.8 1.6 4.5
D 0.8 1.6 9.1
E 0 1.6 4.5
F 0.8 0 18.2
Uncertainty Meter Stick
Length (%) Width (%) Height (%)
A 0 3.8 0
B 0 3.8 0
C 0 0 11.1
D 1.9 0 11.1
E 0 0 11.1
F 0 0 11.1
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Uncertainty is not the same thing as deviation from the mean.
Your eyes can't resolve anything smaller than about .03 millilmeter. So your uncertainty is at least this great. And you can't estimate the unmarked distance between two millimeter marks to the nearest 1/30th of this distance.
What therefore do you think is the uncertainty in your various measurements for each ruler?
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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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Volume for each is calculated by multiplying length by width by height
Dominos Volume (cm_reduced)^3
A 14.88
B 16.10
C 16.97
D 17.42
E 17.40
F 13.51
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`q002. Describe your experience in working with the buoyant balance.
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We taped three paper clips to one end of a plastic piece and balanced it in the middle with the same number of paper clips hanging freely from the opposite side of the plastic piece. When we put the paper clips hanging freely into a cup of water the balance was slightly shifted, lowering the opposite side without the water. The shift shows that water does in fact have a density greater than the paper clips. We then balanced the plastic piece by adding paper clips to the side with the water. We should be able to calculate the bouyant force acting on the paper clips.
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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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`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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If the system moves in the counter clockwise direction then it is said to be in the positive 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 the less because we are just adding mass causing an increase in torque to the other side.
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If the positive torque is greater and the negative torque is smaller then the net torque will change.
The only way a system can be in equilibrium is for the net torque to be zero.
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What caused the torque in the negative direction to change?
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Adding mass on the positive direction will increase torque on that side and decrease torque in the negative direction.
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If the positive torque is greater and the negative torque is smaller then the net torque will change.
The only way a system can be in equilibrium is for the net torque to be zero.
So the magnitude of the torque on the negative side did not decrease.
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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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convert 0.6 grams to kg and multiply by gravity to give you weight, a force. 0.0006 kg * 9.8 m/s^2 = 0.00588 N
then, 0.00588 N (0.2 m) = 0.001176 Nm torque
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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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Once it has come to rest it should be zero.
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This side now has the paper on it. So gravity is exerting more force, resulting in more torque on this side. You previously calculated the difference.
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How is it that the torque on this end changes?
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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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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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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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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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`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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They will meet in the middle, 18 cm from the left wall.
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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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it will still be the same, 18 cm from the left and right walls because they are not losing any speed.
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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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It will still be the same, 18 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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one point in the middle between the two walls becuase they have the same speed.
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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 point would change slightly each time in one direction until both are in sync with each other then it will go back to slightly changing and the system will continue this pattern.
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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 meet 3/4 the way from left to right wall, so, 3/4 of 36 cm = 9 cm, so 36-9=27 cm. They will meet 27 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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Each BB will pass each other from the left wall at measurements of 0 cm (the left wall), 9 cm, 18 cm, 27 cm, and 36 cm (the right 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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the first and second BB will be at the 27 cm mark from the left wall and the third bb will be at the 9 cm mark 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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You could keep firing every time the prevous bb is 9 cm from the left wall or when the previous bb is 27 cm from the left 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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3 units of my squeeze scale is equal to 1% of atm
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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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to raise a column of water 40 cm would be a 4% atm increase, since pressure is equal to the percent change then 4% of 300K is 12. so the change in temperature would be 12 K.
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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 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 divided by 22.4 L (volume in one mole of gas) is equal to 0.0223 moles.
PV=nRT
with the specific heat of air is 1.01 J/g C so the energy is required is
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The specific heat of air is different for constant-volume heating than for constant-pressure heating, which is different from that for a process in which both volume and pressure change.
So there is not one single specific heat for air.
The number you quote, close to 1 J / (g C), is about right for the specific heat, with respect to mass, for a constant-pressure expansion of a diatomic gas. Air is abot 99% diatomic, so it's not a bad value, but it doesn't apply here since pressure isn't constant.
In any case you will be much better served by using the molar specific heat.
The molar specific heat of a gas is related to the gas constant R.
Check previously posted notes for the relationship.
You have the number of moles and the change in temperature, so all you need is the molar specific heat.
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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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I want to go ahead and submit what I have, I will finish these last few Sunday.
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