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course PHY202
2/8 4:20pm
`q001. Include a copy of your data for the domino measurements, including your estimated uncertainty: ****
The 1st set of data is the measurements of the domino with the paper ruler.
Length (cm) Thickness (cm) Width (cm)
Domino A 13 3 7
Domino B 13.5 3 7
Domino C 13.5 3.4 7
Domino D 13.5 3.5 7
Domino E 13.5 3.4 7
Domino F 13.5 3 7
The 2nd set of data represents the dominoes measured with the meter stick
Length (cm) Thickness (cm) Width (cm)
Domino A 5 0.8 2.5
Domino B 5.2 1 2.6
Domino C 5.1 1 2.5
Domino D 5 1 2.5
Domino E 5.2 1 2.6
Domino F 5.1 0.8 2.5
I think my measurements are accurate within 1 or 2mm. When I measured the dominoes I looked from above onto the meter stick or the paper ruler to get a straight on view of what I was looking at.
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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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Did you calculate the volumes?
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`q002. Describe your experience in working with the buoyant balance.
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I was able to get the hanging paper clip to balance with the addition of more water to the cup it was hanging over. Sometimes when it oscillated too “hard” the balance would not stay balanced. It became hard to keep it balanced because the balance kept falling off of the paper clip that was sandwiched in between the 2 dominoes.
The part I enjoyed the most with the balance was the part where we added tape to the ends and generated static electricity in order to get the balance to oscillate. I found that it oscillated more evenly when the tape was placed about 3 or 4 centimeters from the end. If the tape was on the end of the balance, then the oscillations got to be too “heavy.”
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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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We know this because the position of the system has rotated in the positive direction from its original position.
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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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Because the torque is greater with the side opposite the clips hanging in the water, I thought at first the torque would be less on the negative side. However, after some thought, I think that the torque in the negative direction will have a greater PE and therefore produce more torque because that side was raised.
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The basic principle is that once the system comes to equilibrium, the torques, as with every system in equilibrium, must be balanced. Thus the two torques involved here are equal and opposite. So if the torque in one direciton is greater than before, the torque in the other direction must also be greater than before.
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What caused the torque in the negative direction to change?
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The change in position which was raised on that side.
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Raising one side can change the balancing position without significantly changing the torque. The change in the balancing position of a system does not necessarily imply any changes in the torques.
What specifically was it that changed to cause the change in torque?
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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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Torque=force*distance
Here, distance=.2m
Force=.006kg*9.8m/s2 which equals .0588N
Now, .0588N*.2m=.01176N*m=.01176J
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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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The torque would have changed by the force of the paper clips being added to this end.
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Only the paper is being added. The paper clips were already there, and there's no signficant change in the torque they exert.
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How is it that the torque on this end changes?
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From here down, I’m a little stuck.
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This goes back to a question in a previous note. Something changed in the forces acting on the system. On one side it was the addition of the paper. What is it on the other side?
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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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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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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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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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I think on this one we use the Torque=mgy but I am unsure. When solving this way, I get the following:
Torque=.006kg*9.8m/s2*.2m
=.01176J
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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 paperclips wouldn’t have changed would they? I don’t understand how to solve this problem and need a hint.
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You're right, but there are two questions here:
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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18cm because they are traveling at the same velocity and shot at the exact moment the other BB hits the 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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18cm because their velocity is staying the same
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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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Still at 18cm because of no velocity change
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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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2 because they pass when the BB hits the left wall and goes back to the right wall and they pass again when the BB hits the right wall and goes back to the left
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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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If the slower BB was shot from the left wall, the passing point would be gradually be a longer distance from the left wall until it hits the right wall, then the distance would gradually be shorter from the left wall
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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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9cm from the right wall if they are fired in the same direction
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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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Some BBs will pass at the 1 quarter mark, or 9cm, and some will pass at the 3 qtr mark, 27 cm away from the 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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I think that each BB will be 1 quarter (9cm) or 1 half (18cm) or 3 quarters (27cm) away from the wall once we have 3 firings. If more BBs are fired then that ratio will go down.
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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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It would have to be shot at the 1 of the times that the other 2 BBs were shot. If the BB hit the wall and 1 was shot, then the other BB needs to be shot at the time the 1st BB hits the wall.
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If you can fire two BB's in the time it takes to cover the distance, then you could in principle fire three, or four, etc., in this time.
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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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1 unit of my “squeeze scale” corresponds to about a 2.4 cm change in height. So, using the reasoning above, the additional pressure that corresponds to this is .24% of atmospheric pressure.
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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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Because it takes 1% atmospheric pressure to raise a vertical column of water and also a 1% change in temperature is equal to a 1% change in pressure, we can say that to raise a 10cm column of water by 10cm requires a 1% change in temperature. So, for this problem, the water level raises by 40 cm which corresponds to a 4% change in temperature. 4% of 300K is 12K. Therefore, the final temperature would be 312K and the temperature would have changed by 12K.
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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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Because we’re dealing with a diatomic gas with air, then the formula E=5/2nRT where E represents the amount of energy required to achieve a 12 Kelvin change as found in the previous problem.
First we have to find moles. 1 mole of gas=22.4L. So, .5L=.022mole=n; R is the gas constant of 8.31J/mole Kelvin; T=12 Kelvin
5/2*.022 mole*8.31J/mole Kelvin*12 K. The units all cancel out except for J which leaves 5.48J.
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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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On my squeeze scale, a 5 represents a change of about 24cm of the water level. So, 1 unit of change represents a change of about 4.8cm.
Taking the original information from the problem with Temp=300K and replacing the height of the column with 2 cm, we have the following:
1% change represents 10cm in height. So for 4.8cm in height, this represents a .48% change. So taking .48% of 300K=301.4K with a 1.4K temperature change.
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Previously you said that 2.4 cm of column height correspond to a unit of squeeze, which in turn corresponds to .24% of an atmosphere.
Not sure how you got double that result here. But I'm pretty sure you understand what your doing.
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end document
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You've got lots of good answers, but a few that need revision, especially that volumes of the dominoes (which you don't appear to have calculated).
&#Please see my notes and, unless my notes indicate that revision is optional, submit a copy of this document with revisions and/or questions, and mark your insertions with &&&& (please mark each insertion at the beginning and at the end).
Be sure to include the entire document, including my notes.
If my notes indicate that revision is optional, use your own judgement as to whether a revision will benefit you.
Spend a reasonable amount of time on your revision, but don't let yourself get too bogged down. After a reasonable amount of time, if you don't have at least a reasonable attempt at a solution, insert the best questions you can showing me what you do and do not understand, and I'll attempt to clarify further. &#
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