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course PHY201
2/1 8:30pm
`q001. Phy 201 students only: Report the data you took in class on the dimensions of the dominoes. Be sure to use an organized table to report your measurements.****
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
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Indicate the uncertainty in your measurements for the meter stick, and for the paper rulers. Explain why you don’t think your uncertainty is much lower than you estimate, and why you don’t think it’s much higher.
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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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What is your percent uncertainty for the measurement of the length of a domino, for each ruler?
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The average length for each domino with the meter stick was 5.1cm. Because I feel I was accurate within 1mm for the meter stick, then the percent uncertainty would be .2% uncertainty.
The average length for each domino measured with the paper ruler was 13.5cm. I think the uncertainty of this measurement was a little higher because the scale of the ruler was much harder to see with it being so small. The amount of uncertainty for this measurement I estimate to be 2mm. That gives a percentage uncertainty of .015% uncertainty.
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What is your percent uncertainty for the measurement of the width of a domino, for each ruler?
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Using the same uncertainty as the previous 2 being 1mm for the meter stick and 2mm for the paper ruler, the percent uncertainty for the width of the dominoes are as follows:
The average width of a domino measured with the paper ruler was 7cm. With a 2mm uncertainty estimate, the percentage of uncertainty is .3%.
The average width of a domino measured with the meter stick was 2.5cm. With a 1mm uncertainty estimate, the percentage of uncertainty is .4%.
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What is your percent uncertainty for the measurement of the thickness of a domino, for each ruler?
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The average thickness of the domino measured with the paper ruler was 1cm. With a 1mm uncertainty, the percentage is 1%.
The average thickness of the domino measured with the meter stick was 3.2cm. With a 1mm uncertainty, the percentage is .3%
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Calculate the volume of each domino, in the units of each of your rulers.
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When measured with the paper ruler, the volume is as follows:
Domino A= 273 cm3
Domino B= 283.5 cm3
Domino C= 321.3 cm3
Domino D= 330.75 cm3
Domino E= 321.3 cm3
Domino F= 283.5 cm3
When measured with the meter stick, the volume is as follows:
Domino A= 10 cm3
Domino B= 13.5 cm3
Domino C= 13 cm3
Domino D= 12.5 cm3
Domino E= 13.5 cm3
Domino F= 10.2 cm3
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What do you think is the percent uncertainty in your calculation of the volume, for each ruler?
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I think the percent uncertainty for my calculations for the paper ruler is 15% while the meter sick is about 3%.
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Is this the result you get based on the uncertainties in the length, width and thickness?
If so, how is it related to those uncertainties?
If not, can you provide an estimate based on those uncertainties and indicate how it is related to those results?
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`q002. 500 BB’s each of mass 0.15 gram are place in a cylinder 50 cm long and 20 cm in diameter, and the cylinder is shaken rapidly and randomly, giving the BB’s an average velocity of 8 m/s. Assume that their directions of motion are randomized over 3 dimensions of space.
What average pressure is exerted by the BB’s on the ends of the container?
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Using the formula Pave=Fave/A which = N*mv2/V where V=L*A and A=pi*r2*L
A=3.14*(10cm)2*50cm=15700cm3 which =157m3
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157 m^3 would be a volume. I believe you have calculated the volume of the cylinder, not the area of an end.
What is the area of a circle of radius 10 cm?
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N=500
m=.00015kg
v=8m/s
V=L*A=157m3*.5m=78.5m4
So we have 500*.00015kg*(8m/s)2/78.5m4
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The unit m^4 is not a unit of volume, which could tip you off.
See my previous note. The quantity you used for area here appears to be the volume of the cylinder.
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=4.8kg*m2/s2/78.5m4=.06 kg*m/s2=.06N (I think my math is right on this and I know I’m supposed to have a Newton as the unit of measurement but my units didn’t come out this way. I’ve double checked myself and can’t see my mistake)
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You are right that the units don't work out. Fortunately this should be easy to correct (see my preceding notes).
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What is the average force exerted by the BB’s on one of the ends of the container?
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As I was working through this problem, I thought of this question. Why do we multiply by 2 to get the magnitude of the change in force exerted on 1 of the ends? Wouldn’t the momentum go to 0 before starting back toward the other end???
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Momentum does go to 0, which entails a change in momentum. Then the particle rebounds, which entails another change in the momentum, of the same magnitude and in the same direction as the first.
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Momentum of 1BB=mv which =.00015kg*8m/s=.0012 kg*m/s for traveling to 1 end of the container. To get the total force of the trip down and back, we multiply by 2 resulting in .0024 kg*m/s.
In order to get the time it took to travel, we use L/v=.5m by 8m/s=.0625s
To find the momentum change of 500 BBs, we multiply 500*.0012kg*m/s=.6kg m/s
Then Fave=’dp/’dt= .6kg m/s/.0625s=9.6N
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Good.
If you correct the original error in calculating area, and carry the result through, your result should be consistent with this result.
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You have a result for the force, but you haven't yet calculated the pressure.
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Atmospheric pressure is about 10^5 Newtons / meter^2. What percent of atmospheric pressure is achieved by this system?
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9.6/10.5=.0096%------ I’m not sure where the units went on this one.
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What is the total KE of all the BB’s?
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KE of all BBs=500*1/2 mv2=500*.0015kg*(8m/s)2=48J
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To what Kelvin temperature is this situation equivalent?
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Using KE=3/2kT where k=R/N_A=8.31J/(mole Kelvin)/500*6*1023=2.77*10-26J/particle Kelvin
Now we have 48J=(3/2)*2.77*10-26J/particle Kelvin*T
T=1.6*1026
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The kinetic energy should be that of a single particle (note the unit J / (particle Kelvin) ).
The correct answer will still be ridiculously high, but not quite as high.
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This answer doesn’t make sense at all to me.
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`q003. Suppose now that you push on one end of the cylinder, moving it 1 cm closer to the other end, so that the cylinder’s length decreases to 49 cm.
You previously figured out how much average force is exerted on an end of the container by the BB’s. Assuming that the end you are pushing on moves without frictional resistance, how much work must you do to move it through its 1 cm displacement?
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Work=force*distance, so = 9.6N*.1m=.96J
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You just did work on the system. What happened to that energy? Assume no energy loss to friction, and assume that all collisions are perfectly elastic, which means that no kinetic energy is lost in any collision.
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The energy that we used to push the cylinder in 1cm was transferred to the inside of the cylinder by making the total length of the cylinder a shorter distance from 50cm to 49cm.
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Good. The transfer came about in the collisions of the end with the BB's, which causes each BB in each collision to speed up.
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`q004. Two pendulums are released simultaneously. One oscillates at 60 cycles / minute, the other at 56 cycles / minute. So most of the time the oscillations of the pendulums are not synchronized returning to the starting point at different times.
How many times per minute will the two pendulums become synchronized, in the sense that both return to their starting points at the same time?
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The 1st pendulum oscillates every 1 sec
The 2nd pendulum oscillates every 1.07 sec
Would the 2 pendulums synchronize 1.07 times? If so, why?
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Every time the shorter pendulum 'laps' the longer, it goes through an extra cycle.
How many times does this occur in a minute?
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`q005. To the left of the y axis, a line runs parallel to the x axis, 1 centimeter higher than the x axis. At the point (0, 1) the line changes direction so that it passes through the x axis at the point (5 cm, 0). This broken line will constitute path 1.
Another line, which defines path 2, runs below the x axis at a constant distance of ½ cm from the axis.
At what point do these two paths intersect?
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According to my sketch, the 2 paths will intersect at point (5.5cm, -.5cm)
Is there a way mathematically to find the answer to this question?
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A third path consists of a straight line passing through the origin and the point at which the first two paths intersect. At what point to the left of the origin does this third path intersect the first?
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Once again, this looks like it intersects at the point (-4.5cm, 1)
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Your results are consistent with a reasonable sketch.
They aren't exactly right. To get the exact results you would figure this out mathematically using similar triangles.
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`q006. How far is the point (0, 4 cm) from the point (-20 cm, 0)?
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Using the Pythagorean theorem, I understand now how to solve this problem. Using A2+B2=C2 and knowing what they represent, I can solve.
Sqrt(-202+42)=sqrt(416)=20.4cm
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How much further from the point (-20 cm, 0) is the point (0, 5 cm)?
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Sqrt(-202+52)=sqrt(425)=20.6cm
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How far would you have to move along the y axis, starting at (0, 4 cm), in order to move 1 millimeter further from the point (-20 cm, 0)?
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For the point (0,4) and (-20,0), if the difference in 1 point on the graph is is .2cm, this corresponds to 2mm. So, to move 1mm further from the point you would have to move 5mm on the axis.
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Good reasoning, but it appears that it's a 1 cm difference in the y coordinate of the point that results in a .2 cm difference in the distance.
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How far would you have to move along the y axis, starting at (0, 4 cm), in order to move 1 millimeter closer to the point (-20 cm, 0)?
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It seems to me that you would have to move 5mm closer on the axis to move 1 millimeter closer to the point. Same reasoning as above except closer to the axis.
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At what points along the y axis would your distance from the point (-20 cm, 0) differ from the distance between this point and (0, 4) by 2 millimeters?
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If I solved the above problem correctly, then the point would be (0,5cm)
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Take a good look at your results. Do you see any patterns?
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They seem to all be going the same distance either up the y-axis or down the y-axis in order to get a certain distance closer or further from the original point. All distances seem to be the same going up the axis and coming down closer to the origin.
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That will change if you move very far along the y axis, but in the close neightborhood of any point it will be very nearly as you say.
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`q007. A well-insulated container is divided into two compartments.
Suppose the compartments are of equal volume and contain equal amounts of air, with the air in one compartment at 80 degree Fahrenheit and the other at 50 degrees Fahrenheit. If the divider is removed and the air allowed to mix throughly, what would you expect to be the final temperature?
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I would expect the final temperature to be the average of the 2 temperatures which is 65 degrees Fahrenheit.
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Suppose the compartment at 80 Fahrenheit has double the volume of the other and contains twice as much air. If the divider is removed and the air allowed to mix throughly, what would you expect to be the final temperature?
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I would expect the temperature to again be an average of the 3 which corresponds to 70 degrees Fahrenheit.
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The meaning of ‘equal amounts’ is a little vague. ‘Equal amounts’ could possibly mean equal volumes, or equal masses. If the volumes of the compartments are the same and the masses of air are the same, then one of the compartments will be at higher pressure than the other. Which will this be?
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The compartment with the 80 degree temperature will have the highest pressure associated with it because the molecules are moving at a faster rate than the 50 degree temperature.
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If the two compartments have the same volume and are at the same pressure, will the masses of air be equal? If not, which will be greater?
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I think the masses of air would have to be equal because volume and pressure are the same.
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Assuming the temperatures differ, this won't be the case.
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Good effort.
See my notes related to a few errors and discrepancies.
Do spend a reasonable but not excessive amount of time on some revisions.
&#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. &#
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