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course Phy 202
Sorry this is a little late. I will be submitting some more of the questions throughout the week and weekend you should have all of the my submitions no later than Saturday.
`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.
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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
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According to your data some of the dominoes are over three times as wide as others.
However they all have about the same width. Their thicknesses vary significantly, but their widths don't vary by nearly as much as your data would indicate.
Can you clarify?
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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
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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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The paper ruler was minimized down by approximately 50%. The meter stick can give 100% of the measurement.
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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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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%
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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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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%
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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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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%
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Calculate the volume of each domino, in the units of each of your rulers.
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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
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According to your calculations, some dominoes have 10 times the volume of others.
If you place 10 of the thinnest dominoes on one side of a balance and the thickest domino on the other, the balance will tip heavily on the side of the 10 thin dominoes.
Your calculations do follow from your data. I think the problem lies with your recording of the data (see previous note).
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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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What do you think is the percent uncertainty in your calculation of the volume, for each ruler?
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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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`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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50 cm*20 cm= 1,000 cm^2
P= F/A=mg/a= (0.15 g)(9.80 m/s^2)/.010 m^2= 147 g* m/s^2
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To get the force you need to calculate the momentum change of each BB at collision, multiply by the number of BB's to get the total momentum change, find the time between collisions, use the total momentum change and the time between collisions to get the average force, and then divide the average force by the area of the end.
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Gravity is not involved in this model.
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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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.00015 kg * 9.802 m/s^2= .00147 kg m/s^2
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9;8 m/s^2 is not relevant to this model.
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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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1 atm= .01%
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What is the total KE of all the BB’s?
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0.15 g *500= 75 g
KE= 1/2mv^2
= ½(75 g)(8 m/s)^2
= 2,400 g*m/s
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Your calculation is correct, except (8 m/s)^2 = 64 m^2 / s^2 so your final result is 2400 g m^2 / s^2, not 2400 g m/s.
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To what Kelvin temperature is this situation equivalent?
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PV= nRT
(147 g * m/s)(16 m/s)= (1.25 mol/ L)( 0.082 (L* atm)/(mol *k)) (T)/ .1025
2352/.1025= T
T= 22,946.3 K
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You are using the correct relationship. Good.
Your units don't really work out, and your pressure isn't right (see my previous notes). 1.25 mol / L isn't appropriate either; you have 500 BB's and it would take 6.02 * 10^23 BB's to constitute a mole.
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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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W=FD
= .00147 kg m/s^2 * .001 m
= 1.47 * 10^-6 kg * m/s
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Right procedure, but the force isn't correct (see previous notes) and 1 cm is .01 m, not .001 m.
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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 turned into potential energy.
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Nothing moved higher or lower, and no elastic object was stretched or compressed.
What happened was that the BB's were sped up as they coolided with the approaching end of the cylinder. Your work went into the KE of the BB's.
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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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60-56= 4 times/ minutes
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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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Sqrt of(0)^2 + (1)^2= srt of 1 =1
Sqrt of (0)^2 +(5)^2= sqrt of 25 = 5
5-1= 4
Point of intersection (4,0)
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Both paths pass through the x axis at points other than (4, 0), so this can't be the point of intersection. The intersection will occur in the fourth quadrant.
You probably need to show my your drawing for 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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Sqrt of (5)^2 + (0)^2= sqrt of 25= 5
Third path intersection= (0,5)
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`q006. How far is the point (0, 4 cm) from the point (-20 cm, 0)?
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Sqrt of (0)^2 + (4)^2= sqrt of 16= 4
Sqrt of (-20)^2 + (0)^2= sqrt of 400= 20
20-4= 16
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Sketch these points on a set of coordinate axes.
What is the run between the points?
What is the rise?
The rise and the run are legs of a right triangle. How do you use the legs to get the hypotenuse, which will be the distance between the points?
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How much further from the point (-20 cm, 0) is the point (0, 5 cm)?
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Sqrt of (-20)^2 + (0)^2= sqrt of 400= 20
Sqrt of (0)^2 + (5)^2= sqrt of 25= 5
20-5= 15
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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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Sqrt of (0)^2 + (-20)^2= sqrt of 400= 20
20-16=4
-.004 cm to go 1 mm
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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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Work from previous questions
+.004 cm to go 1 mm
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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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Work from 2 questions
4 cm * 2mm/1,000 cm= .008 mm
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Take a good look at your results. Do you see any patterns?
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The distance between the points goes up by .004 mm.
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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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80+50= 130 degrees F
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See my notes on the first question, which should lead you through the process of sketching the correct right triangle and finding its hypotenuse.
Then use the same approach for other pairs of points.
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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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80 degrees/2= 40 degrees F
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None of the air is as cool as 40 F. When mixed the temperature will be somewhere between the temperatures of the two compartments.
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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 at 50 degrees F will have the higher pressure because of the temperature.
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What would happen to the pressure in one of the compartments if you were to keep increasing its temperature?
How is your answer to this question consistent with, or inconsistent with your answer to the original question?
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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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No, the greater air mass would be the compartment with the greater temperature.
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What would happen to the pressure in one of the compartments if you were to keep increasing its temperature?
Would you have to add gas or release gas to get back to the original pressure?
How is your answer to this question consistent with, or inconsistent with your answer to the original question?
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*#&!*#&!
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&#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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