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course Phy 241
12 am 9/16/12
120917Complete and submit the following documents in the same manner as previous documents.
qa_06
query_05
It is worthwhile to become familiar with the PHeT simulations. Just click on the link PHeT to access the complete list of simulations, and follow the instructions for each. You should spend maybe 15 minutes on each one, by which time you'll know what it's good for. Knowing that you'll be able to use the simulation whenever you think it might be beneficial.
PHeT 1.26 Estimation
PHeT 2.24 Moving Man
The Random Problems anticipate some of the problems you are likely to encounter on the Major Quiz. You should select one of the versions of each problem, work it out and submit it.
Assignment 2 (do Problem 1 only)
Assignment 3
Assignment 4
Week 2 Quiz #1
Look at the Memorize This document. It is recommended that you memorize or otherwise be sure you completely understand Ideas 1-4. There's nothing to turn in, but this document will be valuable to many of you.
The page at
http://vhcc2.vhcc.edu/dsmith/geninfo/orientation_maps/testing.htm
includes instructions for testing of distance students away from the VHCC campus (which you can ignore) and testing in the VHCC Learning Lab (which you don't want to ignore since that's where you'll be testing). It also includes a link to the actual randomly-generated tests. You'll want to begin looking at Major Quizzes fairly soon.
`q001. If the area of a 'graph rectangle' is 30 meters and its altitude is 5 meters / second, what is its width?
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6 s.
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If the area of a 'graph rectangle' is 60 Newton * meters and its width is 10 meters, what is its 'graph altitude'?
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6 N.
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If the slope of a 'graph trapezoid' is 40 Newtons / meter and its width is 2 meters, what is its rise?
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80 N.
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If the rise of a 'graph trapezoid' is 50 meters / second and its slope is 200 meters / second^2, what is its width?
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.25 s.
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If the area of a 'graph trapezoid' is 30 meters, its average altitude is 5 meters / second and its slope is 10 meters / second^2, then what is its width and what are its altitudes?
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width = 1 s.
height = 25 m and 35 m.
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This doesn't make its average altitude 5 meters / second.
You acutally have to generalize the concept of a trapezoid to include parts that extend below the horizontal axis, and generalize area so it can be negative.
With ave. altitude 5 m/s the width would have to be 6 s. With that slope, the rise would be way greater than the average altitude and would necessitate that the left altitude be negative.
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Interpret the preceding question in terms of the motion of an object, specifying its initial, average and final velocities, the change in velocity, the displacement, its acceleration and the time interval.
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Acceleration: 10 m/s^2.
V0 = 25 m/s.
VF = 35 m/s.
vAve= 5 m/s.
Displacement = 30 m.
time interval = 0 to 1 second.
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The five quantities associated with a 'graph trapezoid' are its two altitudes, its area, its slope and its width. If we know any three of these quantities we can find the other two. List all possible combinations of these five quantities.
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vf, v0, s, a, t.
vf, v0, s
vf, v0, a
vf, v0, t.
vf, s, a
vf, s, t
vf, a, t
v0, s, a
v0, s, t
v0, a, t
s, a, t
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Explain how you would find the area and slope of a 'graph trapezoid', given its two altitudes and its width.
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height 2 - height 1 = dh.
slope = rise/run = dh/width.
area = length*height1 + 1/2*width*dh.
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Explain how you would find the area and width of a 'graph trapezoid', given its two altitudes and its slope.
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h2 - h1 = dh = rise.
slope = dh/width. width = dh/slope.
area = width*h1 + 1/2 * width* dh.
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Explain how you would find the area and 'right altitude' of a 'graph trapezoid', given its 'left altitude', its slope and its width.
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(right altitude - left altitude)/width = slope. You can solve this for right altitude.
From there you can use A = lw and A triangle = 1/2 lw to get the area.
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Explain how it's more complicated than in previous situations to find the 'right altitude' and width of a 'graph trapezoid', given its left altitude, and its slope and its area.
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Slope = (h2-h1)/w. You still have two unknowns (width, right altitude).
Area = left altitude * width + 1/2 (right altitude - left altitude)*width.
You still have too many unknowns. You would have to get simultaneous equations to solve.
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For each of the preceding four questions, interpret in terms of uniformly accelerated motion, assuming that the graph trapezoid represents velocity vs. clock time.
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Acceleration would be the slope of the trapezoid.
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For each of the preceding four questions, interpret in terms of forces, work and displacement, assuming that the graph trapezoid represents rubber band force vs. length.
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Displacement would be the area of the graph.
Change in force would be right altitude - left altitude. Divide this by two to get average force.
Work would be the slope of the trapezoid.
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`q002. On an Atwood machine, a net force equal to the weight of 6 paperclips accelerates a system of 10 dominoes at 2.5 cm / s^2.
What would be the acceleration of a system consisting of 4 dominoes subject to a net force equal to the weight of 10 paperclips?
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F = number of paperclips.
m = numbers of dominos.
F= ma.
a = F/m. 6/10 does not equal 2.5 If you have F = (ma)/ x where x is a constant, in this case x=4.166 makes the statement 6x/10 = 2.5.
10 = (4*a)/4.166. a = 10.4 cm/s^2.
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If you double the number of dominoes and the number of paperclips, what happens to the acceleration?
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12 = 20a /4.166.
a = 2.5 cm/s^2.
Nothing changes.
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If you double the number of dominoes and halve the number of paperclips, what happens to the acceleration?
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3 = 20a / 4.166.
a = .62 cm/s^2.
This is 1/4 of 2.5 cm/s^2.
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If you double the number of paperclips and halve the number of dominoes, what happens to the acceleration?
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12 = 5a / 4.166.
a = 10 cm/s^2. This is 4 times 2.5 cm/s^2.
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How many paperclips and how many dominoes would result in an acceleration of 40 cm/s^2? There are many possible answers to this question. An answer which involves a fractional number of dominoes and/or paperclips is acceptable for General College Physics, though an answer involving a whole number of paperclips and dominoes is preferable (and is expected for University Physics).
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Fx/m = 40 cm/s^2.
40 is 16 times 2.5.
Double one quantity gives you 2 a.
Quadruple one quantity gives 4a.
Thus 16 times one quantity gives 16a.
96 paperclips = 10 a/4.166.
a = 40 cm/s^2.
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Interesting question: What is the curve in the y vs. x plane, with y = number of clips and x the number of dominoes that result in the given acceleration. (96, 10) would be one point on this curve.
You don't need to answer this. It's not a difficult question and you could. The main thing is that this curve tells us something, perhaps useful and perhaps not, about the system.
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`q003. A small paperclip has a weight of about 4 milliNewtons. A domino has a weight of about 16 grams. The acceleration of 2 dominoes, when subject to a net force equal to the weight of a small paperclip, is observed to be 4 cm/s^2. All results are to be regarded as accurate to within +-5%.
What therefore would be the acceleration of a 1-kilogram mass if accelerated by a net force of 1 Newton?
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After converting units.
.004 N = .016 kg * .04 m/s^2 / x.
x = .16.
1 N = 1kg *a / .16.
a = .16 m/s^2.
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That would be 250 times the force on about 30 times the mass, which would result in a little over 8 times the acceleration, about 32 cm/s^2 or .32 m/s^2.
I think you only counted one domino.
However the given numbers in the problem are just plain wrong. The given acceleration should be about 120 cm/s^2, instead of 4 cm/s^2.
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Based on your previous answer, what would be the acceleration of your mass if accelerated by a net force equal, in Newtons, to your age in years?
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My age = 20 years = 20 newtons.
My mass = 130 lb = 59 kg.
20N = 59 kg * a /.16.
a = .054 m/s^2.
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If based on our results we define a Newton and the force required to accelerate a kilogram at 1 m/s^2,what is the change in your preceding result?
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20 N = 59 kg * a.
a = .34 m/s^2.
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`q004. For each of the following identify each given quantity as v0, vf, a, `dt, or `ds for the interval of uniform acceleration; if the motion is rotational identify instead the corresponding angular quantities omega_0, omega_f, alpha, `dt or `dTheta.
A ball is given a velocity of 30 cm/s at one end of a 60 cm ramp, and accelerates uniformly to the other end in 1.5 seconds.
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dt = 1.5 s.
ds = 60 cm.
v0 = 30cm/s.
(vf+v0)/2 = ds/dt.
2ds/dt -v0 = vf.
vf = 2(60)/1.5 - 30 = 50 cm/s.
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A ball accelerates, starting from rest, down a 60 cm ramp. Its velocity changes with respect to clock time at 40 cm/s^2.
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a = 40 cm/s^2
ds = 60cm.
v0= 0.
v^2 = v0^2 + 2a ds.
v = sqrt( v0^2 + 2a ds).
v= sqrt( 2* 40 * 60) = 69.28 m/s.
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A steel washer is given an upward velocity of 5 m/s at a height of 6 meters above the floor. It ends up on the floor. The acceleration of gravity is approximately 10 m/s^2 (more precisely it is 9.8 m/s^2, but we'll take a 2% 'hit' on our accuracy and use the more convenient number 10).
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No x direction.
In y direction:
s0 = 6m.
sf= 0m.
v0= 5m/s.
vf = 0 m/s.
a = -10m/s^2.
vf = v0 + a t.
(vf - v0)/a = t.
t = (0-5)/-10 = .5 s.
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A rotating strap slows from an angular velocity of 400 degrees / second to 100 degrees/second as it rotates through 1800 degrees.
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dTheta = 1800 degrees.
Omega_0 = 400 degrees/s.
Omega_F= 100 degrees/s.
(Omega_0 + Omega_F)/2 = dTheta/dt.
dt = 2*dTheta/(Omega_0 + Omega_F).
dt = 2*1800/(100+400).
dt = 36/5 = 7.2 s.
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A rotating wheel rotates through 1200 degrees in 15 seconds, starting with angular velocity 100 degrees / second.
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omega_0= 100 deg/s.
dt = 15 s.
dTheta = 1200 deg.
(omega_0 + omega_f)/2 = dTheta/dt.
omega_f = 2*dTheta/dt - omega_0.
omega_f = 60 m/s.
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A rotating sphere coasts to rest as it rotates through 40 pi radians in 20 seconds. You don't have to know what a radian is to answer this.
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dt = 20 s.
dTheta = 40pi radians.
omegaAve = dTheta/dt.
omegaAve = 40pi/20 = 2pi radians/s.
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`q005. For each of the situations in the preceding problem, identify which equation(s) of the four equations of uniformly accelerated motion will yield new information.
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solution in respective problem
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For each equation you identified, algebraically solve the equation for the unknown variable. Don't plug in any numbers, do the algebra with the symbols.
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solution in respective problem
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For each symbolic solution obtained in the preceding question, plug in the values of the given quantities to find the value of the unknown variable.
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solution in respective problem
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`q006. Give the data you obtained in class today, including an explanation of what was measured and how.
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N/A
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For each different system, construct a graph of average angular acceleration vs. average angular velocity. You may at this point use a spreadsheet to analyze your data. Explain how you obtained your accelerations, and describe your graphs.
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N/A
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Does the strap rotating on the threaded rod appear to have reasonably consistent angular acceleration? Is there any evidence that angular velocity has an influence on angular acceleration?
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N/A
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`q006. University Physics
Give your data for today's experiment, along with a brief description of what you did.
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We had a strap with a hole 5 cm from one end, 35 cm from the other. Through this hole a rotating axis (threaded rod) is placed.
A rubberband chain is attached to the 5 cm long arm. A magnet is attached to the end of the 35 cm long arm. Another magnet repels the attached magnet and stretches the rubberband chain.
Chain length, position of small (attached) magnet, position of large magnet.
37, 31, 39.
37, 30, 38.
37.2, 29, 37.
37.3, 28, 36.
37.4, 27.5, 35.
37.5, 27, 34.
37.7, 26.5, 33.
38, 25.5, 32.
38.2, 24.5, 31.
38.3, 24, 30.
38.5, 23.5, 29.
38.8, 22, 28.
39, 21.5, 27.
39.3, 21, 26.
39.5, 20.5, 25.
39.7, 19.5, 24.
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Estimate the uncertainties in your data.
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Chain length. +- .1 cm.
Distance between magnets. +- .3 cm
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Graph rubber band force vs. separation of magnets and compare the shape of this graph to the shape of the 'slope graph' obtained from your graph of coasting angular displacement vs. separation of magnets.
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on paper
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Give your best estimate of rubber band force vs. magnet proximity.
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If chain length = force-centimeters.
37, 31, 39. 37/(39-31) = 4.63 fc/cm.
37, 30, 38. 4.63 fc/cm.
37.2, 29, 37. 4.7 fc/cm.
37.3, 28, 36. 4.7 fc/cm.
37.4, 27.5, 35. 4.9 fc/cm.
37.5, 27, 34. 5.4 fc/cm.
37.7, 26.5, 33. 5.8 fc/cm.
38, 25.5, 32. 5.8 fc/cm.
38.2, 24.5, 31. 5.9 fc/cm.
38.3, 24, 30. 6.4 fc/cm.
38.5, 23.5, 29. 7 fc/cm.
38.8, 22, 28. 6.5 fc/cm
39, 21.5, 27. 7.1 fc/cm.
39.3, 21, 26. 7.9 fc/cm.
39.5, 20.5, 25. 8.8 fc/cm.
39.7, 19.5, 24. 8.8 fc/cm.
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Give your best estimate of magnet force vs. proximity. If you measured the distances at which the rubber band and the magnet acted on the system, use those distances. If not, you can assume for now that the ratio of the latter to the former is either 9 or 5, whichever you think is closer to the ratio for the system as you set it up.
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I'm not quite sure what this question is asking or exactly how it differs from the previous question about the rubber band force vs proximity.
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&#Good responses. See my notes and let me know if you have questions. &#