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course phy201
9/23 7pm
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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30m/5m/s = 6s
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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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60newton/m / 10m = 6 newtons
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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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40n/m * 2 m = 80 newtons
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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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200m/s^2 / 50m/s = 4 seconds
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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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30m / 5m/s = Width is 6 seconds. Altitudes are 4.17 and 5.83 ......... I think this is how to figure out this problem, really got stuck on it.
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The area and average 'graph altitude' can be used to find the width of the trapezoid, which in this case is 30 meters / (5 meters / second) = 6 seconds.
The width is the same as the 'run'. A 'run' of 6 seconds and a slope of 10 meters / second^2 implies rise = slope * run = 6 seconds * 10 meters / second^2 = 60 meters / second.
Half the rise occurs on each side of the middle so the altitudes are
5 m/s - 30 m/s = -25 m/s
and
5 m/s + 30 m/s = 35 m/s.
The object changes direction, starting in the negative direction with positive acceleration. That results in the object first slowing to rest, then accelerating from rest in the positive direction, ending up +30 meters from where it started.
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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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I Don't really understand the previous question. I think the answers would be vave = 5m/s, dt= 6seconds,
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Right. From that you can proceed to reason out the rest.
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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.
****v0,vf, a
v0,vf, dt
v0,vf,ds
v0,dt,a
v0,dt,ds
v0,a,ds
vf,dt,a
vf,dt,ds
vf,a,ds
dt,a,ds
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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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Add the two altitudes together and divide it by 2, then multiply it by its with........ (y2+y1)/2 * Width = Area slope = subtract the altitudes to get the rise and the width would be the run (y2-y1)/width
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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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vf = v0 + a *dt area = ((right altitude + left altitude) / 2) * width
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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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Because area is not one of the 5 quantities that we have been useing in our formulas.
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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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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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`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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10.4 cm/s^2........... I don't know how to do this or the next couple of questions. I have my notes and I can't relate with the info given.
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10.4 cm/s^2 is good. You don't say how you got it, but one way, very likely equivalent to what you did, is as follows:
4/10 as many dominoes would be easier to accelerate, and all other things bein equal would result in 10/4 as much acceleration.
10 clips would provide 10/6 as much acceleration, all else being equal.
Combining the two effects we would get
10/4 * 10/6 = 100 / 24 = 25/6 as much acceleration.
25/6 * 2.5 cm/s^2 = 62.5 / 6 cm/s^2 = 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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The acceleration should stay the same.
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If you double the number of dominoes and halve the number of paperclips, what happens to the acceleration?
****Acceleration should only be about one quarter of the orginal acceleration
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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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Acceleration should double
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It doubles for each of these actions, so it quadruples.
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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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9.6 clips and 1 dominoe
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I agree. You've got this.
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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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8o cm/s ????????????????
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2 dominoes have mass 32 grams.
A force of 4 milliNewtons accelerates 32 grams at 4 cm/s^2.
A 1 kg weight is about 30 times the mass of the 2 dominoes.
A 1 Newton force is about 250 times the weight of a paperclip.
So the question can be reduced to finding the acceleration of 30 times the mass, subject to 250 times the force.
You would end up with 250 / 30 = 8 times the acceleration, roughly.
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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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????????????????
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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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????????????????
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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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v0 = 30cm/s
ds = 60cm
dt = dt
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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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ds = 60cm
v0 = 0
Accel = 40cm/s^2
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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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v0 = 5m/s
ds = 6m
Accel = 10m/s^2
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Initial velocity is in the direction opposite the displacement and the acceleration, so would have sign opposite to the sign of these quantities.
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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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Omega 0 = 400d/s
Omega f = 100d/s
d theta = 1800 degrees
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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 = 100d/s
dt = 15s
dtheta = d theta 1200 degrees
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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 = 20s
ds = 40 pi radian
vf = 0
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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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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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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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The equations of motion are
`ds = (vf + v0) / 2 * `dt
vf = v0 + a `dt
`ds = v0 `dt + 1/2 a `dt^2
vf^2 = v0^2 + 2 a `ds.
For angular motion v becomes omega, `ds becomes `dTheta and a becomes alpha.
For your first situation you've identified
v0 = 30cm/s
ds = 60cm
dt = dt
Which equation(s) contain v0, `ds, and `dt?
For each of your selected equation(s), which variable don't you know? Solve each equation for the unknown variable, and plug in
v0 = 30cm/s
ds = 60cm
dt = dt
Follow the same procedure for each of the above situations.
By the time you're done you'll be in very good shape with the equations of motion.
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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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In class we measured the degrees of change , along with clock times, for a rotating strap. I used a rubberband to launch our strap.
SECONDS Dergees of Change
3.5 900
4.3 1080
2.3 450
2.9 720
5.1 1350
4.7 900
6.8 1530
12.2 3030
12.9 4320
9.1 1800
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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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I divided the dergee of change by the change in time to get there average angular velocity. Since the initial velocity was 0, I doubled the average velocity and divided it by the by the change in time to get the average angular acceleration. My graph was scattered, but did show a rising trend.
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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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Yes it shows a trend
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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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Estimate the uncertainties in your data.
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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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Give your best estimate of rubber band force vs. magnet proximity.
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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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Good application of common sense to mysterious situations. You got a lot of stuff you didn't think you had.
You missed or didn't have time to complete question 5. You'll want to revise that one, and having done so you'll pretty much have a lock on uniformly accelerated motion.
The other questions I think you've got, so only revise other questions if you need my feedback.
&#Please see my notes and submit a copy of this document with revisions, comments 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. &#
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