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course Phy 241
9/11/12 11:30 pm
120905Text assignment: Read the text and work the problems as indicated. If you have questions, you're welcome to submit them using the Submit Question Form.
if you are using the Giancoli text read Chapter 1 and do Text Chapter 1 Problems 1-15.
If you are using the Openstax text read Chapter 1 and complete Text Chapter 1 Problems 1, 4, 7, 10, 12, 13, 14, 19.
Read through the document at self_critique_detail_in_responses . Then work through the qa exercises as described below:
qa exercises: You should work through the documents qa_02 and qa_03. The problems are sequenced in such a way that your understanding should build as you work through the set, but it will be important for you to either be sure you understand each problem in the sequence or self-critique in such a way that your instructor will have all the information needed to help you.
You should work each problem out, then read the given solution. You don't need to type in your solution to a given problem, provided your solution is either correct or your questions are covered in a good self-critique. If your solution doesn't cover everything in the given solution then you should self-critique according to instructions in the document self_critique_detail_in_responses.
Submit the following in the usual manner:
Class followup:
`q001. Explain how you obtained the best data possible for the experiment with the domino and the pendulum.
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What were the maximum height at which the domino hit the floor first, and the minimum height at which the pendulum hit the wall first?
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What was the frequency of your pendulum?
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How much time elapsed from release until the pendulum hit the wall?
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Calculate the acceleration based on this time interval and the maximum height at which the domino hit the floor first. Explain the details of your reasoning.
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Calculate the acceleration based on this time interval and the minimum height at which the pendulum hit the wall first.
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`q002. How far apart were the two most widely separated balancing points for the unloaded steel ramp as it balanced on a domino?
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How far from the balancing point of the unloaded steel ramp was the balancing point when you had the ramp loaded with a domino near one end?
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How far was it between the two most widely separated positions of the domino at which the system remained balanced?
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Based on your results, which do you think weighed more when the loaded ramp was balanced, the part of the ramp on the 'longer' side, or the part of the ramp plus the domino on the other?
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`q003. A ball rolls 58 cm down a ramp, starting from rest, and ends up moving at 30 cm / second. Assuming uniform acceleration, how long did it take to roll down the ramp, and what was its acceleration?
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time = 3.9 seconds.
Acceleration = 7.75 cm/s^2
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`q004. University Physics students: Give your data for the experiment you did today.
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Glass marble. Falling time = .4s. Diameter = 2.6cm. Ramp = 58.3 cm long.
(Dominoes, distance in cm).
(1, 6.7)
(2, 11.3)
(3, 15)
(4, 19.4)
(5, 23.3)
(6, 25.6)
Steel ball. Falling time = .4s. Diameter = 2.6cm. Ramp = 58.3 cm long.
(Dominoes, distance in cm).
(2, 13.6)
(4, 22)
(6, 28.7)
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Determine the acceleration of the ball on each ramp.
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Velocity at the end of ramp = some number x.
Final velocity = 0 when it hits the ground. Flight time (delta t) =.4seconds. Determining horizontal acceleration, not vertical (gravitational). Once the ball leaves the end of the ramp, the acceleration doesnt change.
(x+0)/2 = ds/dt. x = 2* ds/dt.
acceleration = change in velocity over change in time. a = dv/dt. a= x/.4.
Glass marble. Falling time = .4s. Diameter = 2.6cm. Ramp = 58.3 cm long.
(Dominoes, distance in cm).
(1, 6.7) x= 2*(6.7)/.4 = 33.5. a = 83.8.
(2, 11.3) x= 56.5. a = 141.3.
(3, 15) x= 75. a = 187.5.
(4, 19.4) x=97. a = 242.5
(5, 23.3) x= 116.5. a = 291.3.
(6, 25.6) x= 128. a = 320.
Steel ball. Falling time = .4s. Diameter = 2.6cm. Ramp = 58.3 cm long.
(Dominoes, distance in cm).
(2, 13.6) x= 68. a = 170.
(4, 22) x= 110. a = 275.
(6, 28.7) x= 143.5. a = 358.8.
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For the marble, sketch a graph of acceleration on the ramp vs. number of dominoes.
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For the ball, sketch a graph of acceleration on the ramp vs. number of dominoes.
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For the marble, at what average rate do you conclude that the ball's acceleration on the ramp changed with respect to the number of dominoes? Explain how you obtained your result.
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(1, 6.7) x= 2*(6.7)/.4 = 33.5. a = 83.8. 83.8/1. 83.8 cm/s^2 per domino.
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The ball reaches final velocity 33.5 cm/s on the ramp. The ramp is 58 cm long. So the ball requires almost 4 seconds to reach the end of the ramp. Its acceleration will then be around 10 cm/s^2. It looks line you might have used .4 seconds for the time to roll down the ramp.
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(2, 11.3) x= 56.5. a = 141.3. 141.3/2 = 70.6/domino
(3, 15) x= 75. a = 187.5. 187.5/3= 62.5/domino
(4, 19.4) x=97. a = 242.5 242.5/4= 60.6/domino
(5, 23.3) x= 116.5. a = 291.3. 291.3/5= 58.3/domino.
(6, 25.6) x= 128. a = 320. 320/6= 53.3/domino.
Average these together to get 64.9 cm/s^2 per domino.
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For the steel ball, at what average rate do you conclude that the ball's acceleration on the ramp changed with respect to the number of dominoes? Explain how you obtained your result.
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You wouldn't in any case divide the acceleration by the number of dominoes. The average rate definition is change in A / change in B, not A / B.
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(2, 13.6) x= 68. a = 170. 170/2 = 85.
(4, 22) x= 110. a = 275. 275/4= 68.6
(6, 28.7) x= 143.5. a = 358.8. 358.8/6= 59.8.
Average equals 71.1 cm/s^2 per domino.
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`q005. University Physics students: The definitions of velocity and acceleration lead to the following equations for an object accelerating uniformly during an interval:
`ds = (vf + v0) / 2 * `dt
vf = v0 + a `dt.
Eliminate vf from the two equations to get an equation for vf^2. Explain your steps and give the resulting equation.
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If you eliminate vf by plugging v0 + a dt into `ds = (vf + v0) / 2 * `dt
you get
2 ds = (2v0 + a dt) dt
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Solve for `ds to get the usual form
`ds = v0 `dt + 1/ 2 a `dt^2.
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Not quite sure what to do with this...
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Eliminate `dt from the two equations to get an equation for `ds. Explain your steps and give the resulting equation.
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ds = (vf^2-v0^2)/(2a)
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Verify that there is at least one equation with each of the following combinations of variables:
`ds, a, vf
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ds = (vf^2-v0^2)/(2a)
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v0, vf, `ds
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`ds = (vf + v0) / 2 * `dt
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a, `dt, v0
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vf = v0 + a `dt
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There are 10 possible combinations consisting of three of the variables v0, vf, `ds, `dt, a. List all ten:
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I'm assuming this means 10 different equations...
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We're talking about combinations of the five variables.
A few of the combinations would be
v0, vf, `ds
v0, vf, `dt
v0, vf, a
vf, `ds, `dt
There are six more.
There are only four equations, if you don't count all the different rearrangements of the same equation.
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`ds = (vf + v0) / 2 * `dt
ds = (vf^2-v0^2)/(2a)
vf = v0 + a `dt
ds = (2v0 + a dt)/2 * dt
a = (vf -v0)/dt
ds = (2vf - a dt)/2 * dt
dt = (vf- v0)/a
ds = (2v0 + a) dt
I'm not sure how to go about getting the rest, aside from just rearranging the current equations.
The dt is still throwing me off in places. My brain is hitting a road block.
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Is there any combination of three variables which does not appear in at least one of the equations?
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Doesn't look like it.
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Do all equations contain four of the five variables?
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It would seem this would be the case. If you try to eliminate anothe variable, complications ensue.
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You've got a couple of errors, but it won't take you long to fix them.
&#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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