PHY 201
Your 'cq_1_01.1' report has been received. Scroll down through the document to see any comments I might have inserted, and my final comment at the end.
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The problem:
Here is the definition of rate of change of one quantity with respect to another:
The average rate of change of A with respect to B on an interval is
average rate of change of A with respect to B = (change in A) / (change in B)
Apply the above definition of average rate of change of A with respect to B to each of the following. Be sure to identify the quantity A, the quantity B and the requested average rate.
If the position of a ball rolling along a track changes from 10 cm to 20 cm while the clock time changes from 4 seconds to 9 seconds, what is the average rate of change of its position with respect to clock time during this interval?
answer/question/discussion: ->->->->->->->->->->->-> (start in the next line):
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If the velocity of a ball rolling along a track changes from 10 cm / second to 40 cm / second during an interval during which the clock time changes by 3 seconds, then what is the average rate of change of its velocity with respect to clock time during this interval?
answer/question/discussion: ->->->->->->->->->->->-> (start in the next line):
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If the average rate at which position changes with respect to clock time is 5 cm / second, and if the clock time changes by 10 seconds, by how much does the position change?
answer/question/discussion: ->->->->->->->->->->->-> (start in the next line):
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You will be expected hereafter to know and apply, in a variety of contexts, the definition given in this question. You need to know this definition word for word. If you try to apply the definition without using all the words it is going to cost you time and it will very likely diminish your performance. Briefly explain how you will ensure that you remember this definition.
answer/question/discussion: ->->->->->->->->->->->-> (start in the next line):
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You are asked in this exercise to apply the definition, and given a general procedure for doing so. Briefly outline the procedure for applying this definition, and briefly explain how you will remember to apply this procedure.
answer/question/discussion: ->->->->->->->->->->->-> (start in the next line):
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13 minutes
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2.01
1.92
1.95
1.92
1.69
2.14
2.31
2.28
2.00
1.97
10 trials for 1-domino setups
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1.434
1.406
1.453
1.215
1.184
1.328
1.359
1.590
1.402
1.543
10 trials for 2-domino setups
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1.090
1.230
1.000
.984
.906
1.016
1.172
1.109
.980
1.109
10 trials for 3-domino setups
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1.898,.1219
2.14, .1557
1.338, .1284
1.444, .1157
1.042, .1238
1.077, .07779
I followed the directions and gave mean and standard deviation for 1 domino (first 5 and last 5 trials), 2 dominoes (first 5 and last 5 trials), and 3 dominoes (first 5 and last 5 trials).
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28 cm/1.898 sec= 14.752 cm/sec
28 cm/2.14 sec = 13.084 cm/sec
28 cm/1.338 sec = 20.927 cm/sec
28 cm/1.444 sec = 19.391 cm/sec
28 cm/1.042 sec = 26.871 cm/sec
28 cm/1.077 sec = 25.998 cm/sec
I took the distance and divided by the average of the times for the trials or mean.
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28 cm/1.898 sec= 14.752 cm/sec
28 cm/2.14 sec = 13.084 cm/sec
28 cm/1.338 sec = 20.927 cm/sec
28 cm/1.444 sec = 19.391 cm/sec
28 cm/1.042 sec = 26.871 cm/sec
28 cm/1.077 sec = 25.998 cm/sec
I took the distance and divided by the average of the times for the trials or mean.
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13.918 cm/sec
20.159 cm/sec
26.435 cm/sec
I added the right and left velocities and then divided by 2.
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13.868 cm/sec
20.129 cm/sec
26.415 cm/sec
I used the following: 2.019 sec -- 28 cm/2.019 sec, 1.391 sec -- 28 cm/1.391 sec, 1.060 sec -- 28 cm/1.060 sec. I think this may show a range of possible error in the calculations.
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They are close but not exactly the same. No, I honestly cannot explain this.
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.03cm/28 cm, 13.868 cm/sec
.06cm/28 cm, 20.129 cm/sec
.09cm/28 cm, 26.415 cm/sec
.001, .002, .003 I divided out the slope. As the slope was slightly adjusted upwards, the velocity of the marble definitely increased.
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-.0013
6.3
-.0013,0
0, 6.3
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I'm not sure how to figure this out and be accurate with the results.
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I didn't use .10. I used (-.0013,0) and (.003, 26.41).
run .0043
rise 26.415
6143
I think I'm not correct in what I'm doing but I'm trying and spending a lot of time on this.
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1.898, 2.14, 2.019
1.338, 1.444, 1.391
1.042, 1.077, 1.060
This just shows as more dominoes are used to make the incline higher, the time decreases.
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2.019 +-.14= 2.019-.14= 1.879 sec to 2.019 +.14= 2.159 sec
These numbers indicate the range of the answer and helps to show possible error.
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28 cm/1.879 sec, 14.902 cm/sec
28 cm/2.159 sec, 12.969 cm/sec
12.969 cm/sec, 14.902 cm/sec
This reflects the range of velocity.
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1.391 +-.12 = 1.391-.12= 1.271 to 1.391 + .12= 1.511
28 cm/1.271 sec., 22.023 cm/sec
28 cm/1.511 sec., 18.531 cm/sec
18.531 cm/sec, 22.023 cm/sec
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1.060 +-.10 = 1.050 to 1.070
28 cm/1.050 sec, 26.667 cm/sec
28 cm/1.070 sec, 26.168 cm/sec
26.168 cm/sec - 26.667 cm/sec
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.03, 12.969 cm/s
.03, 14.902 cm/s
.06, 18.531 cm/s
.06, 22.023 cm/s
.09, 26.168 cm/s
.09, 26.667 cm/s
As slope increases so does velocity and the rate of acceleration.
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Since I didn't do the first graph right...or at least I don't think so, then this will not work.
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I may try to do this later if time permits and I can see what I've done wrong.
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ok
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Honestly, over 3 hours and I didn't even get everything done.
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You need to submit your lab using the form for the lab. I don't see any apparent problem with your data or any of your answers, but if you don't use the form I don't always know what question matches your answers, so I could be missing something.
Your form for the cq problem appears to have been blank.