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Phy 121
Your 'question form' report has been received. Scroll down through the document to see any comments I might have inserted, and my final comment at the end.
** Question Form_labelMessages **
Rubber BAnd
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If you haven't already done so, include in your report a table of your data for force vs. length for each of the four selected rubber bands.
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I am not sure what to graph I have these data:
7 cm, 7 cm, 6.9 cm, 8 cm, 7 cm, 8 cm-0.19 N
7.20cm,7.20cm,7.50cm,7.50cm,7.50cm,7.50cm- 0.38 N
7.75cm,7.50cm,7.75cm,7.70cm,7.6cm,7.75cm- 0.57 N
7.5cm, 7.5cm, 7.7cm, 7.7cm, 7.8cm, 7.8cm- 0.76 N
7.7cm, 7.7cm,7.8cm,7.9cm, 7.9cm, 7.9cm-0.95 N
8.0 cm,7.9cm ,8.2cm ,8.0 cm, 8.0 cm,8.0 cm-1.14 N
I am not sure what I get my points for my graph. Wouldn't it be (7 cm, 0.19 N) ( 7 cm, 0.19N)...and so on? Sorry I am confused!
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You should do a force vs. length graph for each rubber band.
For the first rubber band the ponts would be (7 cm, .19 N), (7.2 cm, .38 N), (7.75 cm, .57 N) etc..
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#$&*
Phy 121
Your 'question form' report has been received. Scroll down through the document to see any comments I might have inserted, and my final comment at the end.
** Question Form_labelMessages **
Further Analysis
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f the time down the ramp is equal to that of the left-hand boundary of the interval you just sketched, then what would be the average velocity and the acceleration of the ball? Report in comma-delimited format on the first line below.
• Find the same quantities for the right-hand boundary of your interval, and report in similar format on the second line.
• In the third line report the resulting minimum and maximum possible values of acceleration on this interval, using comma-delimited format. Your results will just be a repeat of the results you just obtained.
• Starting on the fourth line, explain what your numbers represent and why it is likely that the actual acceleration of the ball on a 1-domino ramp, if set up carefully so that right-left symmetry is assured, would be between the two results you have given.
------>>>>>> 1-dom vel and accel left boundary of interval, rt boundary, min and max possible accel
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This is what I have so far, I am struggling! This is a VERY long lab!!!
'dt=1.447 s 'ds=28 cm v0=0
28cm/1.447s= 19.35 cm/s
19.35cm/s/1.447 s= 13. 37 cm/s^2
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This calculation appears to be average velocity / time interval, which is not a meaningful calculation. It does not give you the acceleration.
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This should be easy to correct.
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I am unsure of what to do next...
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You have an interval that indicates the maximum and minimum time likely time intervals. You don't say which you are using for this specific calculation.
You would want to calculate the average velocity and acceleration based on the minimum time interval, and on the maximum. You will get maximum and minimum accelerations, not necessarily in that order.
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Your calculation of acceleration isn't quite right, but can be easily fixed.
Presumably before this point you determined the maximum and minimum likely time intervals. You would use these to calculate the maximum and minimum likely acceleration.
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I'll be glad to answer additional questions, if you have more.
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