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The specified scale is 4 cm for 1 N.
You reported forces as great as about 11 Newtons.
However your rubber bands won't be exerting forces of these magnitudes.
Your rubber band calibrations included the following:
7.2cm,5.5cm,5.7cm,5.4cm,7.1cm,6.2cm, .19N
7.4cm, 5.9cm, 5.9cm, 5.6cm, 7.5cm, 6.2cm, .38N
7.6cm, 6.1cm, 6.4cm, 5.8cm, 7.8cm, 6.3cm, .76N
7.7cm,6.6cm,6.4cm, 5.9cm, 8.6cm, 6.6cm, 1.14N
7.9cm, 7.2cm, 6.7cm, 6.1cm,9.5cm, 6.7cm, 1.52N
A force of 11 N would likely break any of these rubber bands.
If you had a more likely force, say a force of 1.3 Newtons, then with a scale of 4 cm per 1 Newton that vector would have a length of
arrow length = 1.3 N * 4 cm / N = 5.2 cm.
A force of 11.23 N would be represented by an arrow about 45 cm long, much too long to sketch on a piece of ordinary paper.
I'll be glad to answer additional questions if you have them.
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Phy 201
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 **
Midpoint clocktime.
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5, .344, .512, .875
Midpoint clocktimes.
I divided .344 by 2= .172
I divided .512 by 2=.256+.172=.428
I divided .875 by 2=.438+.428= .866
Is this the correct way to calculate the midpoint clocktime for Angular Velocity lab.
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If
.344, .512, .875
are time intervals and not clock times, then your explanation and your results are correct.
If
.344, .512, .875
are clock times, then there would be only two intervals, and each midpoint would be halfway between the two clock times.
I suspect that the times you reported here are time intervals rather than clock times. Make sure you know the difference.
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