Precision Question

  • Thread starter quicknote
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  • #1
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I am measuring the position of photogates in a free fall lab experiment.

The question is:
What is the error in s in terms of reading error? Now add in the error in the metal scale, a precision of one part in 4000, in quadrature to give an expression for the total error in s.

My reading error is +/- 0.02cm.
I'm confused about 'adding in the error in the metal scale, 1 part in 4000'.
Does this mean 1/4000 = 0.00025?

If so then would my total error would be [tex] \sqrt{0.00025^2 + 0.02^2} [/tex] =0.02 ... I don't think this makes much sense :confused:


Any help would be greatly appreciated.
Thanks!
 

Answers and Replies

  • #2
andrevdh
Homework Helper
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One calculates an error in a value by taking into account the errors in the variables that the particular value depends on. So my question is do the time measurement depend on the measurement on the metal scale? If so what are the smallest divisions on the metal scale? It is most likely that the error in this variable is so small then that it has virtually no contribution to the error in the time measurement anyway. If the time value do not depend on the measurement on the metal scale it could not have any effect on the error in the time value and should not be included in the error calculation.
 

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