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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

Any help would be greatly appreciated.

Thanks!