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How do you caculate the uncertainty and percentage uncertainty of a time which you have taken to be 26.34s?
Thanks for your help.

2007-05-16 03:57:56 · 2 answers · asked by Anonymous in Science & Mathematics Physics

2 answers

Unless you say otherwise, a measurement is assumed to be + or - 1 in the last significant digit.

So the absolute error would be 0.01s. The percent error would be .01 / 26.34.

Ultimately, the measurer is responsible for stating his accuracy. Just because you clicked your stopwatch and it reads 26.34 does not mean you are accurate to a hundredth of a second. Your thumb speed and various systematic errors in the experiment will bias and introduce randomness into the result that could be greater than that. So the answer is, it depends.

If you can measure the same quantity multiple times, you can get a statistical handle on the random errors. Systematic errors are harder to get hold of unless you can measure the same event using multiple methods.

2007-05-16 04:07:22 · answer #1 · answered by Anonymous · 1 1

Bekki is right of course. But I think it would be a good idea to always state your error assumptions. Do not assume your reader will know that you mean plus or minus 1 of the last significant digit. Even Ph.D.s forget the little things after a while. It serves well to remind them.

2007-05-16 11:43:45 · answer #2 · answered by oldprof 7 · 0 0

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