Anonymous
Anonymous asked in Science & MathematicsPhysics · 10 months ago

If R = 2.400 ± 0.003, what is the relative uncertainty for R−2? Plz also provide explanation/solution.?

Update:

srry its R^-2

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  • 10 months ago
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    There is a rule of thumb method and there is formal analysis.

    The rule of thumb for small errors and for multiplication and division means that you ADD THE FRACTIONAL ERROR for each multiplication.

    Which means that for raising to a power you multiply the fractional error by the absolute value of the power to get the fractional error in the result.

    ie the fractional error would be approximately ( 0.003/2.4)*|(-2)| = 2 * 0.003/2.4 = 0.0025

    Formally you calculate the lowest result and the largest possible result.

    This gives the absolute error which you could THEN turn into a fraction.

    ie ( 2.4 + 0.003)^(-2) < R^(-2) < (2.4 - 0.003)^(-2)

    0.173178 < R ^(-2) < 0.174406

    The EXPECTATION VALUE is 2.4 ^ (-2)= 0.173611

    Now put the differences as a fraction of the expectation.

    (0.1736111- 0.1731779) / 0.1736111 = 0.0025 ( just as we got for the first method )

    The second method is more complex but it is the only TRUE method for finding errors.

    The rule of thumb works fine if the errors are small compared to the target and if the formula is a simple one that contains only one basic process.

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  • NCS
    Lv 7
    10 months ago

    If you raise a value to the nth power, you multiply the uncertainty by n.

    For this problem, n = -2

    and so the relative uncertainty is ±0.006

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  • Anonymous
    10 months ago

    First tell me what the relative uncertainty is for me getting with your mother

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  • 10 months ago

    the same, ± 0.003

    subtraction or addition, the error is by decimal place. Assuming 2 is an exact number, then the error does not change.

    of course if 2 is only accurate to 2±1, then the uncertainty is ±1

    edit, re correction.

    best method is to calculate 1/2.403² and 1/2.397²

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