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Percent Uncertainty homework

  1. Aug 23, 2009 #1
    1. The problem statement, all variables and given/known data

    What, roughly, is the percent uncertainty in the volume of a spherical beach ball whose radius is r = 0.84 + or - 0.06m?
    Answer must have 2 sig figs.

    2. Relevant equations



    3. The attempt at a solution
    I found the volume of the sphere with .84 radius which would be 3.0 m^3 and then I dont know what to do next.
     
  2. jcsd
  3. Aug 23, 2009 #2

    kuruman

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    If y is related to x by a power law,

    y = axn, then the fractional uncertainty is

    [tex]\frac{\Delta y}{y}=n\frac{\Delta x}{x}[/tex]
     
  4. Aug 23, 2009 #3
    Consider what happens to the uncertainty in a variable when you cube the variable (should be a standard treatment in the manipulation of uncertainties in variables)
     
  5. Aug 23, 2009 #4
    So are you saying I should cube the .06 and then divide that by the 3.0m^3?
     
  6. Aug 23, 2009 #5
    Could you show your equations and attempts, please?
     
  7. Aug 23, 2009 #6
    (.84)^3(pi)(4/3) = 2.5 m^3

    .06^3 = 2.2x10^-4

    2.2x10^-4 / 2.5 = 8.6x10^5

    (8.6x10^5) x 100% = .0086%
     
  8. Aug 23, 2009 #7

    kuruman

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    Not how it works. Here, n = 3. Therefore, the fractional uncertainty is

    [tex]\frac{\Delta V}{V} = 3\times\frac{0.06}{0.84}[/tex]

    Multiply by 100 and you have percent uncertainty.
     
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