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Physics AS Specimen Paper. Wtf do I do. This is scaring tf out of me. watch

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    A student has a diffraction grating that is marked 3.5 * 10^3 lines per m.
    Calculate the percentage uncertainty in the number of lines per metre suggested by this marking.

    FREAKING WOT M8. Okay but on a serious note.. how do I know what the uncertainity. I feel like it has something to do with number of sig figs they give the value to. Pls help.!
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    (100/3.5*10^3)*100
    =2.9%
    this should be it, check the mark scheme
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    (Original post by cleggj88)
    (100/3.5*10^3)*100
    =2.9%
    this should be it, check the mark scheme
    Could you explain it? Why is the uncertainity 100?
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    (Original post by Someboady)
    Could you explain it? Why is the uncertainity 100?
    It rounds to 3.5*10^3, so the true value lies in the interval 3.45*10^3<x<3.55*10^3. Therefore the uncertainty is +-50, =100 total
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    (Original post by samb1234)
    It rounds to 3.5*10^3, so the true value lies in the interval 3.45*10^3<x<3.55*10^3. Therefore the uncertainty is +-50, =100 total
    Wait so isnt the uncertainty +- 0.05 = 0.1? :/
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    (Original post by Someboady)
    Wait so isnt the uncertainty +- 0.05 = 0.1? :/
    +-0.05*10^3 =50
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    (Original post by samb1234)
    +-0.05*10^3 =50
    All this for one mark? *cries*
    Thanks a bunch!
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    (Original post by Someboady)
    All this for one mark? *cries*
    Thanks a bunch!
    no problem, let me know if you have any other issues
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    (Original post by samb1234)
    It rounds to 3.5*10^3, so the true value lies in the interval 3.45*10^3<x<3.55*10^3. Therefore the uncertainty is +-50, =100 total
    This is right
 
 
 
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