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    Can anyone please help me with the question 2(d) ? The marking schemes don't help with this. I studied without the help of any teachers and i don't understand this get myself. Please help..
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    (Original post by Zainabr17)
    Can anyone please help me with the question 2(d) ? The marking schemes don't help with this. I studied without the help of any teachers and i don't understand this get myself. Please help..
    Which exam board and specification?
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    (Original post by 16Characters....)
    Which exam board and specification?
    http://www.acethem.com/pastpapers/a-...-5-15323.html/

    I need to know how the error in resistivity is found. The explaination. Even the marking schemes give the answers but i dont get it.
    Thank you so much.
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    Explanation* -_-
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    (Original post by Zainabr17)
    http://www.acethem.com/pastpapers/a-...-5-15323.html/

    I need to know how the error in resistivity is found. The explaination. Even the marking schemes give the answers but i dont get it.
    Thank you so much.
    I'll use a different example. Suppose I was trying to measure g, and I had a value of 9.8 with an error of 0.01. Then I could write that g = 9.8 +/- 0.01. If I then multiplied that by 2, I'd have 2g = 19.6 +/- 0.02. So the error is now 0.02, i.e. it's doubled.

    In general, if we multiply a value by a constant, the error is also multiplied. Hence you can use this to calculate the error in the resistivity from the error in the gradient.
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    (Original post by 16Characters....)
    I'll use a different example. Suppose I was trying to measure g, and I had a value of 9.8 with an error of 0.01. Then I could write that g = 9.8 +/- 0.01. If I then multiplied that by 2, I'd have 2g = 19.6 +/- 0.02. So the error is now 0.02, i.e. it's doubled.

    In general, if we multiply a value by a constant, the error is also multiplied. Hence you can use this to calculate the error in the resistivity from the error in the gradient.
    Thank you!
 
 
 
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