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    Attachment 524873524875Name:  image.jpg
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Size:  496.4 KBLooking for some help on question 5, not sure what to do exactly
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    (Original post by Purple K)
    Attachment 524873524875Name:  image.jpg
Views: 81
Size:  496.4 KBLooking for some help on question 5, not sure what to do exactly
    maybe use F=k \Delta L?? not too sure about the first one

    second one use the formula for strain which = \dfrac{\Delta L}{L}

    3rd obviously use formula for young modulus which is  E=\dfrac{FL}{A \Delta L}
    use values from question and the tension you worked out in part A
    then use formula for strain which is strain= \dfrac{\Delta L}{L}
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    Do you know the answers?
    I don't know if I'm correct but if I had to answer it in an exam I would get

    Tension in both = 49.05N

    Extension = 1.7mm

    Diameter = 8.57x10^-4 m

    But like I said, these are just guesses.
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    Hmm I haven't been able to look through it yet with my teacher but I got some different answers so I'm not entirely sure
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    (Original post by MHD30)
    Do you know the answers?
    I don't know if I'm correct but if I had to answer it in an exam I would get

    Tension in both = 49.05N

    Extension = 1.7mm

    Diameter = 8.57x10^-4 m

    But like I said, these are just guesses.

    I like your answers. One should be able to obtain an exact answer for the diameter if algebraic manipulation is used -  \sqrt{2} \times 0.60 mm.
 
 
 
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