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30 April, 13:00

A force of 5000 n is applied outwardly to each end of a 5.0-m long rod with a radius of 34.0 mm and a young's modulus of 125 x 108 n/m2. the elongation of the rod is:

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  1. 30 April, 13:12
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    Por definicion tenemos que

    (F/A) = E (∆/0)

    Sustituyendo los valores tenemos y despejando ∆:

    ∆ = (F / (πr2 * E)) * 0

    (5000*5) / (3.14 * (34*10^-2) ^2 * (125*10^8))

    5.5*10^-6 m
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