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7 December, 18:33

A load of 5N gives an extension of 0.65cm in a wire which obeys the hookes law. what is the extension caused by the load of 20N

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  1. 7 December, 21:15
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    2.6 cm

    Explanation:

    Hookes Law said

    F = c ∆x

    F = Force

    c = elasticity constant of the spring / wire

    ∆x = change in the length

    5 = c * 0.65

    c = 5/0.65

    if we change the Force to 20 then

    20 = c ∆x

    20 = (5/0.65) * ∆x

    ∆x = 20 * 0.65/5

    ∆x = 2.6 cm
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