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20 December, 02:49

A 2.0 kg hanging mass stretches a coiled spring by 0.15 m. The spring constant, k, is: (A) 0.075 N/m, (B) 2.9 N/m (C) 131 N/m, (D) 1,742 N/m, (E) none of the above.

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  1. 20 December, 03:15
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    C

    Explanation:

    Givens

    m = 2 kg

    F = 2 * 9.81

    F = 19.62 N

    x = 0.15 m

    Formula

    F = k*x

    Solution

    19.62 = k*0.15

    k = 19.62/0.15

    k = 130.8 which rounded to the nearest given answer is C
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