physics

Find the minimum diameter of a 51.5-m-long nylon string that will stretch no more than 1.01 cm when a load of 68.9 kg is suspended from its lower end. Assume that Ynylon = 3.51·108 N/m2

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asked by maria
  1. Hookes Law:

    elongation= force/Y * length/area

    .00101=68.9*9.8/3.51E8 * 51.5/area

    solve for area, then solve for diameter

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