Asked by melisa

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 = 3.51ยท108 N/m2.
Answered by Dominique
51.5(9.8)/(0.0101)3.14(3.51*108)
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