A cable is 50.0 m long and has a diameter of 2.50 cm.A force of 10,000 N is applied to the end of the cable.If the maximum stretch allowed in the cable is 2.00 mm,then what is the minimum tensile modulus allowed? A) 1.02 $\times$ 10⁹ N/m² B) 2.04 $\times$ 10⁹ N/m² C) 2.55 $\times$ 10⁹ N/m² D) 3.06 $\times$ 10⁹ N/m² E) 3.52 $\times$ 10⁹ N/m²

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Quizplus · Accepted Answer

The minimum tensile modulus allowed is 1.02 * 10^9 N/m^2. This is because the maximum stretch allowed in the cable is 2.00 mm, and the force applied to the end of the cable is 10,000 N. The tensile modulus is defined as the ratio of stress to strain, so the minimum tensile modulus allowed is:Tensile modulus = Stress / Strain= (Force / Area) / (Change in length / Original length)= (10,000 N / (π * (2.50 cm / 2)^2)) / (2.00 mm / 50.0 m)= 1.02 * 10^9 N/m^2

A Cable Is 50 ×\times× 109 N/m2 B) 2 ×\times× 109 N/m2 C) 2

A Cable Is 50 $\times$ 10⁹ N/m²
B) 2 $\times$ 10⁹ N/m²
C) 2