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 elastic modulus allowed? A) 5.09 × 10¹¹ N/m² B) 1.27 × 10¹¹ N/m² C) 3.52 × 10⁹ N/m² D) 2.04 × 10⁷ N/m² E) 5.00 × 10⁶ N/m²

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Quizplus · Accepted Answer

The minimum elastic modulus (E) can be calculated using the formula for stress and strain, where stress = force / area and strain = Δlength / original length. The area of the cable's cross-section is πr^2 = π(0.0125 m)^2. The stress is 10,000 N / area, and the strain is 0.002 m / 50 m. Solving for E (E = stress / strain) gives a value close to 5.09 × 10^11 N/m^2.

A Cable Is 50