A rod is 2.40 m long and has a diameter of 2.50 mm.A force of 2000 N is applied to the end to stretch the rod by 1.40 mm.What is the tensile modulus for this rod? A) 6.99 $\times$ 10¹⁰ N/m² B) 6.05 $\times$ 10¹⁰ N/m² C) 5.54 $\times$ 10¹⁰ N/m² D) 4.90 $\times$ 10¹⁰ N/m² E) 4.02 $\times$ 10¹⁰ N/m²

Question

Quizplus · Accepted Answer

The tensile modulus (Young's modulus) is calculated using the formula: E = (F * L) / (A * ΔL), where F is the force applied, L is the original length, A is the cross-sectional area, and ΔL is the change in length. The cross-sectional area A = π * (d/2)^2, where d is the diameter. Substituting the given values, E = (2000 N * 2.40 m) / (π * (0.0025 m/2)^2 * 0.0014 m) = 6.99 * 10^10 N/m^2.

A Rod Is 2 ×\times× 1010 N/m2 B) 6 ×\times× 1010 N/m2 C) 5

A Rod Is 2 $\times$ 10¹⁰ N/m²
B) 6 $\times$ 10¹⁰ N/m²
C) 5