A wire is 1.50 m long and has a diameter of 1.50 mm.The tensile modulus of the wire is 6.20 $\times$ 10¹⁰ N/m².If a force of 400 N is applied to end of the wire,then the increase in length of the wire is A) 5.48 mm. B) 4.28 mm. C) 3.95 mm. D) 3.84 mm. E) 3.45 mm.

Question

Quizplus · Accepted Answer

The correct answer is A.The increase in length of the wire can be calculated using the formula:$$\Delta L = \frac{FL}{AE}$$where:* $\Delta L$ is the increase in length* $F$ is the force applied* $L$ is the original length of the wire* $A$ is the cross-sectional area of the wire* $E$ is the tensile modulus of the wireSubstituting the given values into the formula, we get:$$\Delta L = \frac{(400 N)(1.50 m)}{\pi (0.00075 m)^2 (6.20 * 10^{10} N/m^2)} = 5.48 mm$$Therefore, the increase in length of the wire is 5.48 mm.

A Wire Is 1 ×\times× 1010 N/m2

A Wire Is 1 $\times$ 10¹⁰ N/m²