A wire is 1.50 m long and has a diameter of 1.50 mm. The elastic modulus of the wire is 6.20 × 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 because the increase in length of the wire is given by 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 elastic modulus of the wireSubstituting the given values into the formula, we get:$$\Delta L = \frac{(400 N)(1.50 m)}{\pi (0.00075 m)^2 (6.20 	imes 10^{10} N/m^2)} = 5.48 mm$$

A Wire Is 1