Question

A force of 5000n is applied outwardly to each end of a 5. 0-m long rod with a radius of 34. 0 cm and a young's modulus of 125 × 108 n/m2. The elongation of the rod is: phy101.

Answer 1

The elongation of the rod is $5.51 \times 10^{-4}m$`

Young Modulus of a substance

The young modulus of a substance is the ratio of stress to the strain of material.

$Modulus = \frac{Stress}{Strain}\\ Modulus =\frac{Fl}{Ae}$

where:

• F is the applied force = 5000N
• l is the length of the rod = 5.0m

• A is the cross-sectional area = $\pi r^2 = 0.34^2 * 3.14 =0.362984m^2$

Required

Extension of the rod "e"

Substitute the given parameters into the formula to have;

$e=\frac{Fl}{A \gamma} \\e=\frac{5000 \times 5}{0.362984 \times 125,00,000}\\e=\frac{25000}{45,373,000}\\e= 5.51 \times 10^{-4}m$

Hence the elongation of the rod is $5.51 \times 10^{-4}m$`

Learn more on young's modulus here: brainly.com/question/6864866