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DENIUS [597]
2 years ago
13

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.
SAT
1 answer:
Vitek1552 [10]2 years ago
6 0

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

<h3>Young Modulus of a substance</h3>

The young modulus of a substance is the<u> ratio of stress to the strain</u> 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

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