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# Elasticity

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A phone line 2.0 m in diameter and 0.5 km long is stretched 0.5 m by a force of 4000 N. Determine:
1. the stress on the wire
2. the strain on the wire
3.Young's modulus for the cable material.

https://brainmass.com/physics/classical-mechanics/solving-an-elasticity-problem-38321

## SOLUTION This solution is FREE courtesy of BrainMass!

d=diameter of line= 2 m
Therefore cross sectional Area of line =pi d^2/4= 3.1416 m^2

1. the stress on the wire
Force= 4000 N
Stress= Force/Area= 1273.2366 N/m^2 =4000/ 3.1416

2. the strain on the wire

strain = elongation / original length
elongation= 0.5 m
length = 0.5 km= 500 m

Strain= 0.001 =0.5/ 500

3.Young's modulus for the cable material.

Young's modulus= Stress /Strain= 1,273,237 N/m^2 (or Pascal Pa)
This can also be written as 1.27 Mpa

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