Acable is 100-m long and has a cross-sectional area of 1.0 mm2. a 1000-n force is applied to stretch the cable. young's modulus for the cable is 1.0 × 1011 n/m2. how far does the cable stretch?

1 m

Explanation:

L = 100 m

A = 1 mm^2 = 1 x 10^-6 m^2

Y = 1 x 10^11 N/m^2

F = 1000 N

Let the cable stretch be ΔL.

By the formula of Young's modulus

ΔL = 1 m

Thus, the cable stretches by 1 m.

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= initial pressure

= final pressure = 2

= initial volume

= final volume = /4

= initial temperature = 20 c = 20 + 273 = 293 k

= final temperature = ?

using the equation