Exercise 11.27 A metal rod that is 5.00 m long and 0.49 cm2 in cross-sectional area...

Question

Question

Exercise 11.27

A metal rod that is 5.00 m long and 0.49 cm2 in cross-sectional area is found to stretch 0.13 cm under a tension of 4300 N .

Part A

What is Young's modulus for this metal?

Express your answer using two significant figures.

Y =

Pa

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Answer 1

Answer #1

Young's Modulus is given by

$Y =\frac{ \frac{T}{A}}{\frac{\Delta L}{L}}$

T = 4300 N

L = 5 m

$\Delta L =0.13 cm= 0.13*10^{-2}m$

$A = 0.49 cm^2 = 0.49*10^{-4}m^2$

$Y =\frac{ \frac{4300}{0.49*10^{-4}}}{\frac{0.13*10^{-2}}{5}}$

$Y =3.37*10^{11}N/m^2$

Answer 2

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