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3 years ago

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Consider a steel guitar string of initial length L=1.00 meter and cross-sectional area A=0.500 square millimeters. The Young's m

odulus of the steel is Y=2.0×1011 pascals. How far ( ΔL) would such a string stretch under a tension of 1500 newtons? Use two significant figures in your answer. Express your answer in millimeters.

2 answers:

Reil [10]3 years ago

6 0

Answer:

$\Delta l=0.015m$

Explanation:

We have given initial length of the steel guitar l = 1 m

Cross sectional area $A=0.5mm^2=0.5\times 10^{-6}m^2$

Young's modulus $\gamma=2\times 10^{11}Pa$

Force F = 1500 N

So stress $=\frac{force}{area}=\frac{1500}{0.5\times 10^{-6}}=3000\times 10^{-6}=3\times 10^{9}Pa$

We know that young's modulus $=\frac{stress}{strain}$

So $2\times 10^{11}=\frac{3\times 10^{9}}{strain}$

$strain=1.5\times 10^{-2}=0.015m$

Now strain $=\frac{\Delta l}{l}$

$0.015=\frac{\Delta l}{1}$

$\Delta l=0.015m$

galben [10]3 years ago

4 0

Answer:

Explanation:

L = 1 m

A = 0.5 mm² = 0.5 x 10^-6 m²

Y = 2 x 10^11 Pa

F = 1500 N

ΔL = ?

Use the formula for the young's modulus

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

$\Delta L = \frac{FL}{AY}$

$\Delta L = \frac{1500\times 1}{0.5\times10^{-6}\times 2\times 10^{11}}$

ΔL = 0.015 m

ΔL = 0.02 m

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