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Hoochie [10]
3 years ago
8

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.
Physics
1 answer:
ira [324]3 years ago
6 0

Answer:

\Delta L=15\,mm

Explanation:

Given:

  • length of a steel-string, L=1m
  • area of the string, A=0.5\,mm^2
  • Young's modulus of the steel, Y=2\times 10^{11} Pa
  • force of tension on the string, F=1500\,N

We have the relation for change in length:

\Delta L=\frac{F.L}{A.Y}

\Delta L=\frac{1500\times 1000}{0.5\times 10^{-6}\times 2\times 10^{11}}

\Delta L=0.015m

\Delta L=15\,mm

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Answer:

150000000

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Explanation:

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Force is given by

F_e=Eq\\\Rightarrow F_e=100\times 24\times 10^{-12}\\\Rightarrow F_e=2.4\times 10^{-9}\ N

The ratio is

\dfrac{F_e}{F_g}=\dfrac{2.4\times 10^{-9}}{0.12\times 10^{-3}\times 9.81}\\\Rightarrow \dfrac{F_e}{F_g}=0.00000203873598369

The ratio is \dfrac{F_e}{F_g}=0.00000203873598369

Balancing the forces we get

Eq=mg\\\Rightarrow E=\dfrac{mg}{q}\\\Rightarrow E=\dfrac{0.12\times 10^{-3}\times 9.81}{24\times 10^{-12}}\\\Rightarrow E=49050000\ N/C

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