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Chapter 10: Problem 3

A brass wire with Young's modulus of \(9.2 \times 10^{10} \mathrm{Pa}\) is $2.0 \mathrm{m}\( long and has a cross-sectional area of \)5.0 \mathrm{mm}^{2} .$ If a weight of \(5.0 \mathrm{kN}\) is hung from the wire, by how much does it stretch?

Short Answer

Expert verified
The Young's modulus of brass is \(9.2 \times 10^{10}\ \mathrm{Pa}\). Answer: The brass wire stretches by approximately \(1.09\ \mathrm{mm}\) when the weight is hung from it.

Step by step solution

01

Write down the given values

We are given the following values: - Young's modulus (Y): \(9.2 \times 10^{10} \mathrm{Pa}\) - Length of the wire (L): \(2.0 \mathrm{m}\) - Cross-sectional area (A): \(5.0 \mathrm{mm}^2\) - Weight (W): \(5.0 \mathrm{kN}\)
02

Write down the formula for Young's modulus

The formula for Young's modulus is: \(Y = \frac{stress}{strain} = \frac{F/A}{\Delta L/L}\), where - Y is the Young's modulus - F is the force applied on the wire - A is the cross-sectional area - \(\Delta L\) is the change in length (stretch) of the wire - L is the original length of the wire We will rearrange this formula to solve for the stretch (\(\Delta L\)).
03

Convert cross-sectional area to square meters and weight to Newtons

The cross-sectional area is given in square millimeters, and we need to convert it to square meters: \(A = 5.0 \times 10^{-6} \mathrm{m}^2\) The weight is given in kilo Newtons, and we need to convert it to Newtons: \(W = 5.0 \times 10^3 \mathrm{N}\)
04

Rearrange the formula to solve for the stretch (\(\Delta L\))

From the formula in step 2, we can rearrange it to solve for the stretch \(\Delta L\): \(\Delta L = \frac{F \times L}{Y \times A}\)
05

Plug in the given values and solve for the stretch

Now we can plug in the given values and solve for the stretch: \(\Delta L = \frac{(5.0 \times 10^{3} \mathrm{N}) \times (2.0 \mathrm{m})}{(9.2 \times 10^{10} \mathrm{Pa}) \times (5.0 \times 10^{-6} \mathrm{m}^2)}\) Calculate the stretch: \(\Delta L \approx 1.09 \times 10^{-3} \mathrm{m}\)
06

Convert the stretch to millimeters

To express the stretch in millimeters, we can convert the value of \(\Delta L\) from meters to millimeters: \(\Delta L \approx 1.09 \times 10^{-3} \mathrm{m} \times \frac{1000\ \mathrm{mm}}{1\ \mathrm{m}} \approx 1.09 \mathrm{mm}\) Thus, the brass wire stretches by approximately \(1.09\ \mathrm{mm}\) when the weight is hung from it.

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