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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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Most popular questions from this chapter

An anchor, made of cast iron of bulk modulus \(60.0 \times 10^{9} \mathrm{Pa}\) and of volume \(0.230 \mathrm{m}^{3},\) is lowered over the side of the ship to the bottom of the harbor where the pressure is greater than sea level pressure by \(1.75 \times 10^{6} \mathrm{Pa} .\) Find the change in the volume of the anchor.
(a) What is the energy of a pendulum \((L=1.0 \mathrm{m}, m=\) $0.50 \mathrm{kg})\( oscillating with an amplitude of \)5.0 \mathrm{cm} ?$ (b) The pendulum's energy loss (due to damping) is replaced in a clock by allowing a \(2.0-\mathrm{kg}\) mass to drop \(1.0 \mathrm{m}\) in 1 week. What average percentage of the pendulum's energy is lost during one cycle?
A gibbon, hanging onto a horizontal tree branch with one arm, swings with a small amplitude. The gibbon's CM is 0.40 m from the branch and its rotational inertia divided by its mass is \(I / m=0.25 \mathrm{m}^{2} .\) Estimate the frequency of oscillation.
A 91 -kg man's thighbone has a relaxed length of \(0.50 \mathrm{m},\) a cross- sectional area of \(7.0 \times 10^{-4} \mathrm{m}^{2},\) and a Young's modulus of \(1.1 \times 10^{10} \mathrm{N} / \mathrm{m}^{2} .\) By how much does the thighbone compress when the man is standing on both feet?
The ratio of the tensile (or compressive) strength to the density of a material is a measure of how strong the material is "pound for pound." (a) Compare tendon (tensile strength \(80.0 \mathrm{MPa}\), density $1100 \mathrm{kg} / \mathrm{m}^{3}\( ) with steel (tensile strength \)\left.0.50 \mathrm{GPa}, \text { density } 7700 \mathrm{kg} / \mathrm{m}^{3}\right)$ which is stronger "pound for pound" under tension? (b) Compare bone (compressive strength \(160 \mathrm{MPa}\), density $1600 \mathrm{kg} / \mathrm{m}^{3}$ ) with concrete (compressive strength $\left.0.40 \mathrm{GPa}, \text { density } 2700 \mathrm{kg} / \mathrm{m}^{3}\right):$ which is stronger "pound for pound" under compression?
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