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Norma-Jean [14]
3 years ago
12

An aluminum cylinder with a radius of 2.7 cm and a height of 67 cm is used as one leg of a workbench. The workbench pushes down

on the cylinder with a force of 3.2×104N. What is the compressive strain of the cylinder? Young's modulus for aluminum is 7.0×1010Pa. Express your answer using two significant figures. Compressive strain = nothing
By what distance does the cylinder's height decrease as a result of the forces on it?
Physics
1 answer:
soldier1979 [14.2K]3 years ago
3 0

Answer:

1.9\times 10^{-4}

1.2\times 10^{-4}\ m

Explanation:

r = Radius = 2.7 cm

F = Force = 3.2\times 10^4\ N

A = Area = \pi r^2

\sigma = Stress = \frac{F}{A}

E = Young's modulus = 7\times 10^{10}\ Pa

\epsilon = Strain

L_0 = Original length = 67 cm

\Delta L = Change in length

Young's modulus is given by

E=\frac{\sigma}{\epsilon}\\\Rightarrow \epsilon=\frac{\sigma}{E}\\\Rightarrow \epsilon=\frac{\frac{3.2\times 10^4}{\pi 0.027^2}}{7\times 10^{10}}\\\Rightarrow \epsilon=0.0001996=1.9\times 10^{-4}

Strain is 1.9\times 10^{-4}

Strain is given by

\epsilon=\frac{\Delta L}{L_0}\\\Rightarrow \Delta L=\epsilon\times L_0\\\Rightarrow \Delta L=1.9\times 10^{-4}\times 0.67\\\Rightarrow \Delta L=0.0001273\\\Rightarrow \Delta L=1.2\times 10^{-4}\ m

The cylinder height decreases by 1.2\times 10^{-4}\ m

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Explanation:

In order to calculate the electric field you can use the Gaussian theorem.

Thus, you have:

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If you take the Gaussian surface as a spherical surface, with radius r, the electric field is parallel to the surface anywhere. Then, you have:

\Phi_E=EA=E(4\pi r^2)=\frac{Q}{\epsilon_o}\\\\E=\frac{Q}{4\pi \epsilon_o r^2}

r can be taken as the distance in which you want to calculate the electric field, that is, 0.795m

Next, you replace the values of the parameters in the last expression, by taking into account that the net charge inside the Gaussian surface is:

Q=7.53*10^{-6}C+3.65*10^{-6}C=1.115*10^{-5}C

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hence, the electric field at 0.795m from the center of the spherical shell is 12640.78 N/C

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