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3 years ago

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A gas fills the right portion of a horizontal cylinder whose radius is 5.33 cm. The initial pressure of the gas is 2.29 x 105 Pa

. A frictionless movable piston separates the gas from the left portion of the cylinder that is evacuated and contains an ideal spring, as the drawing shows. The piston is initially held in place by a pin. The spring is initially unstrained, and the length of the gas-filled portion is 18.9 cm. When the pin is removed and the gas is allowed to expand, the length of the gas-filled chamber doubles. The initial and final temperatures are equal. Determine the spring constant of the spring.

1 answer:

const2013 [10]3 years ago

7 0

Answer:

k = 5406.88 N/m

Explanation:

Since the initial and final temperatures are constant while the gas expands, the expansion of the gas obeys Boyle's law

Initial gas pressure, $P_1 = 2.29 * 10^5 Pa$

Let the initial gas volume $V_1$ = V

Volume = Length * Area

Since the cross sectional area does not change and the length of the gas-filled chamber is said to double, then the volume also doubles.

$V_2 = 2V$

The final gas pressure, $P_2$ will be gotten from the Boyle's law equation

$P_1 V_1 = P_2 V_2\\P_2 = \frac{P_1 V_1}{V_2} \\P_2 = \frac{2.29*10^5 * V}{2V}\\P_2 = \frac{2.29*10^5 }{2}\\P_2 = 1.145 *10^5 Pa$

The Cross Sectional Area, $A = \pi R^2$

R = 5.33 cm = 0.0533 m

$A = \pi * 0.0533^2$ = 0.0089 m²

The force exerted on the piston, F = P₂ A

F = 1.145 * 10⁵ * 0.0089

F = 1021.9 N

To get the spring constant, use Hooke's law

F = k Δx

Where Δx = l₂ - l₁

Δx = 2l - l = l ( since the length of the chamber doubles on expansion)

Δx = 18.9 cm = 0.189 m

F = k Δx

1021.9 = k * 0.189

k = 1021.9/0.189

k = 5406.88 N/m

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here we have

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here we have

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A cast-iron flywheel has a rim whose OD is 1 m and whose ID is 0.8 m. The flywheel weight is to be such that an energy fluctuati

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Answer:

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A.

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$C_{s}=$ coefficient of speed fluctuation of the flywheel = 0.222

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I = $\frac{m}{8}\times(d_{o}^{2}- d_{i}^{2})$

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$m=\frac{8I}{(d_{o}^{2}+ d_{i}^{2})}= \frac{8\times 54.78}{(1^{2}+0.8^{2})}=267.22kg$

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$0.03713=\frac{\pi t(1^{2}-0.8^{2})}{4}$

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