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vagabundo [1.1K]
4 years ago
6

A cylinder, which is in a horizontal position, contains an unknown noble gas at 4.00 × 10 4 Pa 4.00×104 Pa and is sealed with a

massless piston. The piston is slowly, isobarically moved inward 16.3 cm, 16.3 cm, which removes 1.50 × 10 4 J 1.50×104 J of heat from the gas. If the piston has a radius of 30.5 cm, 30.5 cm, calculate the change in the internal energy of the system Δ U ΔU .
Physics
1 answer:
alexgriva [62]4 years ago
8 0

Answer:

-13094.55179 J

Explanation:

Q = Heat = -1.5\times 10^{4}\ J

P = Pressure = 4\times 10^4\ Pa

\Delta V = Change in volume = \pi r^2\times -h(negative because it is decreasing)

h = Height = 16.3 cm

r = Radius = 30.5 cm

Entropy is given by

\Delta U=Q-W

Work done is given by

W=P\Delta V\\\Rightarrow W=4\times 10^4\times (\pi 0.305^2\times -0.163)

\Delta U=-1.5\times 10^{4}-(4\times 10^4\times (\pi 0.305^2\times -0.163))\\\Rightarrow \Delta U=-13094.55179\ J

The change in the internal energy of the system is -13094.55179 J

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During a neighborhood baseball game in a vacant lot, a particularly wild hit sends a 0.148 kg baseball crashing through the pane
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Final speed of the ball, v = 11.5 m/s

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The suspension system of a 1700 kg automobile "sags" 7.7 cm when the chassis is placed on it. Also, the oscillation amplitude de
spin [16.1K]

Answer:

the spring constant k = 5.409*10^4 \ N/m

the value for the damping constant \\ \\b = 1.518 *10^3 \ kg/s

Explanation:

From Hooke's Law

F = kx\\\\k =\frac{F}{x}\\\\where \ F = mg\\\\k = \frac{mg}{x}\\\\given \ that:\\\\mass \ of \ each \ wheel = 425 \ kg\\\\x = 7.7cm = 0.077 m\\\\g = 9.8 \ m/s^2\\\\Then;\\\\k = \frac{425 \ kg * 9.8 \ m/s^2}{0.077 \ m}\\\\k = 5.409*10^4 \ N/m

Thus; the spring constant k = 5.409*10^4 \ N/m

The amplitude is decreasing 37% during one period of the motion

e^{\frac{-bT}{2m}}= \frac{37}{100}\\\\e^{\frac{-bT}{2m}}= 0.37\\\\\frac{-bT}{2m} = In(0.37)\\\\\frac{-bT}{2m} = -0.9943\\\\b = \frac{2m(0.9943)}{T}\\\\b = \frac{2m(0.9943)}{\frac{2 \pi}{\omega}}\\\\b = \frac{m(0.9943) \ ( \omega) )}{ \pi}

b = \frac{m(0.9943)(\sqrt{\frac{k}{m})}}{\pi}\\\\b = \frac{425*(0.9943)(\sqrt{\frac{5.409*10^4}{425}) }    }{3.14}\\\\b = 1518.24 \ kg/s\\\\b = 1.518 *10^3 \ kg/s

Therefore; the value for the damping constant \\ \\b = 1.518 *10^3 \ kg/s

5 0
4 years ago
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