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saw5 [17]
3 years ago
6

There is a refrigerator running in a room, the heat flowing into the refrigerator from the outside is 40 J/s, and the refrigerat

or must consume 100 W to maintain the low temperature set in it. (1) What is the heat discharged from the refrigerator at this time?
(2) If the doors and windows of the room are closed, the temperature of the room rises by a few degrees per hour? Assume that the air heat capacity of the entire room is 400 kJ/K.
Physics
1 answer:
Mamont248 [21]3 years ago
8 0

Answer:

(A) 140 j/sec (b) 1.26 K

Explanation:

We have given the heat heat flowing into the refrigerator = 40 J/sec

Work done = 40 W

(a) So the heat discharged from the refrigerator =heat\ flowing\ in\ refrigerator+work\ done=40+100=140j/sec

(b) Total heat absorbed =140 j/sec =140\times 3600=504000j/hour

Let the temperature be \Delta T

Heat absorbed per hour =504000 [tex]=400\times 10^3\times \Delta T

So  \Delta T=\frac{504000}{400000}=1.26K

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Displacement is the straight line distance from the starting position to the final position.

Person X walks halfway around circle.  So her displacement is 100 m.

Person Y walks 3/4 of the way around.  So his displacement is 50√2 m ≈ 70.7 m.

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Therefore:

Z < Y < X

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3 years ago
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A series RLC circuit has a resonant frequency = 6.00 kHz. When it is driven at a frequency = 8.00 kHz, it has an
ANEK [815]

The resistance (R) of the circuit is 707.1 ohms and the inductance (L) is 0.032 H.

<h3>Resistance of the circuit</h3>

For the phase constant of 45⁰, impedance is equal to the resistance of the circuit.

Z= R\sqrt{2} \\\\R  = \frac{Z}{\sqrt{2} } \\\\R = \frac{1000}{\sqrt{2} } = 707.1 \ ohms

<h3>Resonant frequency</h3>

f = \frac{1}{2\pi \sqrt{LC} } \\\\6000 = \frac{1}{2\pi \sqrt{LC} } \\\\2\pi(6000) = \frac{1}{\sqrt{LC} } \\\\\sqrt{LC} = \frac{1}{2\pi (6000)} \\\\LC = (\frac{1}{2\pi (6000)})^2\\\\LC = 7.034 \times 10^{-10} \\\\ C = \frac{7.034 \times 10^{-10} }{L} ---(1)

<h3>At driven frequency</h3>

X_l- X_c = R\\\\\omega L - \frac{1}{\omega C}  = 707.1\\\\2\pi f L -  \frac{1}{2\pi fC} = 707.1\\\\2\pi (8000) L - \frac{1}{2\pi (8000) C } = 707.1\ \ --(2)\\\\

<em>solve 1 and 2 together</em>

2\pi(8000) L - \frac{L}{2\pi (8000)(7.034 \times 10^{-10})} = 707.1\\\\50272L - 28279.48L = 707.1\\\\L = 0.032 \ H

Learn more about impedance of RLC circuit here: brainly.com/question/372577

7 0
3 years ago
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