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Genrish500 [490]
2 years ago
5

The temperature inside a Carnot refrigerator placed in a kitchen at 22 degree Celcius is 2 degree Celcius. The heat extracted fr

om the refrigerator is 89 MJ/h. What power is needed to operate this refrigerator?
Physics
1 answer:
Elanso [62]2 years ago
4 0

Answer: 1.79 kW

Explanation:

Given

The temperature of the refrigerator is T_L=2^{\circ}C\approx 275\ K

The temperature of the kitchen is T_H=22^{\circ}C

COP of the refrigerator is

COP=\dfrac{T_L}{T_H-T_L}=\dfrac{\text{Desired effect}}{\text{Power}}\\\\\Rightarrow COP=\dfrac{275}{22-2}=\dfrac{89}{P}\\\\\Rightarrow P=\dfrac{20}{275}\times 89=6.47\ MJ/h\ \text{or}\ 1.79\ kW

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Explanation:

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3 years ago
In reaching her destination, a backpacker walks with an average velocity of 1.41 m/s, due west. This average velocity results, b
V125BC [204]
The total average velocity v=+1.41 m/s (I assume west as positive direction) is given by the total displacement, S, divided by the total time taken, t:
v= \frac{S}{t}= \frac{S_1+S_2}{t_1+t_2}
where:
-The total displacement S is the algebraic sum of the displacement in the first part of the motion (S_1=6.30 km=6300 m, due west) and of the displacement in the second part of the motion (S_2, due east).
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t_1= \frac{S_1}{v_1}= \frac{6300 m}{2.49 m/s}=2530 s

t_2, instead, can be written as \frac{S_2}{v_2}, where v_2=-0.630 m/s is the average velocity of the second part of the motion (with a negative sign, since it is due east). 

Therefore, we can rewrite the initial equation as:
v=1.41 = \frac{6300+S_2}{2530- \frac{S_2}{0.630} }
And by solving it, we find the displacement in the second part of the motion (i.e. how far did the backpacker move east):
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4 0
3 years ago
How would the gravitational force between the earth and the moon change if the moon had half the mass?
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The moon's gravity, combined with the waltz of Earth and the moon around their center of mass, forces the oceans into an oval shape, with two simultaneous high tides. ... If the moon were half its mass, then the ocean tides would have been correspondingly smaller and imparted less energy to it.

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What is the current in mA through a 500 Ω resistor that is connected to a 1.5 V battery? Show all calculations using Ohm’s law (
Ludmilka [50]

Answer:

3 mA.

Explanation:

The following data were obtained from the question:

Resistor (R) = 500 Ω

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Current (I) =.?

Using the ohm's law equation, we can obtain the current as follow:

V = IR

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Divide both side by 500

I = 1.5 / 500

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Therefore, the current in the circuit is 3×10¯³ A.

Finally, we shall convert 3×10¯³ A to milliampere (mA).

This can be obtained as follow:

Recall:

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3×10¯³ A = 3×10¯³ × 1000 = 3 mA

Therefore, 3×10¯³ A is equivalent to 3 mA.

Thus, the current in mA flowing through the circuit is 3 mA.

3 0
3 years ago
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Explanation:

8 0
3 years ago
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