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otez555 [7]
3 years ago
5

Suppose you want to operate an ideal refrigerator with a cold temperature of − 15.5 °C , and you would like it to have a coeffic

ient of performance of at least 8.25. What is the maximum hot reservoir temperature for such a refrigerator?
Physics
1 answer:
tensa zangetsu [6.8K]3 years ago
8 0

Answer:

15.65 °C

Explanation:

cold temperature (Tc) = -15.5 degree C = 273.15 - 15.5 = 257.65 kelvin

minimum coefficient of performance (η) = 8.25

find the maximum hot reservoir temperature of such a generator (Th)

η = \frac{Tc}{Th-Tc}

Th = Tc x (\frac{1}{η} + 1)

Th = 257.65 x (\frac{1}{8.25} + 1)

Th = 288.8 K

Th = 288.8 - 273.15 = 15.65 °C

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goblinko [34]

Answer: 15.66 °

Explanation: In order to solve this proble we have to consirer the Loretz force for charge partcles moving inside a magnetic field. Thsi force is given by:

F=q v×B = qvB sin α where α is teh angle between the velocity and magnetic field vectors.

From this expression and using the given values we obtain the following:

F/(q*v*B) = sin α

3.8 * 10^-13/(1.6*10^-19*8.9*10^6* 0.96)= 0.27

then  α =15.66°

8 0
3 years ago
What are things that we can do to protect our climate for future generations?
Vilka [71]
<span>Reduce energy use.
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6 0
3 years ago
What gravitational force does the earth exert on a person
lara [203]

The  gravitational force exerted by the earth on a person standing on the earth's surface is 602.74 N.

<h3>What is the gravitational force of the earth on the person?</h3>

The gravitational force exerted by the earth on a person standing on the earth's surface is given below as follows:

  • F = \frac{Gm^{1}m^{2}}{r^{2}}

where

G = 6.67 * 10⁻¹¹

m¹ = 62 kg

m² = 5.97 * 10²⁷ kg

r = 6.4 * 10⁶ m

F = \frac{5.97*10^{24}*62*6.67*10^{-11}}{(6.4*10^{6}){2}} = 602.74\:N

Therefore, the gravitational force exerted by the earth on a person standing on the earth's surface is 602.74 N.

Learn more about gravitational force at: brainly.com/question/940770

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7 0
2 years ago
I just need x isolated
mamaluj [8]

Answer:

x=\frac{-y+\sqrt{y^2+4rt} }{2r}

x=\frac{-y-\sqrt{y^2+4rt} }{2r}

Explanation:

rx+y=\frac{t}{x}\\\\x(rx+y)=(\frac{t}{x})x\\\\rx^2+yx=t\\\\rx^2+yx-t=t-t\\\\rx^2+yx-t=0

Solve using the quadratic formula.

x=\frac{-y+\sqrt{y^2+4rt} }{2r}

x=\frac{-y-\sqrt{y^2+4rt} }{2r}

7 0
3 years ago
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