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valkas [14]
2 years ago
7

The motor in a refrigerator has a power output of 294 W. If the freezing compartment is at 271 K and the outside air is at 310 K

, assuming ideal efficiency, what is the maximum amount of heat (in joules) that can be extracted from the freezing compartment in 81.6 minutes
Physics
2 answers:
Nuetrik [128]2 years ago
7 0

Answer:

Q_{L} = 11.442\times 10^{6}\,J

Explanation:

The ideal efficiency of a refrigerator is given by the model of a Carnot refrigerator, whose indicator is known as Coefficient of Performance:

COP_{R} = \frac{T_{H}}{T_{H}-T_{L}}

COP_{R} = \frac{310\,K}{310\,K-271\,K}

COP_{R}=7.949

Heat rate extracted from the freezing compartment is:

\dot Q_{L} = COP_{R}\cdot \dot W

\dot Q_{L} = 7.949\cdot 294\,W

\dot Q_{L} = 2337.01\,W

The heat that can be extracted in abovementioned period of time is:

Q_{L} = \dot Q_{L}\cdot \Delta t

Q_{L} = (2337.01\,W)\cdot (81.6\,min)\cdot (\frac{60\,s}{1\,min} )

Q_{L} = 11.442\times 10^{6}\,J

dusya [7]2 years ago
5 0

Answer:

Maximum amount of heat = 10002151.38J

Explanation:

Workdone by motor in 86.1 minutes I given by:

W = power × time

W= 294 × 86.1×60

W= 1439424 Joules

W= 1.4 ×10^6Joules

The amount of heat extracted is given by:

/QL /= K/W/ = TL/W/ /(TH - TL)

Where TL= freezing compartment temperature

TH = Outside air temperature

/QL /= 271 × 1439424 / (310 - 271)

/QL/ = 390083904/39

/QL/ = 10002151.38 Joules

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The correct equation for the x component of a vector named A with an angle measured from the x axis would be which of the follow
Lady_Fox [76]

Answer:

Acosθ

Explanation:

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Hence, horizontal, x - component of the vector is :

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8 0
3 years ago
Select the correct answer.
kvasek [131]

Answer:

8.37×10⁻⁴ N/C

Explanation:

Electric Field: This is the ratio of electrostatic force to electric charge. The S.I unit of electric field is N/C.

From the question, the expression for electric field is given as,

E = F/Q.......................... Equation 1

Where E = Electric Field, F = force experienced by the charged balloon, Q = Charge on the balloon.

Given: F = 8.2×10⁻² Newton, Q = 9.8×10 Coulombs = 98 Coulombs

Substitute these values into equation 1

E = 8.2×10⁻² /98

E = 8.37×10⁻⁴ N/C

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Suppose the rocket in the Example was initially on a circular orbit around Earth with a period of 1.6 days. Hint (a) What is its
ruslelena [56]

Answer:

a

The orbital speed is v= 2.6*10^{3} m/s

b

The escape velocity of the rocket is  v_e= 3.72 *10^3 m/s

Explanation:

Generally angular velocity is mathematically represented as

            w = \frac{2 \pi}{T}

Where T is the period which is given as 1.6 days = 1.6 *24 *60*60 = 138240 sec

       Substituting the value

         w = \frac{2 \pi}{138240}

             = 4.54*10^ {-5} rad /sec

At the point when the rocket is on a circular orbit  

   The gravitational force =  centripetal force and this can be mathematically represented as

              \frac{GMm}{r^2} = mr w^2

Where  G is the universal gravitational constant with a value  G = 6.67*10^{-11}

            M is the mass of the earth with a constant value of M = 5.98*10^{24}kg

            r is the distance between earth and circular orbit where the rocke is found

               Making r the subject

                     r = \sqrt[3]{\frac{GM}{w^2} }

                        = \sqrt[3]{\frac{6.67*10^{-11} * 5.98*10^{24}}{(4.45*10^{-5})^2} }

                        = 5.78 *10^7 m

The orbital speed is represented mathematically as

                   v=wr

Substituting value

                  v= (5.78*10^7)(4.54*10^{-5})

                     v= 2.6*10^{3} m/s    

The escape velocity is mathematically represented as

                            v_e = \sqrt{\frac{2GM}{r} }

Substituting values

                             = \sqrt{\frac{2(6.67*10^{-11})(5.98*10^{24})}{5.78*10^7} }

                             v_e= 3.72 *10^3 m/s

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3 years ago
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