Complete Question
A certain refrigerator, operating between temperatures of -8.00°C and +23.2°C, can be approximated as a Carnot refrigerator.
What is the refrigerator's coefficient of performance? COP
(b) What If? What would be the coefficient of performance if the refrigerator (operating between the same temperatures) was instead used as a heat pump? COP
Answer:
a
b
Explanation:
From the question we are told that
The lower operation temperature of refrigerator is 
The upper operation temperature of the refrigerator is 
Generally the refrigerators coefficient of performance is mathematically represented as
=> 
=>
Generally if a refrigerator (operating between the same temperatures) was instead used as a heat pump , the coefficient of performance is mathematically represented as
=> 
=>
Answer:
is it bad if i keep thinking about p ussy
Explanation:
Answer- There are two reasons that we know quotations have been used first is the use of of name of the person who quoted it and secondly the quotation is written inside the quotation marks.
Explanation- Quotation is nothing but using a line that has been already quoted by someone somewhere. Such sentences are normally written inside quotation marks. In the above given paragraph the name of the person who quotes the sentence is also given hence we know that our quotation has been used.
From the calculations, the power expended is 43650 W.
<h3>What is the power expended?</h3>
Now we can find the acceleration from;
v = u + at
u = 0 m/s
v = 95 km/h or 26.4 m/s
t = 6.8 s
a = ?
Now
v = at
a = v/t
a = 26.4 m/s/ 6.8 s
a = 3.88 m/s^2
Force = ma = 850-kg * 3.88 m/s^2 = 3298 N
The distance covered is obtained from;
v^2 = u^2 + 2as
v^2 = 2as
s = v^2/2a
s = (26.4)^2/2 * 3.88
s = 696.96/7.76
s = 90 m
Now;
Work = Fs
Work = 3298 N * 90 m = 296820 J
Power = 296820 J/ 6.8 s
= 43650 W
Learn more about power expended:brainly.com/question/11579192
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One of the useful forns of the formula for electrical power is: Power = (voltage squared) / (resistance). Knowing that power is proportional to (voltage squared), we can see that if the voltage is reduced to 1/2, the power is reduced to 1/4 of its original value. The 220volt/60watt appliance, when operated on 110 volts, dissipates 60/4 = 15 watts.
