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:
Total number of lamps will be 4
Explanation:
We have given power of the lamp W = 400 watt
Potential difference across the lamp V=110 volt
We know that power is equal to 
So 
Total current is given 15 A
As it is given that lamps are connected in parallel so total current is the sum of current through each lamp
So number of lamp will be 
As the lamp can not be in negative
So total number of lamps will be 4
Answer:
–77867 m/s/s.
Explanation:
From the question given above, the following data were obtained:
Initial velocity (u) = 34.5 m/s
Final velocity (v) = –23.9 m/s
Time (t) = 0.00075 s
Acceleration (a) =?
Acceleration is simply defined as the rate of change of velocity with time. Mathematically, it is expressed as:
Acceleration = (final velocity – Initial velocity) /time
a = (v – u) / t
With the above formula, we can obtain acceleration of the ball as follow:
Initial velocity (u) = 34.5 m/s
Final velocity (v) = –23.9 m/s
Time (t) = 0.00075 s
Acceleration (a) =?
a = (v – u) / t
a = (–23.9 – 34.5) / 0.00075
a = –58.4 / 0.00075
a = –77867 m/s/s
Thus, the acceleration of the ball is –77867 m/s/s.
