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:
0.0319 m³
Explanation:
Use ideal gas law:
PV = nRT
where P is pressure, V is volume, n is amount of gas, R is the gas constant, and T is temperature.
Since P, n, and R are held constant:
n₁ R / P₁ = n₂ R₂ / P₂
Which means:
V₁ / T₁ = V₂ / T₂
Plugging in:
0.0279 m³ / 280 K = V / 320 K
V = 0.0319 m³
In order to change the frictional force between two solid surfaces, it can be changed by shorter distances and by the amount of weight it has or the amount of force that is pushing that object to go however distance it can.
Answer:
Explanation:
Component of force perpendicular to stick
= F Sin 60°
=√3 / 2 F.
Taking torque about the other end
= √3 / 2 F x 1 Nm
Weight of stick = 60 gm
= 60 x 10⁻³ kg
= 60 x 10⁻³ x 9.8 N
= .588 N
This weight will act from the middle point of stick so torque about the
other end
= .588 x 1 Nm
Balancing these two torques we have
.588 = √3 /2 F

F = 0.679 N