Answer:
0.265
Explanation:
Draw a free body diagram. There are four forces:
Normal force Fn pushing up.
Weight force mg pulling down.
Tension force T at an angle θ.
Friction force Fn μ pushing left.
Sum the forces in the y direction:
∑F = ma
Fn + T sin θ − mg = 0
Fn = mg − T sin θ
Sum the forces in the x direction:
∑F = ma
T cos θ − Fn μ = 0
Fn μ = T cos θ
μ = T cos θ / Fn
μ = T cos θ / (mg − T sin θ)
Given T = 164 N, θ = 10.0°, m = 65.0 kg, and g = 9.8 m/s²:
μ = (164 N cos 10.0°) / (65.0 kg × 9.8 m/s² − 164 N sin 10.0°)
μ = 0.265
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
=>
=>
Kinetic energy=1/2mv^2
=1/2(142*10^-3)(42.9)^2=130.6=131J
Answer:
Explanation:
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