The refrigerator's coefficient of performance is 6.
The heat extracted from the cold reservoir Q cold (i.e., inside a refrigerator) divided by the work W required to remove the heat is known as the coefficient of performance, or COP, of a refrigerator (i.e., the work done by the compressor). The required inside temperature and the outside temperature have a significant impact on the COP.
As the inside temperature of the refrigerator decreases, its coefficient of performance decreases. The coefficient of performance (COP) of refrigeration is always more than 1.
The heat produced in the cold compartment, H = 780.0 J
Work done in ideal refrigerator, W = 130.0 J
Refrigerator's coefficient of performance = H/W
= 780/130
= 6
Therefore, the refrigerator's coefficient of performance is 6.
Energy conservation requires the exhaust heat to be = 780 + 130
= 910 J
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Answer:
Explanation:
The artificial gravity generated by the rotating space station is the same centripetal acceleration due to the rotational motion of the station, which is given by:
Here, r is the radius and v is the tangential speed, which is given by:
Here 
is the angular velocity, we replace (2) in (1):
Recall that 
.
Solving for :
Answer:
7.82 s
Explanation:
Given:
Δy = 300 m
v₀ = 0 m/s
a = 9.8 m/s²
Find: t
Δy = v₀ t + ½ at²
(300 m) = (0 m/s) t + ½ (9.8 m/s²) t²
t = 7.82 s
