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:
Heat can travel from one place to another in three ways: Conduction, Convection and Radiation. ... Thermal energy is transferred from hot places to cold places by convection. Convection occurs when warmer areas of a liquid or gas rise to cooler areas in the liquid or gas.
Answer: 459.14 N
Explanation:
from the question, we have
diameter = 10 m
radius (r) = 5 m
weight (Fw) = 670 N
time (t) = 8 seconds
Circular motion has centripetal force and acceleration pointing perpendicular and inwards of the path, therefore we apply the equation below
∑ F = F c = F w − Fn ..............equation 1
Fn = Fw − Fc = mg − (mv^2 / r) ...................equation 2
substituting the value of v as (2πr / T) we now have
Fn = mg − (m(2πr / T )^2) / r
Fn= mg − (4(π^2)mr / T^2) ..........equation 3
Fw (mass of the person) = mg
therefore m = Fw / g
m = 670 / 9.8 = 68.367 kg
now substituting our values into equation 3
Fn = 670 - ( (4 x (π^2) x 68.367 x 5 ) / 8^2)
Fn = 670 - 210.86
Fn = 459.14 N
Answer:
The correct answer is The plates smash together with no subduction. I just took this on Edge. Glad I could help!
