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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a) For the motion of car with uniform velocity we have ,
, where s is the displacement, u is the initial velocity, t is the time taken a is the acceleration.
In this case s = 520 m, t = 223 seconds, a =0 
Substituting

The constant velocity of car a = 2.33 m/s
b) We have 
s = 520 m, t = 223 seconds, u =0 m/s
Substituting

Now we have v = u+at, where v is the final velocity
Substituting
v = 0+0.0209*223 = 4.66 m/s
So final velocity of car b = 4.66 m/s
c) Acceleration = 0.0209 
Answer:
option A
Explanation:
given,
For exerted by the worker = 245 N
angle made with horizontal = 55°
we need to calculate Force which is not used to move the crate = ?
Movement of crate is due to the horizontal component of the force.
Crate will not move due to vertical force acting on the it.



hence, worker's force not used to move the crate is equal to 200.69
The correct answer is option A
Answer:
10.93m/s with the assumption that the water in the lake is still (the water has a speed of zero)
Explanation:
The velocity of the fish relative to the water when it hits the water surface is equal to the resultant velocity between the fish and the water when it hits it.
The fish drops on the water surface vertically with a vertical velocity v. Nothing was said about the velocity of the water, hence we can safely assume that the velocity if the water in the lake is zero, meaning that it is still. Therefore the relative velocity becomes equal to the velocity v with which the fish strikes the water surface.
We use the first equation of motion for a free-falling body to obtain v as follows;
v = u + gt....................(1)
where g is acceleration due to gravity taken as 9.8m/s/s
It should also be noted that the horizontal and vertical components of the motion are independent of each other, hence we take u = 0 as the fish falls vertically.
To obtain t, we use the second equation of motion as stated;

Given; h = 6.10m.
since u = 0 for the vertical motion; equation (2) can be written as follows;

substituting;

Putting this value of t in equation (1) we obtain the following;
v = 0 + 9.8*1.12
v = 10.93m/s
Боже, как это сложно! Ну ладно.
Между прочим это ты сам должен делать, а то не куда не поступишь!