In each cycle of its operation or refrigerator removes 17 J of heat from the inside of the refrigerator and releases 40 J of heat into the room how much work per cycle is required to operate this refrigerator?

23 J

Explanation:

According to the law of conservation of energy, the total heat given off to the surroundings by a refrigerator is equal to the work done by the refrigerator and the heat removed from the inside of the refrigerator.

So we can write:

where

is the heat removed from the cold reservoir (the inside of the refrigerator)

W is the work  done

is the work released into the hot reservoir (the surroundings)

Here we have:

(heat removed from the inside of the refrigerator)

(heat released into the surroundings)

Solving the equation for W, we find the work done:

powerful computers now, not then

the minimum constant speed at which she can catch the train is 2.7 m/s

given:

acceleration = 0.6

time = 4.5 s

to find:

velocity = ?

formula used:

from 1st kinematic equation we get,

v = u + at

u = initial speed of train = 0 m/s

v = final speed of the train

a = acceleration of the train

t = time

solution:

from 1st kinematic equation we get,

v = u + at

u = initial speed of train = 0 m/s

v = final speed of the train

a = acceleration of the train

t = time

v = 0 + (0.6 × 4.5)

v = 2.7 m/s

the minimum constant speed at which she can catch the train is 2.7 m/s