During each cycle, a heat engine ejects 75 j of thermal energy for every 115 j of input thermal energy. this engine is used to lift a 345-kg load a vertical distance of 27.0 m at a steady rate of 52.5 mm/s. part a how many cycles of the engine are needed to accomplish this task?

no of cycle required is 1885.3

Explanation:

given data

thermal energy Q1 = 75 J

thermal energy Q2 = 115 J

load = 3454 kg

distance = 27 m

steady rate = 52.5 mm/s

to find out

How many cycles of the engine are need

solution

we find here first work done that is express as

work done W = Q2 - Q1     1

put here all these value

W = 115 - 75

work done is 40 J

so energy required is

Energy required E = m×g×h     2

put here value

E = 285 × 9.8 × 27

energy required = 75411 J

so here no of cycle is

no of cycle = energy required / work done

put here value

no of cycle required = 75411 / 40

so no of cycle required is 1885.3

trig 15.3 cos