The power output of a car engine running at 2500 rpm is 500 kW . You may want to review (Pages 574 - 577) . Part A How much work is done per cycle if the engine's thermal efficiency is 20.0 %

Explanation:

2500 rpm = 2500 / 60 rps ( revolution per second )

= 41.67 rps .

power output = 500 kW

thermal efficiency = 20 %

power input = (100 /20) x 500 kW

= 2500 kW .

work done per second = 2500 kJ

work done per cycle = 2500 / 41.67

= 60 kJ .

r=rho l/a where a is the area, l is the length and rho is resistivity so the greater the area the lesser the resistance thats why the earth pin is thicker bcos current follow the shortest path, so less resistive .

in order to follow the instructions and complete the statements, i need to use data from table c of your student guide. can you think of a way to let me see your student guide for a while ? i can't answer the question without the data in table c.