When an automobile moves with constant velocity, the power developed is used to overcome the frictional forces exerted by the air and the road. If the engine develops 30 hp, what total frictional force acts on the car at 140 mph? One horsepower equals 746 W, and one mile is 1609 m. Answer in units of N.

Explanation:

Given:

Power, p = 30 hp

1 hp = 746 W

= 22.38 kW

Velocity, v = 140 mph

1 mile = 1609 m

Converting mph to m/s,

140 mile/hour × 1609 m/1 mile × 1 hour/3600 s

= 62.572 m/s

Summation of forces = 0

Power, p = energy/time

Energy = force × distance;

Velocity = distance/time;

Power = force × velocity

Force, fr = 22380/62.572

= 357.67 N

Frictional force, fr = 0.358 kN.

answer: 802.3 kilo joules of energy represents 191.75 kilo calorie of energy.

explanation:

energy released during combustion of one mole methane = 802.3 kilo joules

if 1 kilo joule = 0.239006   kilo calorie

then ,802.3 kilo joule will be = 802.30.239006 kilo calorie

802.3 kilo joule = 191.75 kilo calorie

hence, the energy in kilo calorie is 191.75 kilo calorie.