The vehicle runs 48km at the speed of 72km / h, and the efficiency of gasoline engine is 25%（ Q gasoline = 4.6 * 10 ^ 7J / kg

The vehicle runs 48km at the speed of 72km / h, and the efficiency of gasoline engine is 25%（ Q gasoline = 4.6 * 10 ^ 7J / kg

The total energy consumed by the engine is 8 * 4.6 * 10 ^ 7J = 36.8 * 10 ^ 7J
The total energy output is 36.8 * 0.25 * 10 ^ 7 = 9.2 * 10 ^ 7J
The running speed of the car is 72 km / h = 20 m / s
The total running time is 48000 / 20 = 2400 seconds
The average power is 9.2 * 10 ^ 4 / 2400 = 38.33kw

The resistance of the car on the straight road is 8000n, the uniform speed is 72km / h, the consumption of gasoline is 6kg (1) car power (2) car efficiency
Q gasoline = 3x10 seventh power J / kg

1)72km/h=20m/s
P=Fv=fv=8000*20=160000W
2) Efficiency = w / Q = Pt / MQ
=160000*2*3600/6*3*10^7=
=16 * 2 * 36 / 6 * 46 = 6.4 is wrong. The efficiency must be less than 1,
this

The speed of a car is 50 kilometers per hour. There is a clock which is 3 minutes per hour fast. If you use the meter to read the time, what is the speed of the car?

Speed v = distance S / time T. If the time is calculated by this meter, it will change from t to t * 60 / 57, then the speed V will become V * 57 / 60 = 50 * 57 / 60, which is 47.5, which is equivalent to slowing down. In fact, it's very easy to understand that when the car has been driving for 57 minutes, the meter has already walked for 1 hour, but the distance is only 57 minutes