The rated power of the engine P = 6.0 x 10 of the fourth power W, the mass of the car M = 5.0 x 10 of the third power kg, The rated power P of an automobile engine is the fourth power W of 6.0 x 10, and the mass m is the third power kg of 5.0 x 10 (1) it is necessary for the vehicle to move at a constant speed while maintaining rated power,

The rated power of the engine P = 6.0 x 10 of the fourth power W, the mass of the car M = 5.0 x 10 of the third power kg, The rated power P of an automobile engine is the fourth power W of 6.0 x 10, and the mass m is the third power kg of 5.0 x 10 (1) it is necessary for the vehicle to move at a constant speed while maintaining rated power,

1. V = P / F, f = 0.1mg, so v = P / 0.1mg = 12m / s

The rated power of automobile engine is p = 6.0 × 10 fourth power tile, the mass of automobile is m = 5 × 10 third power kilogram, the resistance of automobile driving on the horizontal straight road is 0.1 times of the vehicle weight, take 10 to find out the speed of automobile moving at constant speed under the rated power? (2) if the automobile runs at a constant speed of 8m / s, what is the power of automobile engine?

(1)F=f=0.1G=0.1*5000kg*10N/kg=5000N
v=P/F=60000W/5000N=12m/s
(2)P1=Fv1=5000N*8m/s=40000W

When a vehicle with a mass of 2.0 x 10 cubic kg and a rated power of 6.0 x 10 quartic W is driven on a horizontal highway at rated power, the resistance it suffers is a certain value. At a certain moment, the speed of the vehicle is 20 m / s and the acceleration is 0.50 M / S square
What is the maximum speed a car can reach?
2. What is the acceleration when the speed of the car is 10m / S?
3. When the car starts to move in a straight line with uniform acceleration from standstill, and the acceleration is a = 1m / S ^ 2, how long can this process last?

This kind of problem is relatively easy, but emphasizes the format 1, from, P = f * V to get f = P / v = 3x10 & # 179; n due to the constant resistance in the process, from Newton's second law to get: f-f = ma = 2.0x10 & # 179; n when the car reaches the maximum speed: P = FV, then V (max) = P / F = (6.0x10 ^ 4 / 2.0x10 & # 179

A vehicle with a power of 6.615 * 10 to the fourth power W uses a gasoline engine with a calorific value of 4.6 * 10 to the seventh power J / kg as fuel, and the engine efficiency is 20%
There is 50kg gasoline in the mailbox, which can be used to calculate the average speed of the car through 100km distance

Suppose the speed of the vehicle is V, and the time required for the vehicle to pass s = 100km is t
Then v = s / T (1)
From w = Pt (2)
Combining (1) and (2), we get
v=sP/W=1*10^5*6.615*10^4/50*4.6*10^7*20%=14.38m/s