There is a car with a mass of 3 x 10 cubic kg, driving at a uniform speed of 15 m / s on a straight road. It is known that the resistance of the car is 0.06 times of its weight What is the power of automobile engine? (g = 10N / kg)

Resistance f = 0.06mg = 0.06 * (3 * 10 ^ 3) * 10 = 1.8 * 10 ^ 3 N
Drive at a constant speed with balanced force, traction f = resistance F
Engine power P = w / T = FS / T = FV = (1.8 * 10 ^ 3) * 15 = 27 * 10 ^ 3W = 27kw

The friction force of a truck weighing 8 × 10 to the fourth power n is 0.02 times of the weight of the truck when driving at a constant speed on a horizontal road. The traction force of the truck is calculated

Friction force at constant speed = traction force on truck
Therefore, the traction force F = 0.02mg = 0.02 * 8 * 10 ^ 4 = 1600n

When a 104n car moves in a straight line at a constant speed on the road, its resistance is 0.03 times of the weight of the car. What is the traction force of the engine?

Because the car moves in a straight line at a constant speed, the traction and resistance of the car are a pair of balance forces, so the traction force of the car is f = f = 0.03g = 0.03 × 104n = 300N. A: the traction force of the car engine is 300N

For a train with a mass of 10 cubic T, the traction force of the locomotive is 3.5 * 10 quintic n, and the resistance in motion is 0.01 times of the vehicle weight
How long does it take to start the uniform acceleration linear motion and change the speed to 180km / h? How many meters did you advance in the process? (g take 10m every second power)

The third power of M = 10 and the sixth power of T = 10
Ψ g = mg = 10 to the sixth power × 10N / kg = 10 to the seventh power n
F = the seventh power of 10 × 0.01 = the fifth power of 10 n
∵F=ma
A = f / M = (f traction-f resistance) / M
=(3.5 × 10 fifth power N-10 fifth power n) / (10 sixth power kg)
=(2.5 × 10 fifth power n) / (10 sixth power kg)
=0.25m/s²
∵ the train starts to make uniform acceleration linear motion from static state, 180km / h = 50M / s
∴v=v0+at
∴t=(v—v0)/a
=(50-0)/0.25
=200s
s=v0t+(1/2)at²
=0×200+(1/2)×0.25×200²
=5000m
Therefore, it takes 200 seconds for the train to make uniform acceleration straight-line movement from static to 180 km / h;
In the process, 5000 meters were advanced
By ivy

There is a car with a mass of 3 x 10 cubic kg, driving at a uniform speed of 15 m / s on a straight road. It is known that the resistance of the car is 0.06 times of its weight What is the power of automobile engine? (g = 10N / kg)