A locomotive with a mass of 80000 kg runs at a constant speed on the horizontal rail, and the resistance is 0.1 times of the vehicle weight. The locomotive runs at a constant speed s = 100 m After that, the engine is turned off, and the locomotive continues to move forward. L = 60m later, it comes to a standstill This is my homework tonight. Now it's the second day. I have to hand in my homework,

A locomotive with a mass of 80000 kg runs at a constant speed on the horizontal rail, and the resistance is 0.1 times of the vehicle weight. The locomotive runs at a constant speed s = 100 m After that, the engine is turned off, and the locomotive continues to move forward. L = 60m later, it comes to a standstill This is my homework tonight. Now it's the second day. I have to hand in my homework,

Mass m = 80000kg, vehicle weight g = mg = 800000n, when driving at a constant speed, the resultant force is 0, traction force = resistance = 0.1 times of vehicle weight = 800000 × 0.1 = 80000n. Formula: w = FS. Traction work: w = FS = 80000n × 100m = 8 × 10 * 6J (only 100m distance has traction force) resistance work: w = FS = 80000n × 16m = 1

A truck with a total weight of 6x10 fourth power n drives up a 100m long slope at a speed of 36km / h, and the known traction force is 4.8x10 third power n
Find (1) the time required for the car to drive from the bottom of the slope to the top of the slope at a constant speed,
(2) The work done by the traction of a car
(3) The power of automobile engine

1. 1 / 36 = a
2. Work, force * displacement: 4.8 * 10 ^ 3 * 100 * 10 ^ (- 3) = b
3. Power, work / time = B / A
The above formulas are in kilometers and hours

A car is going to the destination 180 km away from the starting place. It will drive at the same speed as originally planned within the first hour after departure. One hour later, it will drive at 1.5 times of the original speed and arrive at the destination 40 minutes ahead of the original plan. The average speed of one hour will be calculated

Let the average speed of one hour be X
There are: original time required - total time of this acceleration = 40 minutes (2 / 3 hours)
The total time of accelerating this time = 1 hour at the original speed + 1.5 times the remaining distance
The equation is as follows
180/x-[1+(180-x)/1.5x]=2/3
The solution is: x = 60
The original average driving speed was 60km / h

A car goes to the destination 180 km away from the starting place, drives 60 km at the original planned speed, and then drives at 1.5 times of the original speed. As a result, it reaches the destination 40 minutes earlier than the original plan, and calculates the original planned speed

Suppose the original planned driving speed is x km / h, then: 180 − 60x − 180 − 601.5x = 4060, the solution is x = 60, and the test shows that x = 60 is the solution of the original equation and conforms to the meaning of the problem, so x = 60. Answer: the original planned driving speed is 60 km / h