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Direction of friction force on running automobile tire A 4WD vehicle runs at a uniform speed on a horizontal road, regardless of air resistance. Its tires are pulled by the forward friction, but because there is still rolling friction, its tires are also subjected to the backward friction. However, the friction is only opposite to the relative movement trend. How can the friction be both forward and backward?
Are you sure it's a 4WD? According to the following statement, this car should not be a 4WD car. If it is not a 4WD car, there are driving wheels and driven wheels. The friction of the driving wheel is the same as the moving direction, and the friction of the driven wheel is opposite to the moving direction
When a car turns on a horizontal curve, which forces act on it and who provides the centripetal force
Friction, gravity, support, traction, air resistance
But usually the air resistance is ignored
The centripetal force required by a car turning on a horizontal highway is determined by___ Force to provide
When turning on a horizontal highway, the vehicle is subjected to gravity, road support force and static friction. If gravity and support force balance, the centripetal force required is provided by static friction
So the answer is: static friction
For a vehicle traveling at the speed of 20m / s, the acceleration after emergency braking is 5m every 2 power seconds. What is the speed at the end of 2S and 6S after braking?
Speed after 2S V2 = 20m / s-5m / S ^ 2 * 2S = 10m / S
After 4S, the speed is 20m / s-5m / S ^ 2 * 4S = 0m / s
That is, the last train in 4S has stopped
It means that it has stopped at 6S and the speed is 0
The speed of uniform acceleration of the vehicle is 10m / s, and the acceleration of emergency braking in case of red light is 5m per second, then what is the speed at the end of 3S after braking Sorry. It should be "the speed of uniform motion"
When the car stops, t = V / a = 2S, that is, the time required for the car to stop braking is 2S
Therefore, the vehicle speed is zero at the end of 3 seconds
A car is running at a uniform speed of 20m / s, and an acceleration of 4m / S2 is obtained after emergency braking. What is the displacement within 8s after braking? Such as title
The parking speed is 0, and the time from deceleration to parking is t
0=20-4t t=5/s
That is, the car stops in 5S
x=vo*t+1/2at ²
=20*5-2*25
=50m
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