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How big is the frictional force when a car stops on a flat road?
Wiki User
∙ 10y ago
Very big because it has no grip on the surface for the frictional force to stop and it takes time to stop the car.
Wiki User
∙ 10y ago
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Describe the size and direction of the frictional forces when a car stops on a flat road?
describe the size and direction of the frictional forces when a car stops on a falt road?
When you are running on a road you are acted upon by frictional force of the road.Is there another force of friction acting on you?
When you are running on a road ,you are acted upon by frictional force of the road.Is there another force of friction acting on you ? name and explain
When you are running on a road you are act Upon by A frictional force of the road. is there another force acting on you?
Yes. Gravity. Air resistance (friction again).
What is the frictional force between a tire and a road?
the energy to change the shape of the tire were it is crunched is a type of drage
When the road is wet it takes times longer to stop?
Yes. There is less frictional force between the car tyres and a wet road surface than with a dry road surface.
What is the condition of skidding on circular track?
when the centripetal force is greater than frictional force,skidding occurs.if mu is the coefficient of friction between the road and tyre,then the limiting friction .
How does banking of roads reduce wear and tear of the tires?
Motion of a car on a level road Forces acting on the car : 1.The weight of the car, mg 2.Normal reaction, N 3. Frictional force, f N--mg = 0 N = mg The centripetal force required for circular motion is along the surface of the road, and is provided by the component of the contact force between road and the car tyres along the surface. This force is nothing but the frictional force. The static friction that provides the centripetal acceleration. Motion of a car on a banked road We can reduce the contribution of friction to the circular motion of the car if the road is banked N cos θ = mg + f sin θ The centripetal force is provided by the horizontal components of N and f. N sin θ + f cos θ =mv2/R As you can see, the magnitude of frictional force has decreased by banking the road. Hence the wear and tear of the tyres will be less
A 1500 kg car accelerates at 4.4 ms2 on a level stretch of road If there is a wind resistance force of 67 N what must be the frictional force between the tires and the road?
The car is accelerating due to the net force acting on it which is a sum of the frictional force and wind resistance F = Fr - Fw. The net force can be found from Newton's 2nd law F = m x a and hence F = 6600 N. Since the wind acts to slow the car the frictional force must be greater than the wind and so Fr = F + Fw = 6600 + 67 = 6667. Which is alarmingly close to the Devil's number, is your teacher a satanist?
Why does a piece of chalk wear out as it is used on a blackboard?
With every stroke of the chalk piece it leaves some of the chalk on the board. This abrasion of the piece of chalk causes it to wear down. This is similar to why the rubber tyres of cars wear down when they are driven over a road surface.
Which surface would you slide further on road covered in ice or straw?
A road covered in ice; the frictional force that would slow you down would be much greater on a road covered in straw than one covered in ice.
Which surface would you slide further on-road covered in ice or straw?
A road covered in ice; the frictional force that would slow you down would be much greater on a road covered in straw than one covered in ice.
The coefficient of static friction between the tires of a car a dry road is 62 the mass of the car is 1500 kg what maximum force is obtainable?
Frictional force, F = coefficient of friction, u x normal reaction, R Maximum normal reaction = Weight => Maximum frictional force = 62 x 9.8 x 1500 = 911400 N
