Acar of mass 1200 kg drives around a curve with a radius of 25.0 m. if the driver maintains a speed of 20.0 km/hr, what is the force of friction between the tires and the road? what is the minimum coefficient of static friction required to keep the car in this turn?

F=1487.52 N

μ = 0.12

Explanation:

Given that

m= 1200 kg

r= 25 m

v= 20 km/hr = 20 x 5/18 m/s

v= 5.55 m/s

To maintain the speed 5.55 m/s ,friction force should be equal to the radial force

Friction force = Radial force(F)

By putting the values

F=1487.52 N

So friction force = 1478.52 N

We also know that

Friction force = μ m g

1478.52 N = μ  x 1200 x 10                         ( take g =10 m/s²)

μ = 0.12

This is  minimum coefficient of static friction required to keep the car in this turn.

correct it is c.) b b

beta is electron from nucleus

something from nucleus has changed, but charge must be same for cons of charge