A 1500-kg car drives at 30 m/s around a flat circular track 300 m in diameter. Since the net force is equivalent to the force of static friction, your answer to Part D is the magnitude fs. Based on this value, what is the minimum coefficient μs of static friction between the road and the car?

The minimum coefficient μs of static friction between the road and the car μs is 0.6116

Explanation:

Here we have

Angular acceleration = v²/r = 30²/150 = 6 rad/s²

Therefore, the force of the car away from its center of motion =

Force, F = Mass × Acceleration = 1500 kg × 6 rad/s² =  9000 N

The centripetal force is balanced by the frictional force so the car does not skid off the track

Frictional force Ff = Normal force (Weight of car) × Coefficient of friction, μs)

Weight of car = Mass × Acceleration due to gravity = 1500 kg × 9.81 m/s²

= 14715 N

Therefore, for equilibrium

Frictional force Ff of car = Centripetal force on car

14715 N × μs = 9000 N

Therefore, μs = 9000 N/(14715 N) = 0.6116

The minimum coefficient μs of static friction between the road and the car μs = 0.6116.

wavelength variability is the right one

ill do it for 100, also this isnt 80 its 40