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.

molecules are free to separate from each other and will take on the shape of the container. molecules move completely at random and high speeds. they vibrate in a fixed position and do not move around much. molecules expand to the edge of their container.

initial speed when it is dropped is 150 ft/s

now at the same moment the acceleration due to gravity is perpendicular to its velocity

so as we know that

now we have centripetal acceleration a = 32 ft/s^2 as its perpendicular to velocity

while tangential acceleration at the same moment will be zero

so

now radius of curvature is given as

so the radius of curvature just after it is dropped is 703.125 ft