A race car rounds a curve at 60 m/s. The radius of the curve is 400 m, and the car’s mass is 600 kg. (a) What is the centripetal acceleration of the car? (b) What is the centripetal force of the car?

A race car rounds a curve at 59 m/s. The radius of the curve is 417 m, and the car's mass is 620 kg. (Assume the car's speed remains constant. Take the radially inward direction to be positive. Indicate the direction with the sign of your answer.) (a) What is the car's (centripetal) acceleration? _______________m/s2What is it in g's? __________g(b) What is the centripetal force acting on the car? ______________N

A snowboarder drops from rest into a halfpipe of radius R and slides down its frictionless surface to the bottom (see the figure below). Show that (a) the snowboarder's speed at the bottom of the halfpipe is v = V 2gR, (b) the snowboarder's centripetal acceleration at the bottom is a, = 2g, and (c) the normal force on the snowboarder at the bottom of the halfpipe has magnitude 3mg.

A 1,669 kg automobile is moving at a maximum speed of 27 m/s on a level circular track of radius 88 m. What is the coefficient of friction?

A child moves with a speed of 1.50 m/s when 9.0 m from the center of a merry-go-round. Calculate the centripetal acceleration of the child and the net horizontal force exerted on the child (m = 25.0 kg).

The friction force between the tires of a car and the road cause the centripetal acceleration of the car. If the coefficient of static friction is 0.5 and the car has a mass of 1200 kg, what is its centripetal acceleration?

What is the maximum speed with which a 1300 kg car can round a turn of radius 100.0 m if the coefficient of static friction between the tires and the road is 0.70? Does the result depend onthe mass of the car? What is the centripetal acceleration of the car at the maximum speed it can go around the turn? What is the corresponding centripetal force on the car?

A race car travels 38 m/s around a banked (45° with the horizontal) circular (radius = 0.39 km) track. What is the magnitude of the resultant force (in newtons) on the 75-kg driver of this car?

A 3.00 kg mass is placed on an incline of 35 degrees. What is the parallel component of the gravitational force directed down the incline?

Because of Earth’s rotation about its axis, a point on the equator has a centripetal acceleration of 0.034 0 m/s2 , whereas a point at the poles has no centripetal acceleration. (a) Show that, at the equator, the gravitational force on an object (the object’s true weight) must exceed the object’s apparent weight. (b) What are the apparent weights of a 75.0-kg person at the equator and at the poles? (Assume Earth is a uniform sphere and take g 5 9.800 m/s2.)