Aroller coaster car is loaded with passengers and has a mass of 500 kg along with a speed of 18 meters/second at the dip. the radius of curvature of the track at the bottom point of the dip is 12 meters (gravity = 9.8 meters/second2). what force is exerted on the roller coaster car by the track at the bottom of the dip?

The roller coaster is moving in a circular path, so the force that must be computed is the centripetal force. The centripetal force is the force acting on an object undergoing circular motion and is directed towards the center of the circular path. This is computed using:
F = mv²/r
F = (500 * 18²) / 12
F = 13,500 N

Now, at the bottom of the track, the track is also supporting the weight of the car and its passengers, which is:
W = mg
W = 500 * 9.81
W = 4,905 N

The total reactive force exerted by the track to counter the centripetal force and the weight of the car is:
F = 13,500 + 4,905
F = 18,405 Newtons

Ibelieve the answer is either b or c, but i'm leaning more towards c.hope my answer has you : )