After coming down a steep hill at a constant speed of 43 m/s, a car travels along the circumference of a vertical circle of radius 618 m until it begins to climb another hill. r x what is the magnitude of the net force on the 34 kg driver of the car at the lowest point on this circular path? answer in units of kn.

F=0.101 kN

Explanation:

Newton's 2nd law, F = ma, but this is circular path, the acceleration (a) is the centripetal acceleration.

a = (v²) / r  

F = (m×v²) /r

F=(34 kg)×(43 m/s)² / 618 m

F=101.72 N

To convert Newtons into kilo-Newtons divide it 1000.

F=0.101 kN

the answer is f1 and d2 because if you put equal force on those two points the beam would balance trust me on this