A car travels at a constant speed around a circular track whose radius is 3.22 km. The car goes once around the track in 248 s. What is the magnitude of the centripetal acceleration of the car

a = 1,838.17m/s²

Explanation:

centripetal acceleration a = v²/r

v is velocity of the car

r is the radius = 3.22km = 3220m

time t = 248

First is to find to velocity using the equation of motion:

v = u+gt

v = 0+(9.81)(248)

v = 2,432.88m/s

a = 2,432.88²/3220

a = 5,918,905.0944/3220

a = 1,838.17m/s²

The magnitude of the centripetal acceleration of the car is 1,838.17m/s²

air exerts forces on falling objects near earth’s surface.

explanation:

the term 'free falls' refer to object that are falling towards the earth, in a situation in which gravity is the only force acting on the object.

for an object in free fall, the acceleration is constant (g, the gravitational acceleration), and so the velocity of the object constantly increases.

for objects near earth's surface, there is always another force acting on the object: the resistance due to the air. this force is opposed to the direction of fall of the object, and it is proportional to the velocity of the object, so as the object falls down, this force increases up to a point when it balances the force of gravity, and the object continues its motion at constant velocity (called terminal velocity). since in this case gravity is not the only force acting on the object, we are not in a situation of free fall.

i think the answer is a and b

i wish you luck