The radius of the loop is 17.6 m. We need to find the minimum speed must the car traverse the loop so that the rider does not fall out while upside down at the top.

We know that, mg be the weight of car and rider, which is equal to the centripetal force.

$mg = \dfrac{mv^2}{r}\\v = \sqrt{rg}\\v = \sqrt{17.6\times 9.8}\\v = 13.13\ m/s\\$

So, the minimum speed must the car must be 13.13 m/s.

5 0

2 years ago

A truck is using a hook to tow a car whose mass is one quarter that of the truck. If the force exerted by the truck on the car i

strojnjashka [21]

Answer:

The car exerts a force on the truck which is c.) 6000 N.

Explanation:

If the truck exerts a force of 6000 N on the car then according to Newton's Third Law of Motion that every action has an equal and opposite reaction we can say that the force exerted by the car on the truck is 6000 N.

The only reason the truck is able to move the car is because of the frictional forces. That is frictional force of the trucks tires on the surface of the road and the frictional forces of the cars tire on the surface of the road, and since these forces are not equal this therefore gives rise to motion.

7 0

3 years ago