Look at Concept Simulation 5.2 to review the concepts involved in this question. Two cars are identical, except for the type of tread design on their tires. The cars are driven at the same speed and enter the same unbanked horizontal turn. Car A cannot negotiate the turn, but car B can. Which tread design, the one on car A or the one on car B, yields a larger coefficient of static friction between the tires and the road?

Tread design on car B would yield a larger coefficient of static friction between the tires and the road

Explanation:

The car model using the coefficient of static friction doesn't work well with tires. A higher coefficient of static friction would require more force to cause a loss of attraction.

The static frictional force helps to keep the unbanked horizontal turn. This means that the frictional force is the centripetal force.

The tread design of car B ensures that the centripetal force is enough to negotiate the turn. On the other hand, the tread design of car A does not provide the necessary centripetal force, hence car A is unable to negotiate the turn.

Therefore, tread design on car B would yield a larger coefficient of static friction between the tires and the road.

the answer is c.)

explanation:

snell's law

refractve index = sin incidence / sin refraction