On an icy winter day, the coefficient of friction between the tires of a car and a roadway is reduced to 1/4 its value on a dry day. as a result, the maximum speed vmax dry at which the car can safely negotiate a curve of radius r is reduced. the new value for this speed is what percentage of its value on a dry day?

50%

Explanation:

For the particular case we assume that the centripetal force is equal to the frictional force. Starting from this, we realize that while the car takes a curve, which allows it to not leave the track, it is the friction force that adheres it to the floor, in this way,

The values of these equations are given by,

Equating the terms

Both gravity and radius are values that during the trajectory with constants, so

The difference comes in winter and dry, so,

\mu_{dry} = \mu

as in winter it is a quarter, you have to,

\mu_{winter} = \frac{1}{4} \mu_{dry} = \frac{1}{4} \mu

Performing the proportion we have to

We can conclude that the new value for this speed is 50% of its value on a dry day.

temp rises, and expansion

answer; equivalent to a one(1) horsepower motor;