A 13,900 N car traveling at 40.0 km/h rounds a curve of radius 1.80 ✕ 102 m.
(a) What is the centripetal acceleration of the car?
m/s2
(b) What is the centripetal force?
N
(c) What is the minimum coefficient of static friction between the tires and the road that will allow the car to round the curve safely?

a)

b) 964.5 N

c) 0.069

Explanation:

a)

When an object is moving in a circular motion, the direction of its velocity is changing - therefore, it has an acceleration towards the center of the circle, called centripetal acceleration.

The magnitude of the centripetal acceleration is given by

where

v is the speed of the object

r is the radius of the circle

For the car in this problem:

v = 40.0 km/h = 11.1 m/s is the speed

r = 180 m is the radius of the circle

Substituting, we find the acceleration:

b)

The centripetal force is the force that keeps the object along its circular motion. It also acts towards the center of the circle, and it is given by

where

m is the mass of the object

a is the centripetal acceleration

Here the weight of the car is

where

is the acceleration due to gravity

So the mass is

Therefore, the centripetal force is

c)

In this case, the force of static friction between the tires and the road provides the required centripetal force to keep the car in circular motion. This force is given by:

where

is the coefficient of friction

Equating the frictional force to the centripetal force,

So we get:

And substitutng:

(centripetal acceleration)

We find:

im srry but ur wrong the answer is d a smooth rock, some red dirt, and an empty bird's nest.

305.7

explanation:

density =m/v

density =917/3.00

density = 305.

approximately = 305.7

in four significant figures