For a car moving along an unbanked turn, the frictional force provides the centripetal force required to keep the car in circular motion. Therefore, we can write:

$\mu mg = m\frac{v^2}{r}$

where the term on the left is the frictional force while the term on the right is the centripetal force, and where

$\mu=0.40$ is the coefficient of friction between the tires and the road

m is the mass of the car

$g=9.8 m/s^2$ is the acceleration of gravity

v is the speed of the car

r = 150 m is the radius of the curve

Solving for v, we find the (maximum) speed at which the car can move along the turn:

$v=\sqrt{\mu gr}=\sqrt{(0.40)(9.8)(150)}=24.3 m/s$

For speed larger than this value, the frictional force is no longer enough to keep the car along the turn.

Learn more about circular motion:

brainly.com/question/2562955

brainly.com/question/6372960

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8 0

3 years ago

A 31 cm tall object is placed in front of a concave mirror with a radius of 34 cm. The distance of the object to the mirror is 9

Setler [38]

Focal length is the distance between the center of a convex lens or a concave mirror and the focal point of the lens or mirror — the point where parallel rays of light meet, or converge. From the optics the focal length of the mirror can be defined as the radius of the mirror divided between two, or in other words, half the radius of the mirror.

$f = \frac{R}{2}$

$f = \frac{34}{2}$

$f = 17cm$

Therefore the focal length of the mirror is 17cm

5 0

3 years ago