I have some physics questions, can I get some help?

In the first two problems, the centripetal force is equal to the friction force.

Fc = m * v^2 ÷ r

Ff = μ * m * g

m * v^2 ÷ r = μ * m * g

v^2 = r * μ * g

v = √( r * μ * g)

For the first problem, v = √(81 * 0.6 * 9.8) = √476.28

This is approximately 21.8 m/s.

How large must the coefficient of static friction be between the tires and the road if a car is to round a

level curve of radius 135 m at a speed of 115 km/h ?

One km = 1000 meters

One hour = 3600 seconds

One km/h = 1000/3600 = 5/18 m/s

v = 115 * 5/18 = 575/18

This is approximately 31.9 m/s.

v^2 = r * μ * g

μ = v^2 ÷ (r * g)

μ = (57.5/18)^2 ÷ 1,323

This is approximately 0.77.

What is the magnitude of the acceleration of a speck of clay on the edge of a potter's wheel turning at 47 rpm (revolutions per minute) if the wheel's diameter is 39 cm ?

During one minute, the speck of clay moves a distance that is equal to the circumference of a circle.

d = 0.39 * π

Total distance = 47 * 0.39 * π = 18.33* π

One minute is 60 seconds.

v = 18.33* π ÷ 60 = 0.3055 * π

The velocity is approximately 0.097 m/s.

Centripetal acceleration = v^2 ÷ r = (0.3055 * π)^2 ÷ 0.195

The velocity is approximately 4.72 m/s^2.

A horizontal force of 340 N is exerted on a 2.5-kg ball as it rotates (at arm's length) uniformly in a horizontal circle of radius 0.90 m.Calculate the speed of the ball.

Fc = 340 ÷ 2.5 = 136 m/s

Fc = v^2 ÷ 0.9

v^2 ÷ 0.9 = 136

v = √(122.4

The velocity is approximately 11.1 m/s

A child sitting 1.60 m from the center of a merry-go-round moves with a speed of 1.65 m/s . Calculate the centripetal acceleration of the child.

Centripetal acceleration = 1.65^2 ÷ 1.60

This is approximately 1.7 m/s^2

Calculate the net horizontal force exerted on the child. (mass = 24.5 kg )

This is the centripetal force,

Fc = 24.5 * (1.65^2 ÷ 1.60)

This is approximately 41.7 N. I hope this is helpful for you in the future.