To test the performance of its tires, a car
travels along a perfectly flat (no banking) circular track of radius 333 m. the car increases
its speed at uniform rate of

(use image)

until the tires start to skid.
if the tires start to skid when the car reaches
a speed of 31 m/s, what is the coefficient of
static friction between the tires and the road?
the acceleration of gravity is 9.8 m/s^2

0.45

Explanation:

Sum of forces in the y direction:

∑F = ma

N − mg = 0

N = mg

There are friction forces in two directions: centripetal and tangential.  The centripetal acceleration is:

ac = v² / r

ac = (31 m/s)² / 333 m

ac = 2.89 m/s²

The total acceleration is:

a = √(ac² + at²)

a = √((2.89 m/s²)² + (3.32 m/s²)²)

a = 4.40 m/s²

Sum of forces:

∑F = ma

Nμ = ma

mgμ = ma

μ = a / g

μ = 4.40 m/s² / 9.8 m/s²

μ = 0.45

the chart show the masses and velocities of two colliding objects that stick together after a collision.

according to the law of conservation of momentum, the momentum of the object after the collision is  

answer

1,500

kg • m/s.

so easy. just google the electromagnetic waves and it’ll pop up.