Acar accelerates uniformly from rest and reaches a speed of 21.9 m/s in 9.01 s. assume the diameter of a tire is 59.0 cm.(a) find the number of revolutions the tire makes during this motion, assuming that no slipping occurs. rev(b) what is the final angular speed of a tire in revolutions per second? /s

To solve this problem we will apply the linear motion kinematic equations. There we will find the acceleration, which will allow us to find the total displacement. From the relationship given between the circumference of a circle and the total length, we can determine the number of revolutions.

For the second part we will use the relationship between angular and linear velocities given through the radius of the circumference to calculate the final value.

PART A) Calculate the acceleration of the car

Where,

a = Acceleration

= Initial velocity

v = Final velocity

t = time

Replacing we have that

Calculate the displacement

Replacing

Calculate the circumference of wheel:

Where,

d = Diameter

= Circumference

Calculate the number of revolutions

Therefore the number of revolutions is 53.32rev

PART B )

Calculate the angular velocity

Therefore the final angular speed is 74.23rad/s

when insulating materials rub against each other, they may become electrically charged . electrons , which are negatively charged, may be 'rubbed off' one material and on to the other. the material that gains electrons becomes negatively charged. the material that loses electrons is left with a positive charge.

the answer is a. at the equator