Calculate the linear acceleration of a car, the 0.270-m radius tires of which have an angular acceleration of 15.0 rad/s2. Assume no slippage and give your answer in m/s2. m/s2 (b) How many revolutions do the tires make in 2.50 s if they start from rest? rev (c) What is their final angular velocity in rad/s? rad/s (d) What is the final velocity of the car in m/s? m/s

A) 4.05 m/s²

B) 7.46 revolutions

C) 37.5 rad/s

D) 10.125 m/s

Explanation:

A) Radius; r = 0.27m

Angular acceleration; α = 15 rad/s²

Now, the formula for linear acceleration is given by;

a = rα

Where r is radius and α is angular acceleration while a is linear acceleration.

Thus,

a = 0.27 x 15

a = 4.05 m/s²

B) First, Let's find the final velocity in rev/s using the equation of motion; V = u + αt

Where α is angular acceleration.

V = 0 + (15 x 2.5) = 37.5 rad/s

Now, we are calculating in revolutions, so let's convert the velocity to rev/s

1 rad/s = 1/2π rev/s

Thus, 37.5 rad/s = 37.5/2π revs/s = 5.968 rev/s

Now,average velocity = (v + u)/2 = (5.968 + 0)/2 = 5.968/2 = 2.984 revs/s

Thus, number of revolutions will be= average velocity x t =

2.984 rev/s x 2.5 s = 7.46 revolutions

C) we will get the final angular velocity from;

ω_f = ω_o + αt

Since it starts from rest, initial angular velocity ω_o = 0

Thus,

ω_f = 15 x 2.5 = 37.5 rad/s

D) The final velocity is given by;

v_f = ω_f•r = 37.5 x 0.27 = 10.125 m/s

weight of the box is given as

here this is placed on the inclined of angle 42 degree

now this weight will have two components one is along the incline other is perpendicular to it

so here we need to apply a force parallel to incline opposite to its component of weight

so here the component of weight is given as

here we have

so it requires 75 n force to move it upwards