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

f=gm1m2/d^2 it d

f is force g is gravitaiton constant, m1 is mass of first object, m2 is mass of second object. d is distance between two object

jupiter is the largest planet in our solar system, having a mass and radius that are respectively, 318 and 11.2 times that of earth. suppose that an object falls from rest near the surface of each planet and that the acceleration due to gravity remains constant during the fall. each object falls the same distance before striking the ground. determine the ratio of the time of fall on jupiter to that on earth.