Acord is wrapped around the rim of a solid uniform wheel 0.250 m in radius and of mass 9.20 kg. a steady horizontal pull of 40.0 n to the right is exerted on the cord, pulling it off tangentially from the wheel. the wheel is mounted on frictionless bearings on a horizontal axle through its center. (a) compute the angular acceleration of the wh

The angular acceleration of the wheel is 4.33m/s^2

Explanation:

F=40N, r=0.25m, m=9.20kg

V = √(Fr ÷ m) = √(40×0.25 ÷ 9.2) = √1.09 = 1.04m/s

Angular acceleration (a) = V^2/r = 1.04^2/0.25 = 1.0816/0.25 = 4.33m/s^2

the minimum constant speed at which she can catch the train is 2.7 m/s

given:

acceleration = 0.6

time = 4.5 s

to find:

velocity = ?

formula used:

from 1st kinematic equation we get,

v = u + at

u = initial speed of train = 0 m/s

v = final speed of the train

a = acceleration of the train

t = time

solution:

from 1st kinematic equation we get,

v = u + at

u = initial speed of train = 0 m/s

v = final speed of the train

a = acceleration of the train

t = time

v = 0 + (0.6 × 4.5)

v = 2.7 m/s

the minimum constant speed at which she can catch the train is 2.7 m/s

the answer is 100. i just did this lesson and got a 100% on it. : )

-fishtank