Nelectric motor rotating a workshop grinding wheel at a rate of 94 rev/min is switched off. assume constant negative angular acceleration of magnitude 1.70 rad/s. (a) how long does it take for the grinding wheel to stop? (b) through how many radians has the wheel turned during the interval found in (a)? rad

a) It takes 5.79 seconds for the grinding wheel to stop.

b) In an interval of 5.79 seconds it rotates 28.48 rad.

Explanation:

a) We have equation of motion v = u + at

  Here v = 0 rad/s, a = -1.70 rad/s², u = 94 rev/min = 9.84 rad/s

  Substituting

         0 = 9.84 - 1.70 x t

          t = 5.79 seconds.

It takes 5.79 seconds for the grinding wheel to stop.

b) We have equation of motion v² = u² + 2as

      v = 0 rad/s, a = -1.70 rad/s², u = 9.84 rad/s

  Substituting

         0² = 9.84² - 2 x 1.70 x s

          s = 28.48 rad

 So in an interval of 5.79 seconds it rotates 28.48 rad.

i think it’s a. this is because if they measured the mass of something with a mass of 150g, then their final answer would have to be approximately 150g. though all of their answers are almost 150, rosa and tran have the closest answers (or the exact answer) to the accurate one.

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