Acentrifuge in a biology laboratory rotates at an angular speed of 3,400 rev/min. when switched off, it rotates 48.0 times before coming to rest. find the constant angular acceleration of the centrifuge (in rad/s2). consider the direction of the initial angular velocity to be the positive direction, and include the appropriate sign in your result.

- 210 rad/s²

Explanation:

n = frequency of rotation = 3400/60 = 170/3 per sec.

angular velocity  ω ( 0 ) at time 0  = 2π n = 2π x 170/3

angular velocity at time t = ω(t) = 0

now, ω²( t) = w²(o) + 2α Φ ( α = angular acceleration and Φ = angular displacement) = 2π x 48 rad.

0 = ( 2π x 170/3 )² + 2α x 48 x 2π

α = - (2π x 170 x 170 )/ (3 x 3 x 2 x 48 ) = 210 rad / s²

the answer to this question is 60 n

0.863333

86%

hope this