**Angular Motion Problem**Dario, a prep cook at an Italian restaurant, spins a salad spinner and observes that it rotates with a constant speed of 20.0 times in 5.00 seconds before he stops spinning it. The salad spinner rotates 6.00 more times before it comes to rest. Assume that the spinner slows down with constant angular acceleration.**Question:** What is the magnitude of the angular acceleration of the salad spinner as it slows down? *Express your answer numerically in radians per second per second.***Answer Input:** There is a text box provided to enter the answer for angular acceleration (α) in radians per second squared (radians/s²).**Hints Available:** You can view available hints by expanding the "View Available Hint(s)" section.This exercise is aimed at understanding angular motion and calculating angular acceleration in physics.

The constant angular speed of the salad spinner is, w=(20×2pi)/5 rad/s= 8pi rad/s.( 1 rotation= 2pi rad, here pi=22/7) By applying a constant angular acceleration of A on the salad spinner, it comes to rest by traversing an angular displacement "theta". So its final angular speed, wf=0. Acording to the motion, wf2=w2-2×A×theta. So, 0=w2-2×A×theta Or, A=w2/(2×theta)Here, w=8pi rad/s, theta=6×2pi rad= 12pi rad So, A=(8pi)2/(2×12pi)= 8pi/3 So the constant acceleration which slows down the salad spinner is 8pi/3 rad/s2 , or 8.38 rad/s2.(approximately)