Afloor polisher has a rotating disk that has a 19-cm radius. the disk rotates at a constant angular velocity of 1.6 rev/s and is covered with a soft material that does the polishing. an operator holds the polisher in one place for 43 s, in order to buff an especially scuffed area of the floor. how far (in meters) does a spot on the outer edge of the disk move during this time?

Arc length(S) = 82.14m

Step-by-step explanation:

Given: Radius=19cm= 0.19m , angular velocity =1.6 rad/s , time = 43secs

As the disk is rotating,angular displacement=s/r

Where r=radius and s=arc length

The rate of change of the angular displacement represents the angular velocity .w=angular displacement/ time

w=(1.6)(2×3.142)

w=10.054 rad/s

Angular displacement =at

Angular displacement =10.054 ×43=432.32rad

S=angular displacement × radius

S= 432.32 ×0.19

S= 82.14m

the answer is 5636.7

the answer is 62