Afloor polisher has a rotating disk that has a 15-cm radius. the disk rotates at a constant angular velocity of 1.3 rev/s and is covered with a soft material that does the polishing. an operator holds the polisher in one place for 59 s, in order to buff an especially scuffed area of the floor. how far does a spot on the outer edge of the disk move during this time?

d = 7228.8 cm

Explanation:

Because the  disk  uniformly rotates at a constant angular velocity we apply the following formula to calculate the tangential velocity of a spot on the outer edge of the disk :

v = ω*R   Formula 1

Where:

v :  tangential velocity (m/s)

ω : angular velocity (rad/s)

Data

ω =  1.3 rev/s =1.3* 2π rad/s =8.16 rad/s

R = 15 cm

v = ω*R

v = 8.16 * 15 = 122.5 cm/s

and We apply the following formula to calculate the linear displacement of a spot on the outer edge of the disk :  

d = v*t Formula 2

Where:

d : linear displacement (m/s)

v :  tangential velocity (m/s)

t = time interval (s)

Data

v = 122.5 cm/s

t = 59 s

d = v*t

d = (122.5 )*(59) = 7228.8 cm

answer

c, on a nylon carpet in a dry area. give me brainliest

explanation:

the answer is b i think