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: the exterior temperature of commercial jet planes as they leave the stratosphere and return to the ground increases steadily.

explanation:

the lower most layer of atmosphere is troposphere which extends from 0 to 12 km above the surface of the earth. the temperature initially decreases as distance is increased from the earth's surface. commercial jets fly in lower part of the stratosphere. the stratosphere extends upto 50 km. the temperature in stratosphere starts rising again due to presence of ozone which traps the uv rays.

thus, when the planes leave the stratosphere and return to the ground, the exterior temperature increases steadily.

the displacement produced by the body.  

the amount of force subjected to the body.  

the angle between the direction of force and displacement