Jack and Jill are maneuvering a 3100 kg boat near a dock. Initially the boat's position is < 2, 0, 3 > m and its speed is 1.6 m/s. As the boat moves to position < 6, 0, 1 > m, Jack exerts a force of < -440, 0, 220 > N, and Jill exerts a force of < 170, 0, 340 > N.

How much work does Jack do?
How much work does Jill do?

(a) work done by Jack is -2200J

(b) work done by Jill is 0

Explanation:

Given;

Initially boat's position =  < 2, 0, 3 > m

Final boat's position = < 6, 0, 1 > m

change in position = < 4, 0, -2 >

Force exerted by Jack =  < -440, 0, 220 > N

Force exerted by Jill = < 170, 0, 340 > N

Part (a) How much work does Jack do?

work done = force x distance

Jack's work = < -440, 0, 220 > N x < 4, 0, -2 > m

                    = [(-440*4) + (0*0) + (220*-2)]

                    = -1760 J - 440 J

                    = -2200J

Part (b) How much work does Jill do?

Jill's work = < 170, 0, 340 > N x < 4, 0, -2> m

                 = [(170*4) + (0*0) + (340*-2)]

                  = 680 J - 680 J

                   = 0

6/12=1/2

step-by-step explanation:

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.