A 15.0 kg crate, initially at rest, slides down a ramp 2.0 m long and inclined at an angle of 20.0° with the horizontal. Using the work-kinetic energy theorem and disregarding friction, find the velocity of the crate at the bottom of the ramp. (g = 9.81 m/s?)

The component of the crate's weight that is parallel to the ramp is the only force that acts in the direction of the crate's displacement. This component has a magnitude of

F = mg sin(20.0°) = (15.0 kg) (9.81 m/s^2) sin(20.0°) ≈ 50.3 N

Then the work done by this force on the crate as it slides down the ramp is

W = F d = (50.3 N) (2.0 m) ≈ 101 J

The work-energy theorem says that the total work done on the crate is equal to the change in its kinetic energy. Since it starts at rest, its initial kinetic energy is 0, so

W = K = 1/2 mv ^2

Solve for v :

v = √(2W/m) = √(2 (101 J) / (2.0 m)) ≈ 10.0 m/s

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.