Assume a small 11 kg crate is attached to a parachute. the chute pulls up on the crate with 150 n.
a. find fnet
b. find a
m/s²
c. find v 5 seconds after release
m/s
i'm really stuck on this.

Explanation:

a. The net force is the upward force of the chute minus the weight of the crate.

∑F = F − mg

∑F = 150 N − (11 kg) (9.8 m/s²)

∑F = 42.2 N

b. From Newton's second law, the net force equals the mass times acceleration:

∑F = ma

42.2 N = (11 kg) a

a = 3.84 m/s²

c. Acceleration is the change in velocity over change in time.  Assuming the crate is released from rest:

v = at + v₀

v = (3.84 m/s²) (5 s) + (0 m/s)

v = 19.2 m/s

i need the answer to this too but idk it ; -;

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.