A 1502.7 kg car is traveling at 33.1 m/s when the driver takes his foot off the gas pedal. It
takes 4.7 s for the car to slow down to 20 m/s.
How large is the net force slowing the car?
Answer in units of N.

By definition of average acceleration,

a = (20 m/s - 33.1 m/s) / (4.7 s) ≈ -2.78 m/s²

Vertically, the car is in equilibrium, so the net force is equal to the friction force in the direction opposite the car's motion:

∑ F = (1502.7 kg) (-2.78 m/s²) ≈ -4188.38 N ≈ -4200 N

If you just want the magnitude, drop the negative sign.

a and c

explanation:

both a and c are correct answers to my knowledge, it's been a bit since i've done anything related to magnetic fields, but it should be right.

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.