A 9.0 kg potted plant slides down a 25.0º incline with an acceleration of 2.4 m/s2. What is the coefficient of kinetic friction between the potted plant and the incline?

0.196

Explanation:

According to Newton's second law

\sumFx = ma

Fm-Ff = ma

Fm is the moving force

Ff is the frictional force

m is the mass = 9kg

a is the acceleration = 2.4m/s²

Fm = Wsin theta

Fm = 9(9.8)sin25°

Fm = 37.28N

Ff is the frictional force = nmgcos theta

Ff = n(9)(9.8)cos25°

Ff = 79.94n

Substitute the given values into the formula

37.28-79.94n = 9(2.4)

-79.94n = 21.6-37.28

-79.94n = -15.68

n = 15.68/79.94

n = 0.196

Hence the coefficient of kinetic friction between the potted plant and the incline is 0.196

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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.