A car accelerates from 14.0m/s to 21m/s in 6.0s. What was its acceleration? How far did it travel in this time? Assume constant acceleration.

Acceleration (a) = 1.167 m/s²

Distance travel = 105 m (Approx)

Explanation:

Given:

Initial speed (u) = 14 m/s

Final speed (v) = 21 m/s

Time taken (t) = 6 sec

Find:

Acceleration

Distance travel

Computation:

v = u + at

21 = 14 + a(6)

7 = 6a

Acceleration (a) = 1.167 m/s²

S = ut + (1/2)(a)(t²)

S = (14)(6) + (1/2)(1.167)(6²)

S = 84 + 21

S = 105 m (Approx)

Distance travel = 105 m (Approx)

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.

the answer is f1 and d2 because if you put equal force on those two points the beam would balance trust me on this