A block with a mass of 10 kg is pulled at a 20 degree angle with a force of 75 N. The force of friction acting on the block is 15 N. What is the acceleration of the block in m/s?? Round your answer to the nearest tenth. (Numbers Only!
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There are 4 forces acting on the block:

• its weight (magnitude w = m g = 98 N, where m = 10 kg and g = 9.80 m/s²), pointed downward

• the normal force (mag. n), pointed upward

• friction (mag. f = 15 N), pointing opposite the direction of motion

• the pulling force (mag. p = 75 N), pointing at a 20° angle with the positive horizontal direction

Let the direction in which the block is being pulled be the positive horizontal direction, and upward the positive vertical direction. Split up each force into their horizontal and vertical components, then apply Newton's second law. We have a net horizontal force of

∑ F = p cos(20°) - f = m a

where a is the acceleration of the block. The net vertical force is 0, since the block gets dragged along the surface and doesn't move up or down (presumably).

Solve for a :

(75 N) cos(20°) - 15 N = (10 kg) a

a ≈ 5.4 m/s²

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 a. hope this .