A 1.00-kg object A is connected with a massless string across a massless, frictionless pulley to a 2.00-kg object B. The smaller object rests on an inclined plane which is tilted at an angle of θ= 30.0°

a. What is the acceleration of the system? (Express your answer to three significant figures.)
b. What is the tension in the string? (Express your answer to three significant figures.)

Tension of the object is = 12.25

Acceleration of the system= 7.35 m/s²

Explanation:

Given

mass of A= 1kg

mass of B= 2kg

angle = 30°

let us consider the second mass m₂ of object B

m₂g - T = m₂a

T + 2a= 2 × 9.8

T + 2a=  19.6

where T denotes the tension of the cord and "a" being the acceleration.

Now considering mass m₁ of object A

T - m₁gsin θ  = m₁a

T - (9.8 × sin30°) = a

T - a = 4.9

on solving the above equations we get,

T + 2a = 19.6

T - a = 4.9

The tension T of the object is = 12.25

Acceleration a of the system= 7.35 m/s²

f=ma a/f/m a= 480/60 a = 8m/s/s that is quick wow

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.