A vertical spring with a spring constant of 430 N/m is mounted on the floor. From directly above the spring, which is unstrained, a 0.30 kg block is dropped from rest. It collides with and sticks to the spring, which is compressed by 2.2 cm in bringing the block to a momentary halt. Assuming air resistance is negligible, from what height (in cm) above the compressed spring was the block dropped

3.54 cm

Explanation:

From the question,

The potential energy of the block = energy needed to compress the spring

mgh = (ke²)/2 Equation 1

Where m = mass of the block , h = height from which the block was dropped, g = acceleration due to gravity, k = spring constant of the spring e = compression.

Firstly, we make h the subject of the equation

h = ke²/2mg Equation 2

Given: k = 430 N/m, e = 2.2 cm = 0.022 m, m = 0.3 kg, g = 9.8 m/s²

Substitute these values into equation 2

h = 430(0.022²)/2×0.3×9.8

h = 0.20812/5.88

h = 0.0354 m

h = 3.54 cm

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.

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