A1.1 g marble is fired vertically upward using a spring gun. the spring must be compressed 6.2 cm if the marble is to just reach a target 16 m above the marble's position on the compressed spring. (a) what is the change δug in the gravitational potential energy of the marble-earth system during the 16 m ascent

ΔU = 0.17 J

Explanation:

m = 1.1 g = 1.1*10⁻³ Kg

h₁ = 6.2 cm = 0.062 m

h₂ = 16.062 m

ΔU = ?

We apply the formulas

U₁ = m*g*h₁

U₂ = m*g*h₂

ΔU = U₂ - U₁ = m*g*h₂ - m*g*h₁ = m*h*(h₂ - h₁)

⇒  ΔU = (1.1*10⁻³ Kg)*(9.81 m/s²)*(16.062 m - 0.062m)

⇒  ΔU =  0.17 J

what is the question because that is not a question.

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.