A metal worker wants to heat commercial bronze (a copper alloy) by using an air cannon pointed downward. If she wants to increase the temperature of the bronze by 1.5 o F and the cannon is 20 feet above the ground, what would be exit velocity from the cannon need to be (in ft/s)? Neglect air resistance. [V = 78.9 ft/s]

exit velocity from the cannon is 78.9 ft/s

Explanation:

Given that;

increase the temperature of the bronze by 1.5 o F

cannon is 20 feet above the ground.

we know that

( Kinetic energy + potential energy )at state1 = Internal energy

1/2mv² + mgh = mCpΔT

V²/2 + gh = CpΔT

V²/2 = CpΔT - gh

V² = 2(CpΔT - gh)

V = √(2(CpΔT - gh))

So we substitute

V = √(2((0.1 × 778.16 × 32.174 × 1.5) - (32.17 × 20)))

V = √(2(3755.4779 - 643.4))

V = √(2 × 3112.0779)

V = √6224.1558

V = 78.89 ≈ 78.9

Therefore exit velocity from the cannon is 78.9 ft/s

the gas collects in coils that then becomes a liquid that then flows through the coils inside the refrigerator

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.