The propeller of a World War II fighter plane is 2.30 m in diameter and spins at 1200 rev/min. What is the centripetal acceleration of the propeller tip? Calculate it in meters per second squared and convert to multiples of g.

Ac = 36320 m/s^2 = 3702 g

Explanation:

First let's find the linear velocity of the propeller tip.

The angular velocity is 1200 rev/min, which is 20 rev/s

One rev is 2*pi radians, so 20 rev/sec = 40*pi rad/s

To find linear velocity, we multiply the angular velocity by the radius, so:

V = 40*pi * 2.3 = 92*pi m/s

The centripetal acceleration is given by Ac = V^2/r, being r the radius. So:

Ac = (92*pi)^2/2.3 = 36320 m/s^2

If g is the gravity acceleration = 9.81 m/s^2, we can find Ac in multiples of g by dividing their values:

Ac = 36320/9.81 = 3702 g

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.