A jet plane flying 600 m/s experiences an acceleration of 4.0 g when pulling out of a circular dive. What is the radius of curvature of the circular part of the path in which the plane is flying?

9188 m is the radius of curvature of the circular part of the path in which the plane is flying.

Explanation:

When an air-plane dives, to exit the dive, it must follow a circular path to move from vertical downward to horizontal motion. To draw this circular movement, the pilot manipulates the wings so that the air force and the lifting forces create a force centered in the center that accelerates the plane towards the center of the wheel.

Given data:

speed of the plane = 600 m/s

acceleration = 4 g

We need to find radius of curvature of loop (r)

we know, angular acceleration ----> eq 1

v = r ω ===> ---- eq 2

putting eq 2 in equation (1), we get

Substitute the given values, we get

-----eq 3

Substitute value in eq 3 to eq. 2, we get

as we know our atmosphere or air have different levels. every level has other gasses, and every gas has another permeability. when the light comes from lower permeability to higher permeability, it changes its direction. so that means that when the light comes from the moon to the earth, it changes it's direction many more times. it basically means that the shape of the light part of the moon is changed on the earth.

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