Question

A jet plane flying 600 m/s experiences an acceleration of 4g when pulling out of the dive. What is the radius of curvature of the loop in which the plane is flying?

Answer 1

Answer:

91.84 m/s²

Explanation:

velocity, v = 600 m/s

acceleration, a = 4 g = 4 x 9.8 = 39.2 m/s^2

Let the radius of the loop is r.

he experiences a centripetal force.

centripetal acceleration,

a = v² / r

39.2 x r = 600 x 600

r = 3600 / 39.2

r = 91.84 m/s²

Thus, the radius of the loop is 91.84 m/s².