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 th
e loop in which the plane is flying?
1 answer:
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.
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