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Scott, MIT Graduate
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A stunt pilot who has been diving her airplane vertically pulls out

Resolved Question:

A 53.0 kg stunt pilot who has been diving her airplane vertically pulls out of the dive by changing her course to a circle in a vertical plane.
a) If the plane''s speed at the lowest point of the circle is 100 m/s, what should the minimum radius of the circle be in order for the acceleration at this point not to exceed 4.00g ?
b) What is the apparent weight of the pilot at the lowest point of the pullout?

Submitted: 9 years ago.
Category: Homework
Expert:  Scott replied 9 years ago.

Hi there!

Part A:

The formula for this is:

a = v^2/r

Plug in our values:

9.81*4 = 100^2/r

Multiply by r:

r*9.81*4 = 100^2

Divide by 9.81*4:

r = 100^2/(9.81*4)

r = 254.84 meters

Part B:

Use F = ma:

F = ma(dive) + ma(gravity)

F = 53*4*9.81 + 53*9.81

F = 2599.65 N

Let me know if you have any questions.


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Customer: replied 9 years ago.
How do you know for part b to add ma(dive) and ma (gravity) to find apparent weight? I don't understand.
Expert:  Scott replied 9 years ago.

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