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Radda [10]
3 years ago
5

An airplane flies in a loop (a circular path in a vertical plane) of radius 200 m . The pilot's head always points toward the ce

nter of the loop. The speed of the airplane is not constant; the airplane goes slowest at the top of the loop and fastest at the bottom
(a) What is the speed of the airplane at the top of the loop, where the pilot feels weightless?
(b) What is the apparent weight of the pilot at the bottom of the loop, where the speed of the air-plane is 280 km/h? His true weight is 700 N.
Physics
1 answer:
Naya [18.7K]3 years ago
8 0

Answer:

Explanation:

a ) for the pilot to feel weightless , his weight will provide the centripetal force . The reaction force from airplane will be zero.

mg = mv² / r

g = v²/r

v² = gr

= 9.8 x 200

v = 44.27 m / s .

b )

v = 280 km / h

v = 77.77 m /s

If R be the reaction force of floor of airplane on him

R - mg = mv² / r

R = mg + m v² / r

= 700 + 70 x  77.77² / 200  ( m = mg / g = 700/ 10 = 70 )

= 700 + 2116.86

= 2816.86  N .

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Complete question:

The exit nozzle in a jet engine receives air at 1200 K, 150 kPa with negligible kinetic energy. The exit pressure is 80 kPa, and the process is reversible and adiabatic. Use constant specific heat at 300 K to find the exit velocity.

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