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kicyunya [14]
3 years ago
7

A jet airplane is in level flight. The mass of the airplane is m=8950 kg. The airplane travels at a constant speed around a circ

ular path of radius R=9.33 mi and makes one revolution every T=0.123 h. Given that the lift force acts perpendicularly upward from the plane defined by the wings, what is the magnitude of the lift force acting on the airplane?
Physics
1 answer:
Mice21 [21]3 years ago
7 0

Answer:

The net force is 91780.8 N.

Explanation:

mass, m = 8950 kg

Radius, R = 9.33 miles = 15015.2 m

Time, T = 0.123 h = 442.8 s

There are two forces acting on the plane.

Horizontal force is the centripetal force and the vertical force is the weight.

Fx =m R w^2\\\\Fx = m R \frac{4\pi^2}{T^2}\\\\Fx = 8950\times 15015.2\times \frac{4\times 3.14\times 3.14}{442.8\times 442.8}\\\\Fx = 27030.8 N \\\\Fy = m g \\\\ Fy = 8950\times 9.8 \\\\Fy = 87710 N

The net force is

F = \sqrt{Fx^2 + Fy^2}\\\\F = \sqrt {27030.8^2 + 87710^2}\\\\F = 91780.8 N

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The momentum p of an object is equal its mass m times its velocity v. That is: \vec{p} = m \cdot \vec{v}.

Assume that the air resistance on this ball is negligible. If that's the case, then the ball would accelerate downwards towards the ground at a constant g \approx -10\; \rm m \cdot s^{-2}. In other words, its velocity would become approximately 10\; \rm m \cdot s^{-1} more negative every second.

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3 years ago
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