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Neporo4naja [7]
3 years ago
14

Freeeeeeeeeeeeeeeeeeeeeeeee points

Physics
2 answers:
Klio2033 [76]3 years ago
5 0
Thanks!!!!!!!!!!!!!!!
andreyandreev [35.5K]3 years ago
3 0

Answer:

yehyehhfujghnnkoohu

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An airplane is traveling at 250 m/s in level flight. If the airplane is to make a change in direction, it must travel is a horiz
Nonamiya [84]

Answer:

The radius of curvature of the curved path of the airplane is 23784.356 meters (23.784 kilometers).

Explanation:

We assume that airplane can be represented as a particle. The free body diagram of the vehicle is presented below as attachment, whose variables are:

W - Weight of the airplane, measured in newtons.

F - Lift, measured in newtons.

\theta - Banking angle, measured in sexagesimal degrees.

The equations of equilibrium associated with the airplane are, respectively:

\Sigma F_{r} = F\cdot \sin \theta = m\cdot \frac{v^{2}}{R} (Eq. 1)

\Sigma F_{z} = F\cdot \cos \theta - W = 0 (Eq. 2)

From (Eq. 2):

F = \frac{W}{\cos \theta}

In (Eq. 1):

W\cdot \tan \theta = m\cdot \frac{v^{2}}{R}

By using the definition of weight, we eliminate the mass of the airplane:

g\cdot \tan \theta = \frac{v^{2}}{R}

Where:

g - Gravitational acceleration, measured in meters per square second.

v - Speed, measured in meters per second.

R - Radius of curvature, measured in meters.

Lastly, we clear the radius of curvature with the expression:

R = \frac{v^{2}}{g\cdot \tan \theta}

If we know that v = 250\,\frac{m}{s}, g = 9.807\,\frac{m}{s^{2}} and \theta = 15^{\circ}, the radius of curvature is:

R = \frac{\left(250\,\frac{m}{s} \right)^{2}}{\left(9.807\,\frac{m}{s^{2}} \right)\cdot \tan 15^{\circ}}

R = 23784.356\,m

The radius of curvature of the curved path of the airplane is 23784.356 meters (23.784 kilometers).

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