**Flight Physics Problem****Problem 6:**An airplane is flying in a horizontal circle at a speed of 490 km/h. If its wings are tilted at an angle \( \theta = 45^\circ \) to the horizontal, what is the radius of the circle in which the plane is flying? Assume that the required force is provided entirely by an "aerodynamic lift" that is perpendicular to the wing surface.---To solve this, we need to understand the relationship between the forces acting on the airplane and the radius of the circular path it takes.1. **Aerodynamic Lift**: The lift force \( L \) acts perpendicular to the wing surface.2. **Centripetal Force**: A portion of the lift force provides the necessary centripetal force \( F_c \) to maintain circular motion.The centripetal force \( F_c \) needed for circular motion is given by:\[ F_c = \frac{mv^2}{r} \]Where:- \( m \) is the mass of the airplane- \( v = 490 \, \text{km/h} \) (convert this to meters per second, \( v = 136.11 \, \text{m/s} \))- \( r \) is the radius of the circleThe component of the lift force providing the centripetal force is \( L \sin\theta \). Since the plane is in horizontal flight, the vertical component \( L \cos\theta \) balances the weight \( mg \) of the airplane.By equating the centripetal force to the horizontal component of the lift force:\[ \frac{mv^2}{r} = L \sin\theta \]And since \( L \cos\theta = mg \):\[ L = \frac{mg}{\cos\theta} \]By substituting \( L \) into the centripetal force equation:\[ \frac{mv^2}{r} = \left(\frac{mg}{\cos\theta}\right) \sin\theta \]The mass \( m \) cancels out:\[ \frac{v^2}{r} = \frac{g \sin\theta}{\cos\theta} \]Simplify using \( \tan\theta \):\[ \frac{v^2}{r} = g \tan\theta \]Solving for \( r

As we know, Let F₁ is the force of aerodynamic lift. We also note that | a | = v2/R. Applying…

6. An airplane is flying in a horizontal circle at a speed of 490 km/h. If its wings are tilted at angle e = 45° to the horizontal, what is the radius of the circle in which the plane is flying? Assume that the required force is provided entirely by an "aerodynamic lift" that is perpendicular to the wing surface.

6. An airplane is flying in a horizontal circle at a speed of 490 km/h. If its wings are tilted at angle e = 45° to the horizontal, what is the radius of the circle in which the plane is flying? Assume that the required force is provided entirely by an "aerodynamic lift" that is perpendicular to the wing surface.

Physics for Scientists and Engineers with Modern Physics

10th Edition

ISBN:9781337553292

Author:Raymond A. Serway, John W. Jewett

Publisher:Raymond A. Serway, John W. Jewett

Chapter6: Circular Motion And Other Applications Of Newton's Laws

Section: Chapter Questions

Problem 44AP: A model airplane of mass 0.750 kg flies with a speed of 35.0 m/s in a horizontal circle at the end...

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