An airplane is flying in a horizontal circle at a speed of 360. km/h (see the figure). If its wings are tilted at angle 0 = 43.0° to thehorizontal, 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. Use g = 9.81 m/s?.%3DNumberiUnits

Answered: Image /qna-images/answer/45608695-8325-4592-9674-82ee00628fad.jpg

Answered: An airplane is flying in a horizontal…

College Physics

11th Edition

ISBN:9781305952300

Author:Raymond A. Serway, Chris Vuille

Publisher:Raymond A. Serway, Chris Vuille

Chapter1: Units, Trigonometry. And Vectors

Section: Chapter Questions

Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...

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Topic: Vectors and Scalars
In the figure, a stuntman drives a car (without negative lift) over the top of a hill, the cross section of which can be approximated by a circle of radius R = 281 m. What is the greatest speed at which he can drive without the car leaving the road at the top of the hill? Number i Units
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ng An airplane flies in a horizontal circle at a speed of 485 km/h. The wings of the aircraft are at an angle of 41 degrees to the horizontal (as shown in the figure). Assume the aircraft wings generate an aerodynamic lift force perpendicular to the wings. What is the radius of the circle in which the plane is flying? Your answer should be in meters:
The force F has a magnitude of 480 Ib and acts along the diagonal of the parallelepiped as shown. Express Fin terms of its magnitude times the appropriate unit vector and determine its x-, y-, and z-components. A 6' F = 480 b 8' 3' B Answers: i+ i j+ i F= ( i Ib)( i Fx Ib i Fy Ib F2 Ib i
V-axis X-axis A Figure 1. An object (yellow point) travels from point A to point B along the arc of a circle that is centered at point O (red point). velocity of an object, either due to a change in speed or direction, means the object is accelerating. According to Newton's second law, an object will accelerate only if there is a non-zero total force acting on the object. For uniform circular motion, it turns out that this force must always be pointing towards the center of the circle that the object is travelling around and we say that there is a centripetal force present. You can use Newton's second law and a little bit of geometry to understand why an object undergoing uniform circular motion must be experiencing a force that is pointing towards the center of the circle the object is travelling on. If the acceleration vector pointed above or below the plane of the circle, the velocity vector would also end up pointing out of the plane that the circle is on. Since the velocity vector…
A 55.8kg ice skater is moving at 4.03 m/s when she grabs the loose end of rope, the opposite end of which is tied to a pole. She then moves in a circle of radius 0.797 m around the pole . A.)Determine the force exerted by the horizontal rope on her arms. B.) what is the ratio of this force to her weight
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