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A position vector that belongs to the xy plane is described by the components (x, y) - (-3.00 m, 5.00 m) measured from the origin of the center of Cartesian coordinates. If we represent the vector in its polar coordinates, that is, in terms of its magnitude and angle formed with the positive horizontal "x" axis, we would obtain as a result: a) 0 = 239.0° r= 5.83 m b) 0 = 59.0° r= 2.82 m c) 0 = 239.0° r= 8.00 m d) 0 = 59.0°T=5.83 m || in february. it had to decelerate in the Martian

A position vector that belongs to the xy plane is described by the components (x, y) - (-3.00 m, 5.00 m) measured from the origin of the center of Cartesian coordinates. If we represent the vector in its polar coordinates, that is, in terms of its magnitude and angle formed with the positive horizontal "x" axis, we would obtain as a result: a) 0 = 239.0° r= 5.83 m b) 0 = 59.0° r= 2.82 m c) 0 = 239.0° r= 8.00 m d) 0 = 59.0°T=5.83 m || in february. it had to decelerate in the Martian

College Physics
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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A position vector that belongs to the xy plane is described by the components (x, y) - (-3.00 m, 5.00 ri) measured from the origin of the center of Cartesian coordinates. If we represent the vector in its polar coordinates, that is, in terms of its magnitude and angle formed with the positive horizontal "x" axis, we Would obtain as a result: a) 0 = 239.0° r = - 5.83 m |3D b) 0 = 59.0°r= 2.82 m c) 0 = 239.0° r= 8.00 m d) 0 = 59.0°T=5.83 m bed at Mars in february, it had to decelerate in the Martian
Transcribed Image Text:A position vector that belongs to the xy plane is described by the components (x, y) - (-3.00 m, 5.00 ri) measured from the origin of the center of Cartesian coordinates. If we represent the vector in its polar coordinates, that is, in terms of its magnitude and angle formed with the positive horizontal "x" axis, we Would obtain as a result: a) 0 = 239.0° r = - 5.83 m |3D b) 0 = 59.0°r= 2.82 m c) 0 = 239.0° r= 8.00 m d) 0 = 59.0°T=5.83 m bed at Mars in february, it had to decelerate in the Martian
Expert Solution
Step 1

Cartesian coordinates(X,Y) in polar coordinates are given by

X=rcos(theta)

Y=r sin(theta)

Where r is position vector and theta is angle of position vector with horizontal x axis

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