problem03_50

# University Physics with Modern Physics with Mastering Physics (11th Edition)

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3.50: The bird’s tangential velocity can be found from m/s 05 . 10 s 5.00 m 27 . 50 s 00 . 5 ) m 00 . 8 ( 2 rotation of time nce circumfere = = = = π x v Thus its velocity consists of the components m/s 05 . 10 = x v and m/s 00 . 3 = y v . The speed relative to the ground is then m/s 10.5 or m/s 49 . 10 002 . 3 052 . 10 2 2 = + = + = y x v v v (b) The bird’s speed is constant, so its acceleration is strictly centripetal–entirely in the
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Unformatted text preview: horizontal direction, toward the center of its spiral path–and has magnitude 2 2 2 2 c m/s 12.6 or m/s 63 . 12 m 00 . 8 m/s) 05 . 10 ( = = = r v a x (c) Using the vertical and horizontal velocity components: ° = =-6 . 16 m/s 10.05 m/s 00 . 3 tan 1 θ...
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