The ends of the 0.63-m bar remain in contactwith their respective support surfaces. End Bhas a velocity of 0.40 m/s and an accelerationof 0.47 m/s² in the directions shown.Determine the angular acceleration a(positive if counterclockwise, negative ifclockwise) of the bar and the acceleration ofend A (positive if up, negative if down).28°105⁰Answers:a =SaA=0.63 mBi 0.560.2-ag=0.47m/s²UB=0.40 m/srad/s²m/s²

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The ends of the 0.63-m bar remain in contact with their respective support surfaces. End B has a velocity of 0.40 m/s and an acceleration of 0.47 m/s² in the directions shown. Determine the angular acceleration (positive if counterclockwise, negative if clockwise) of the bar and the acceleration of end A (positive if up, negative if down). A 28° 105⁰ Answers: a = aA = i 0.63 m B 0.56 i 0.2 -ag=0.47m/s² UB=0.40 m/s rad/s² m/s²

The ends of the 0.63-m bar remain in contact with their respective support surfaces. End B has a velocity of 0.40 m/s and an acceleration of 0.47 m/s² in the directions shown. Determine the angular acceleration (positive if counterclockwise, negative if clockwise) of the bar and the acceleration of end A (positive if up, negative if down). A 28° 105⁰ Answers: a = aA = i 0.63 m B 0.56 i 0.2 -ag=0.47m/s² UB=0.40 m/s rad/s² m/s²

Elements Of Electromagnetics

7th Edition

ISBN:9780190698614

Author:Sadiku, Matthew N. O.

Publisher:Sadiku, Matthew N. O.

ChapterMA: Math Assessment

Section: Chapter Questions

Problem 1.1MA

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