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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Concept explainers

Power Transmission Elements

It is defined as the energy transfer from one location to another location. The energy will start moving from the energy generation source to the energy applied position to obtain the work output. From this method, a large amount of electric energy moves from the power station to various electrical substations.

Question

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 a
(positive if counterclockwise, negative if
clockwise) of the bar and the acceleration of
end A (positive if up, negative if down).
28°
105⁰
Answers:
a =
S
aA
=
0.63 m
B
i 0.56
0.2
-ag=0.47m/s²
UB=0.40 m/s
rad/s²
m/s²

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