What is the tangential speed of a point on the rim of the turntable at t =0,207 s? Express your answer in meters per second. V= 1.11 m/s Previous Anawers v Correct

Elements Of Electromagnetics
7th Edition
ISBN:9780190698614
Author:Sadiku, Matthew N. O.
Publisher:Sadiku, Matthew N. O.
ChapterMA: Math Assessment
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Part C
What is the tangential speed of a point on the rim of the turntable at t= 0,207 s?
Express your answer in meters per second.
V=1.11 m/s
Previous Answers
Correct
Part D
What is the magnitude of the resultant acceleration of a point on the rim at t= 0.207 s?
Express your answer in meters per second squared.
阳AE4
a=7.95
m/s
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Transcribed Image Text:Part C What is the tangential speed of a point on the rim of the turntable at t= 0,207 s? Express your answer in meters per second. V=1.11 m/s Previous Answers Correct Part D What is the magnitude of the resultant acceleration of a point on the rim at t= 0.207 s? Express your answer in meters per second squared. 阳AE4 a=7.95 m/s Submit Previous Anywers Request Answer x Incorrect; Try Again; One attempt remaining
An electric turntable 0.740 m in diameter is rotating about a fixed axis with an initial angular velocity of
0.290 rev/s and a constant angular acceleration of 0.909 rev/s
Part A
Compute the angular velocity of the turntable after 0.207 s
Express your answer in revolutions per second.
W= 0473 rev/s
Previous AnsWers
Correct
Part B
Through how many revolutions has the turntable spun in this time interval?
Express your answer in revolutions.
N= 795-102 rev
Transcribed Image Text:An electric turntable 0.740 m in diameter is rotating about a fixed axis with an initial angular velocity of 0.290 rev/s and a constant angular acceleration of 0.909 rev/s Part A Compute the angular velocity of the turntable after 0.207 s Express your answer in revolutions per second. W= 0473 rev/s Previous AnsWers Correct Part B Through how many revolutions has the turntable spun in this time interval? Express your answer in revolutions. N= 795-102 rev
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