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M Wiley PLUS C Chegg C edugen.wileyplus.com/edugen/student/mainfr.uni EE Apps HP Connected Watch NHL Live Stre... watch NBA Live Str Practice Assignment Gradebook ORION ignment FULL SCREEN PRINTER VERSION BACK NEX Question 18 A cat rides a merry-go-round turning with uniform circular motion. At time t 1.90 s, the cat's velocity is (3.30 m/s) (3.60 m/s) J measured on a horizontal xy coordinate system. At t2 5.20 s, its velocity is (-3.30 m/s) (-3.60) J. What are (a) the magnitude of the cat's centripetal acceleration and (b) the magnitude of the cat's average acceleration during the time interval t2 ti, which is less than a period of the motion? (a) Number Unit (b) Number Unit SHOW HINT Question Attempts: 0 of 8 used SAVE FOR LATER SUBMIT ANSWER Copyright 2000-2016 by John Wiley & Sons, Inc. or related companies. All rights reserved. 1:12 PM 2/13/2016

Explanation / Answer

(a) Centripetal acceleration = V2 / R
where R is the radius of the circle in which object is rotating.
So first of all we will calculate the value of V
V = (VX2 + VY2)1/2 = 4.8836 m/s
If R is the radius of the circle then
Centripetal acceleration = V2 /R = 23.85 /R

(b)We know that the

Acceleration = Change in velocity / time
So the average acceleration = (1 / (5.2-1.9)) *((3.3-(-3,3)) i + (3.6 - (-3.6)j)
= (6.6 i +7.2 j) / (3.3) = 2i + 2.182 j
Resultant = 2.9599 m/s2

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