post Plz answer the question -5 Chapter 04, Problem 063 robiem At t-3.00 s, the

Question

post Plz answer the question -5 Chapter 04, Problem 063 robiem At t-3.00 s, the acceleration of a partide moving at constant speed in counterdockwise aircular motion is Problem Problem Problem ProblemAt t emblem | | = (9.00m/s2); + (4.00m/s2)/ 4.00 s (less than one period later), the acceleration is a2 = (4.00m/s2)i-(9,00m/s2); The period is more than 1.00 s. What is the radius of the cirde? At s by Number the tolerance is +-2% Click if you would like to Show Work for this question: Units Open Show Work

Explanation / Answer

the acceleration = a = sqrt(92 + 42) =sqrt(97)
a = 9.848 m/s2

Use the standard relationship:
a = v2/r

If A is the angle between the acceleration vectors then:
distance/time = Ar/(4 - 3) = velocity = v

a = [Ar/1]^2/r
a = A2r
r = (a/A2)

Find A:
dot product = SQRT(97)SQRT(97)cos(A) = 97cos(A)
dor product = (9)(4) + (4)(-9) = 36 - 36 = 0
So cos(A) = 0 and A = pi/2
But the particle is moving in a counterclockwise direction so the angle we need is 2pi - A = 3pi/2 = 4.712

r = 16(9.848)/(4.712)^2
r = 7.0967m

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