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Physics 10 Online
OpenStudy (anonymous):

Can someone explain how this answer was obtained

OpenStudy (anonymous):

Im wondering how they got the answer to part b

OpenStudy (mrnood):

This is not a static problem - i.e. the forces are not in equilibrium The tension to support the mass = mg The tension to produce the circular acceleration = mV^2/r So total tension T = m(g +(v^2/r)) This evaluates to your answer

OpenStudy (anonymous):

how did they get the answer to (a)? :/

OpenStudy (mrnood):

It says it is held horizontal. Therefore you know that it is 2.19 m above the bottom position when released. The loss of potential energy in falling to the bottom =mgh The gain of kinetic energy in the same path = 0.5 mv^2 These are equal so: mgh = 0.5mv^2 so v = sqrt(2gh)

OpenStudy (anonymous):

oh ...thanks :)

OpenStudy (anonymous):

What does r stand for @MrNood

OpenStudy (mrnood):

sqrt = 'squareroot'

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