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@ghazi @Aperogalics

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it is already solved lol

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dw:1351178590831:dw \[T=\frac{ mv^2 }{ r }\]

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i think you are done

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Tmgcosx = mv^2/r

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but hw Do i Find the angle....)

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\[\sin \theta =\frac{ a \cos \theta }{ T }\]

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a  tangential acceleration component

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mgsinx = ma ....?

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nope, just from the diagram i took it

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i aksed mgsinx = ma...is it correct

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dw:1351179063684:dw

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yea your equation seems correct but why to go for that when things are done here