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Yahoo! Group TitleBest ResponseYou've already chosen the best response.1
@ghazi @Aperogalics
 one year ago

ghazi Group TitleBest ResponseYou've already chosen the best response.3
it is already solved lol
 one year ago

ghazi Group TitleBest ResponseYou've already chosen the best response.3
dw:1351178590831:dw \[T=\frac{ mv^2 }{ r }\]
 one year ago

ghazi Group TitleBest ResponseYou've already chosen the best response.3
i think you are done
 one year ago

Yahoo! Group TitleBest ResponseYou've already chosen the best response.1
Tmgcosx = mv^2/r
 one year ago

Yahoo! Group TitleBest ResponseYou've already chosen the best response.1
but hw Do i Find the angle....)
 one year ago

ghazi Group TitleBest ResponseYou've already chosen the best response.3
\[\sin \theta =\frac{ a \cos \theta }{ T }\]
 one year ago

ghazi Group TitleBest ResponseYou've already chosen the best response.3
a  tangential acceleration component
 one year ago

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mgsinx = ma ....?
 one year ago

ghazi Group TitleBest ResponseYou've already chosen the best response.3
nope, just from the diagram i took it
 one year ago

Yahoo! Group TitleBest ResponseYou've already chosen the best response.1
i aksed mgsinx = ma...is it correct
 one year ago

ghazi Group TitleBest ResponseYou've already chosen the best response.3
dw:1351179063684:dw
 one year ago

ghazi Group TitleBest ResponseYou've already chosen the best response.3
yea your equation seems correct but why to go for that when things are done here
 one year ago
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