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anonymous
 4 years ago
A bead slides without friction around a looptheloop. The bead is released from a height
20.3 m from the bottom of the looptheloop
which has a radius 7 m. The acceleration of gravity is 9.8 m/s^2. How large is the normal force on it at point A if its mass is 3 g?
anonymous
 4 years ago
A bead slides without friction around a looptheloop. The bead is released from a height 20.3 m from the bottom of the looptheloop which has a radius 7 m. The acceleration of gravity is 9.8 m/s^2. How large is the normal force on it at point A if its mass is 3 g?

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anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I already calculated speed. 11.112 m/s. To find the answer to this problem do I use the formula. F=mv^2/R = ((.005kg)(11.112 m/s)^2)/5m = 0.123N ?

JamesJ
 4 years ago
Best ResponseYou've already chosen the best response.0A is the top of the loop?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0if point "A" in top normal force is N=(mv^2/r)mg if "A" in bottom N equal to:dw:1329244817817:dw(mv^2/r)+mg

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0hence in bottom 2nd law of newton is Nmg=mv^2/r & on top is N+mg=mv^2/r
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