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![[Post New]](/templates/default/images/icon_minipost_new.gif) 20 Apr 2008 20:08:51 IST
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Two identical balls each having charge of 2.00 x 10-7 and mass 100 g are suspended from a common point by two insulating string each 50 cm long.The balls are held at a separation 5 cm apart and then released.Find the components of the resultant force on it along and perpendicular to the string,also find the tension in the string ,and the acceleration of ome of the balls.Answer are to be obtained only for the instant just after the release.
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 20 Apr 2008 20:32:02 IST
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Please reply me.rates assured
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 20 Apr 2008 20:43:12 IST
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is tension in the string is 28.8x10^ -5 N???
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 20 Apr 2008 20:46:00 IST
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ans is .986 N.please explain
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 20 Apr 2008 20:55:48 IST
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you see in the figure that the vtical forces are cancells out.so net force in the verticle direction is zero. In prependicular forces, net force is Fe cosatheta- mgsintheta as shown in the figure. now put the values and you will get net force prependicular to the rope.
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 20 Apr 2008 20:59:57 IST
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Equate mgcostheta and T you will get 0.986 N if g=9.8m/s^2 I have equate these because the components are shown in the figure.
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 20 Apr 2008 21:00:39 IST
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resolve tension in the string into itsd components .Then... Tcos  =mg T=mg/cos  cos(theta)= [ ] (1-sin 2 ) sin (theta)=2.5/50......(perp./hypo) calculate T
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 20 Apr 2008 21:08:21 IST
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Divide the net force in horizontal direction with its mass you will get its accn.
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 20 Apr 2008 22:16:55 IST
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also find component of resultant force and acceleration of the ball
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