Recent content by clavin

ok tnx for that i forgot about the coriolis force anyways that's not in my sylabus so i am going to edit the qs a bit please refer to the first post

ell let's say there's a cabin and its rotating about its axisand we have a block of mass m placed (not at the the centre of the cabin) and there's some friction between the the block and the surface now let's say the speed of the cabin is increased a lil high so that friction can no more...

well if the charge was placed on the outer surface then no electric field inside at any point due to electrostatic shielding in case it is placed inside, then r is the distance of the straight line joining the charge and the point at which to calculate the electric field again that point...

well my first qs is lets say i have a pulley and rope system and two blocks each of mass m hanging from either end of the rope which goes over a pulley the pulley has a mass of 1 kg, frictionless and rope is massless and frictionless now if i were to analyze the force on the pulley it would...

is there any difference in oh- from a strong base than from a weak base apart from concentration? second lets say we have this reaction ch3cooh + oh- => ch3coo- + h2o in this reaction my book says the reaction will go to completion but isn't acetic acid a weak acid. so shouldn't the...

ok maybe i am getting it a bit i know now this is getting annoying but can u answer one more circuit problem lets say i have a 5 v battery and a 3 micro farad capacitor and a resiter 2 ohm all in parallel to each other now i believe i have to think as if no current passes thru the 3 micro...

well in case of gravity power isn't dissipated but changed into another form but i think u can stil apply that formula(actually more proper would be to apply f.v though it is very much derived from e/t) and u will know the ke that converted into potential well not sure about ur calculation...

thank u rlbhatt and well thnx for the solution tim but i don't exactly want the solution i want to know why you have taken the assumption that the current in middle branch is zero or why a charged capacitor works as a broken wire and lastly if i replace the capacitor with a similar emf...

the circuit is as shown. http://img186.imageshack.us/img186/6627/circuitg.jpg now i have to find the current from the battery. now the book says since the capacitor is charged its branch will no more have any current and we can neglect that branch. now my qs is. after the capacitor is...