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Mechanics
28 Mar 2007 11:19:23 IST
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28 Mar 2007 13:28:42 IST
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A ball falls frm height "h" and hits the earth inelastically then..............
height of ball after nth collision = e^2n . h
total dist. travelled by ball bfore cmng to rest = h ( 1 + e^2 / 1 - e^@2)
total time taken by ball to com to rest = ( 1 + e / 1 - e). (2H/g)^1/2
height of ball after nth collision = e^2n . h
total dist. travelled by ball bfore cmng to rest = h ( 1 + e^2 / 1 - e^@2)
total time taken by ball to com to rest = ( 1 + e / 1 - e). (2H/g)^1/2
28 Mar 2007 13:58:18 IST
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1)Suppose two particles have position vectors r2 and r1 and their velocities are v2 and v1 (note that r2, r1 , v2, v1 are vectors)..collide is:
r2-r1/ [r2-r1] = v2-v1/[v2-v1]
where [x] denotes magnitude of x.....
2) in the general river- swimmer case....
let Vmg => vel of man w.r.t ground
Vmw=> vel of man w.r.t water
Vwg => vel of water w.r.t ground..
then....the conditions for:
(i) zero drift is: Vwg is perpendicular to Vmg
(ii) minimum time is: Vmw is perpendicular to Vmg
(iii) non-zero min drift is: Vmw is perpendicular to Vmg
the case (iii) arises when vel of man is less than vel of water flw..
28 Mar 2007 22:17:08 IST
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NEWTON'S LAWS OF MOTION
MOTION OF CONNECTED BODIES
Consider three masses M1,M2,M3 conncted to each other wih strings.A Tension T1is applied on M1. The system lies on a frictionless surface.the acceleration produced by the system is given by,
a= T1
----------
M1+M2+M3
The tensions in da strings between M1 and M2 is T2=(M2+M3)a
similarly tension in which M3 is pulled is given by
T3=M3a= M3T1
----------
M1+M2+M3
here we kno T1......
thank u!
i will post more of dese....
dont forget to rate me!!!!
29 Mar 2007 11:02:22 IST
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Desi funda.................................
in the type......................
Q) A body of mass 2kg lying on an inclined plane of incination 30' . If coeff. of limiting friction b/w 2 surfaces is 0.7($) then find force of friction b/w the body and surface of plane
Ans ) though we always use F = $mg cos#
but sumtimes ans is nt appropriate as in this qs we hav to use
F = mgsin#
so now the qs is how can we know wich formula is to b applied
ts simple apply both and in wich F comes out to b lesser , that is the ans
hope u like it
in the type......................
Q) A body of mass 2kg lying on an inclined plane of incination 30' . If coeff. of limiting friction b/w 2 surfaces is 0.7($) then find force of friction b/w the body and surface of plane
Ans ) though we always use F = $mg cos#
but sumtimes ans is nt appropriate as in this qs we hav to use
F = mgsin#
so now the qs is how can we know wich formula is to b applied
ts simple apply both and in wich F comes out to b lesser , that is the ans
hope u like it
30 Mar 2007 12:32:32 IST
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ELECTROSTATICS
MISCONCEPTION-1
WE THINK THAT RESISTORS/CAPACITORS IN A CIRCUIT EITHER HAVE TO BE IN PARALLEL OR IN SERIES. THIS IS NOT TRUE.
REMEMBER THE CRITERIA FOR TWO RESISTORS/CAPCITORS TO BE IN PARALLEL ARE THAT TH POTENTIAL AT THE STARTING AND THE ENDING POINT OF THE RESISTOR MUST BE THE SAME .
CONSIDER THE CIRCUIT
IN THIS CIRCUIT NONE OF THE ELEMENTS ARE IN PARALLEL OR IN SERIES, WE HAVE TO USE KIRCHOFF?S LAWS OR SYMMETRY LAWS TO WORK OUT TH EQUIVALENT RESISTANCE.
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T = M1M2 ( 1 + sin $ ) ( g +- a)
Tension in any string = mass draged * F/M