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here magnetic flux is zero ..
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we can find initial velocity i.e 10m/s
power is F.V = 1 force acting on it is in oppsite direction of velocity since final velocity is zero.
so F = -1/V
MdV/dt = -1/v or
M*ds*dV/dt = -ds/V
MVdV = -ds/v
MV2dV = -ds on integrating v from 10m/s to 0 and s from 0 to s
s = 6000mts
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yes in above post...in V1 and
V2 equations there shouldnt be square of r and R.. By mistake it is written by me..
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force on man
mg - T = ma
for 'a' to be min. T should be max and T can be 2mg/3
so a = g/3
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u will get +30 if the magnification is -2...at that time our object will be virtual object..light will not diverge from a source but they apeear converging at a point... right side of pole..
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here lateral magnification
m=height of image/height of object = -v/u = 2
v = -2u
1/u + 1/v = 1/f
1/u - 1/2u = - 1/20
u = -10 cm..
if u know the value ,substitute it with sign as we do for focal length .
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line no. 5 in above post
copmaring volumes
4PiR3/3 = 27*4Pir3/3
so R = 3r
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potential at the surface of drops....
V1 = q/4(Pi)er2
when all drop form a bigger drop
V2 = Q/4(pi)eR2
Q = 27*q and 4(Pi)R3 /3 = 27*4(pi)r3
R = 3r
V2= 27q/4(pi)e(3r)
= 27 * 220 /3 = 1980 v
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Use gauss's Law ..draw a cylendrical gaussian surface around line charge of radius r and length L..since all point are same on line charge bcz it is infinite and have constant linear charge density.. try to find it now...
if you dont want to use gauss's law take any point and draw a perpendicular on line charge take small two elements on both side at equal distance find electric field at the point and integrate it...
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Momemnt of inertial of rod
I1 = ML2 /12
Moment of inertia of ring
I2 = MR2
= 12I1/L2 * L2 /4(Pi)2
since 2PiR = L
I 1/I 2 = (Pi)2 /3
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if collision is elastic K.E of rolling body will not change..
and K.E of a rolling body is equal to the
Sum of Translation K.E and Rotaional K.E ...
if collision is not perfect then coefficent of restitution will be given..
If possible post some questions ...
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If pencil draw the circle with constant speed ..
Then frictional force act towrds the center of circle on the pencile and
F.ds will be zero bcz displacement is perpendicular to frictional froce...
so work done by frictional force would be zero.
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Take center of platform on origin (0,0) & men on platform along x-axis, A on (3,0) & B on (-3,0)
Rcm = (40*3 - 60*3 + 50*0 )/150
= -2/5
when both men walk and reach the center of platform..
let center of platform displaced X mts along x-axis ..
since external forces on the system(man A , man B & platform)is zero so c.m of system remain on same place ..
Rcm = (40*X + 60*X + 50*X)/150
since men meet at the center of platform..
X = -2/5 mts
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SEE
FG = KM1 M2/R2
K is gravitational constant i.e G.... gravitational force depends on particle's masses and distance bw them...
Fe = kQ1 Q2/R2
K is 1/4pi(eo)here electric force depends on charge..and distance bw them
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F = K/S where S distance covered in time t
if 'dr' is displacement in dt time then
mod dr is distance dS covered in dt time
dW = F.dr = (K/S)ds
W = K*ln(s/s1)
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If mechanical energy of a system is not conserved use theorom
Work done by all the forces = change in K.E
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momentum is a vector quantity and work is scalar so you cant define it in terms of work done by force..
But we can find a relation bw work done by forces on particle and magnitude of its change in momentum ..
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Guys who said to you that Gauss Law is applied to closed charged bodies...
We can use the Gauss Law for any type of charge distribution ..
Gauss's Law states that the flux of the net electric field through a closed surface equals to the net charge enclosed by the surface divided by Eo
here closed surface is gaussian surface not closed charged surface...
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If you want to raise the potential diff. of two counductors placed near to each other connect them to the battery. Both counductor will be charged and the potential diff. across conductors wil ultimately become equal to the emf of battery.
Potentail diff. bw two points is defined as the change in potential energy per unit test charge in taking test charge from one point to other point without changing the position of other charges..
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Vriti Edustore
