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goiit is one of the sweeeeeeeeeeeeeeeeeeeeeeeeetest places of my lyf.......an onlyn heaven......wat i call..............

it has been ruined by some people....but d ss cene is diff......a y
ear ago................it was jussssssssssssstttttt born........everyone new to d place.....easy to manage ...............n very few users.......
now........users mounting up................50000......

so lets be calm .......things'll be back on track soon......n we'll bring dem back................:):):)

edinburgh

u got d eqn of projectile...........
put dere x=32
.....y=16
fyn n u=24m/s
dis gives d ans .........

but i will............at least for d efforts u made:)

look,
lim{(a-n)nx-tan x}/x X (sin nx)/x
lim{a-n)n-tanx/x}Xnsin nx/nx
=lim(a-n)n-1)X n .....
dis gives........
a=n+1/n
:)

look,
a^2 + b^2 + c^2 + d^2=CQ^2+CP^2+PB^2+BS^2+AS^2+AR^2+RC^2+CQ^2<=
AB^2+BC^2+CD^2+AD^2
THIS GIVES
a^2 + b^2 + c^2 + d^2 =<4

hmmm.......especially for a month or so .......
fake ids disputes over rates.....act of
.
publicly insulting people has become .....an everyday occasion.......

n now all dis is actually very bad on d part of d only girl expert of d community...........

meaning less queries....posted here r actually attention seeking stunts(wont point out neone)..n d fake users behind that succed in 95% of d cases.....

such threads get immense popularity as well
sir goiit needs moderators more........not experts!

i feel ............

let it attians speed v
den
total distance=dis...1+dis...2+dis....3
4250=1/2a10^4+100a1x300+100a1 x150-1/2 a2 (150)^2
eqn 1
also
(100 a1)=a2 x 150
solve dis n u'll get d ans

bhaiyya
u mean to prove R=a/rt 2
fyn......
i'll try:)

now for part c )
at a dis tance of 2a +r
net charge=Q-Q/(vol of shell))x(vol of shell formed within req distance)
=Q-(Q/19a^3) x(r^3-8a^3)
=Q(11a^3-r^3)
therefore
Ex 4 pi (2a+r)^2=Q(11a^3-r^3)/e not
E=Q(11a^3-r^3)/4 pi (2a+r)^2 e not

within the sphere charge is zero so ...........
field is zero
b/w
hrcharge is q so atdistance r field is
using guass law
E x 4 pi r^2=Q/e not
or
E=Q/4pi r^2e not

c)outside d shell
again total charge is zero
so field is zero

yeah ......spidey n mansi di:)

really nice !!!!!!!!!!

here a>b>0

let a side X forms triangle wid a and b den

dis gives

a^2+b^2-2abcosx=X^2

ryt

now

(sinx/X)^2=k(constant)................constant=(1/4R^2)

thus given integral is integral of a constant

integral=pi x constant

thats it !!!!!!1

rest ........still m solving

tangent is
4x+3y=25......ryt
normal is
3x-4y=0
so
area =1/2x base x hgt
=1/2 x25/4 x3.................
=75/8 sq units
draw d fig .it will make it clear