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no vertical acc is independent of the normal force
when the object is in the air the normal force from the ground is 0..
accleration in the vertical direction is mg-n
so if n=0
then acc will not cease

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if the force is parael to incline then
f=mg sin 30

WHICH BOOK TO FOLLOW FOR MATHS OLYMPAID

WHICH BOOK TO FOLLOW FOR MATHS OLYMPAID

IF IT IS WINDING AROUND THE PEG THEN THE RADIUS AT THE BOTTOM MI=OST POINT WILL BE L-X
AND AT THE HIGHEST POINT WILL BE L+X
PLS EXPLN

YEP
BUT HOW IS THE RADIUS
L-X
IF IT IS SO THEN THE DIAMETER IS COMING AS 2L-X
WHICH SHOULD BE 2L
PLS EPLN

HOW IS THE RADIUS L-X
PLSSSS EXPLN

UR EXTREMELY RIGHT

A SMALL HEAVY BOCK IS ATTATCHED TO A LIGHT ROD OF LENGTH l WHICH CAN BE ROTATED ABUOT ITS CLAMPED UPPER END.WHAT MINIMUM HORIZONTAL VELOCITY SHOULD BE GIVEN SO THAT IT MOVES IN A COMPLETE CIRCLE

ANS UNDER ROOT 4GL

MY STEP IS THAT THE BLOCK MOVES IN A CIRCLE OF RADIUS l SO AT THE TOPMOST POINT MV^2/L=MG

V^2=LG

THE POTENTIAL ENERGY IS 2MGL

SO

1/2MV1^2=1/2MV^2+MGH

SOLVING IT WE GET V

AS UNDER ROOT 5LG

BUT THE ANSWER IS WRONG

PLS EXPLN THE FLAW IN MY WORKING

2MG-T=2MA
T-MG=MA
MG=3MA
A=G/3
T=2M(G-a)
16=2M(2G/3)
16=4MG/3
16*3/4=MG
12=MG
M=12/G
CONSIDER THE MOTION OF THE BIGGER BLOCK
S=1/2AT^2
=G/6
POTENTIOL ENERGY DECREASED=24/G*G/6*G=4G
CONSIDER THE MOTION OF THE SMALLER BLOCK
DISTANCE WILL BE SAME
POTENTIOL ENERGY INCREASED
=12/G*G/6*G=2G
THEREFORE POTENTIOL ENERGY DECREASED=2G=19.6
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BUT THIS TIME WE CANT BE BACKSTABBED
UU SEE US IS TRYING TO MAKE FRIENDSHIP WITH US TO MAKE USE OF THE GROWINGG ECNOMY

FRAZER UR RIGHT WE HV THE RESPONSIBIITY AND THE DUTY TO LEAD INDIA TO MAKE IT A BETTER PLACE
I ESPECIALLY WANT THE INDO-US NUCLEAR DEAL TO PASS
WHAT DO U DUDES THINK

pls expln in detail

1)

tan a +tan b=A

cot a +cot b= B

prove-

cot(a+b)=1/a-1/b

2)

sin(a+b)=1 sin(a-b)=1/2 where 0<a , b<90

find tan(a+2b) answer -root 3

and tan(b+2a) ans-1/root3

3)

tan x+tan(x+60)+tan(x+120)=3

prove that

3 tan x -tan^3 x =1

1-3 tan^2 x

pls expln in detail

rates assured

1-3tan^2x