Messages posted by celestine

None of the liquids will rise

Effect of magnetism is negligible . If your ans is that one of the liquid rises its like keeping a stone close to rubber and telling that it gets attracted due to gravitation!!

extensive

and how did u draw the graph

k banagar how are u tellin its 8

(x-y)^2 - 4(x+y)^2 =2

3x^2 + 3y^2 + 10xy +2 =0

Im sure tis is correct

mod( Z + 2\Z) <= modZ + mod2\Z <= 2+1<= 3]

mod( Z + 2\Z) >= modZ - mod2\Z >= 2-1>= 1

1. im gettin 32\3 m check ur question please

2. Its C . dv\dt = -kv^2 = v dv\dx
solve it to get ans

IT was a nice question
from where did u get it

, Ab = gsin37 ? Awcos16 = -0.275 g
is actually Ab = gsin37 - Awcos16 = -0.275 g

Hey allamraju your substitution as pi + some Tan-1 is meaningless bcos tan-1 7 is within pi/2 and 0

Considering FBD

The equations involved are

N- reaction btw wedge and blk

N=.6gcos37 = .48g

For acc of wedge, Aw along incline
Nsin(53-37) + 1.2gsin53 = 1.2Aw

Aw = .91 g

Now acc of blk wrt wedge , Ab = gsin37 ? Awcos16 = -0.275 g

Net Horizontal acc, A = Absin53 + Awsin37 = 0.326 _~ .3251_~ 144/443 g

Note: The actual ans is approximated bcos of the various initial approx that I made in the sin and cos values . But the method of approach is same and genuine.

2) !Y!= sinx

sinx> 0 X $\epsilon (0,\Pi) also Y= cos-1(cos x) = x becos x$ $\epsilon (0,\Pi)$

eq bcoms sinx =!x! which has 2 solutions ( draw graph and check for yourself)

IF YOUR QUESTION IS CORRECT THE ANS IS 0

bcos taking tan on both sides assuming a,b>0 gives

7 =( a+b )/(1- ab)

but ab is >= 1 and the RHS is always negative contradicting RHS=7

mastercelestinepreetham@yahoo.com

could you mail it to me

I pondered over this Question and found that my previous ans was wrong

I m sure the ans is V / ( R\r - sin $\alpha$ ).

Let V* be velocity of spool

The working method is that $\Omega$ R= V* in rolling where $\Omega$ is ang.vel

also at point of pulling V = $\Omega$ r - V*sin $\alpha$ .

Solve these eq to get ans

consider plane along incline as X-axis and the perpendicular plane to incline as Y-axis.

The working method is splitting the velocity into 2 components_ one along incline and other perpendicular to incline

Let V= root.(2gh)

IN X axis
Sx = VsinOT + .5 gsinO T^2
IN Y axis
T= 2 VcosO\gcosO= 2V\g
put T as 2V\g in first eq to get 4V^2 sinO\g = 8hsinO

If this explanation isnt sufficient do tell me

you can scan the diagram and save it in a file and upload it or draw it in paint by yourself and send it

Its Arnie cracker bcos he is aware that he is suspected on account of his dress

V x (R/r + sin $\alpha$ )

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