Recent content by Wiz

hi, we all know tht when a body moves in a circular path,it experiences an outward force(eg:a coin on rotating disc gets thrown outward).Theoritically fine.But can anyone give me the LOGICAL explanation for this?Wht does it get thrown outward? Also consider a person in a rotar standing with...

tends to 1 from?? i am confused on this one...my friend told me that as x->0 ,(sinx)/x tends to 1 from left hand side...similarly he said something abt tanx/x approaching 1 from right hand side...can anyone please explain this to me?? thanks.. wiz

if x and y are rational then irrational terms on both sides of the eq must be equal adn also rational terms on both sides must be equal.hence y=x=0.5...get it?

hey check ur working again...i think it is r^2 + x^2 in the denominator and not wht u hv written...

maybe because in the first case the force contributed by each element is the same where as the in second case the force exerted by each ring of thickness dr is different.so u hv to integrate to get the total force by the disc(or all the rings of thickness dr)..hence the field...does this suffice??

if i understand u right then tht implies x=y=1/2...so find wht u want...

i understood absolutely nothing in ur previous post...could u please explain??...

ok thtz correct...but is there a way to generalise it??...say selection of r persons out of n ppl??.(in the same prob)

hmm...i get ur method..but ur ans is wrong..check it again..in the meanwhile i'll give it another shot..do post ur corrected ans...thanks.

i don't get u...wht are u trying to say??.are u challenging the question or asking me a question which i myself posted on the forum?...what??

I am stuck on this prob...16 men are seated along a round table...in how many ways can i select 7 out of em such tht none of the selected individual were sitting next to each other?... thanks

Hey i solved the prob, The fact we forgot was tht since the body is moving towards the orgin due to -ve value of acc, the correct relation wud be v= - sqrt[(1-x)/x] ----(1) Now put x = sin^2($) hence dx/d$ = sin2$ hence dx = sin2$*d$ --------(2) By (1), - dx/sqrt[(1-x)/x] = dt => -dx...

Oh,my god...i am sooooo sorry the correct eq is F(x)= -k * 0.5 * x^-2 where k= 10^-2 Nm^2 the mass of the body is 10^-2 kg Now itz 100% correct. Wht i did was, i got a= -0.5 * x^-2 as k and m canceled out so i got dv/dt=(-0.5 * x^-2) hence dv/dx*dx/dt=(-0.5 * x^-2) hence dv*v=(-0.5 *...

The prob i posted is the same, All i did was...i found out acc. from the force.. Then integrated it and i found velocity...the constant in tht case was c=-1 Btw i hv mentioned tht the particle is initially at x=1 The prob is same as given in the book, This is an iit-jee question meant for...

forgot I forgot...the mass of the particle is 10^-2 kg...