Recent content by sic

no, you multiply by k every time, meaning that outside it would turn into k^k

Actually i didn't pay attention to the condition T(x,y,z,0)=0, which really makes the problem trivial (T=0 always), but if it's so, the boundary conditions have absolutely no meaning.. are you sure there's no typo?

yes, you take the first order approximation: dP(y)/dy=Whatever hence dP(y)=Whatever*dy F=dP(y)*A=Whatever*A*dy (harmonic oscillator equation) and k/m=omega^2=Whatever*A/m now pay attention that PO=PA+mg/A

basically, there is no meaning to the function Dirac delta is multiplied by (only to its value at 0), because in any point except for 0, delta is 0 by itself anyway..

Actually it is simplier to begin from (2) Explanation of the terms: first term is the vector parallel to e, second orthogonal to e A x e is normal to both A and e and has magnitude of the projection of A on the plane normal to e, and e x ( A x e ) has the same magnitude and is normal to e and...

what? try to sketch the function. obviously, the function doesn't have a maximum for any constant R. what is the question from the beginning?

Omega={A,B} M(gurnisht)=0 M({A})=1/2 M({B})=1/2 M({A,B})=1 then M^2({A,B})!=M^2({A})+M^2({B}) (Either it's that simple or i don't understand the problem)

HallsofIvy: it is, but it's the trivial one.. it's a linear equation, the solution can be multiplied by a constant, and knowing that 0 is a good solution doesn't give you anything.. danai_pa: try the following steps: 1) solve the boundary. you'll see that on the three of the faces of the...

Projecting the forces onto axes: say we choose two axes parallel to the sides of the square, then if the square side length is a, each one of the two charges on the axes will apply force equal to q^2/a^2 on its axis and 0 on the second one, and the charge on the diagonal will apply...