In order for a dilute magnetic alloy to show the kondo effect (resistance minimum at low T), the magnetic impurites must be far apart and non-interacting.
In the Kondo lattice, the magnetic impurities are much closer togeather (one impurity per unit cell), so why does this system show the...
If two objects of equal mass are dropped....one from h1 and the other fro h2.....will they hit the ground with the same force?
Because the only force acting on the objects is gravity so we can say:
F=ma
and the acceleration is constant...so the forces must be the same.
Is this...
Hello everyone,
I am an undergratude student doing to research and I am making dirty Au films. At certian concentrations of Fe (say >.5%) this Au-Fe is showing charateristics of a sping glass. I am attempting to learn what exactly a spin glass is and what its trends look like (resistance vs...
i need help coming up with a way to solve for the eigen vectors of schrodinger's time independant equation in c++.
so i want to write a class that uses the shooting method, but i am not sure how to do that.
ohh thats right...ok so now if i calcuate what E is
E=Kq/r^2
now r is at the the center of the cube so r=.064/2
E=(8.99x10^9)(2.1x10^-6)/.032^2=1.84x10^7
A=.064*.064=.004096
now 1/6flux=(1.84x10^7)(.004096)
flux=4.53x10^5....but that still isn't the right answer
help me please
A 2.10- mC charge is placed at the center of a cube of sides 6.40 cm. Determine the electric flux through each of the sides.
ok so electric flux=E(A)
so i figure that the electic flux will be the same for all six sides of the cube so i am just calculating the flux on one side then X6...
(2x+3)/(x+1)^2
so this is what i am thinking...but it does not make sense
=(A/x+1)+(B/(x+1)^2)
so then 2x+3=A(x+1)^2+B(x+1)
2x+3=Ax^2+A2x+A+Bx+1
so that would make...
0=A
2=2A+B
3=A+B
this solution does not make any sense becuase if A=0 then according to the second...
Positive point-charges of +16.0 mC are fixed at two of the vertices of an equilateral triangle with sides of 1.30 m, located in vacuum. Determine the magnitude of the E-field at the third vertex.
so...all the particals are in equalibrum so that means F=ma=0
so if i call the third vertex...
As shown in the figure above, a ball with a mass of 0.180 g and positive charge of q=31.0
mC is suspended on a string of negligible mass in a uniform electric field. We observe
that the ball hangs at an angle of q=16.0o from the vertical. What is the magnitude of
the electric field...
find the integral of (cosx)^2(tanx)^3
so first of all i rewrote tan to be (sinx)^3/(cosx)^3 so that i could cancle out the (cosx)^2 on top
and now i have the integral of (1/cosx)(sinx)^3
i rewrote that to be (tanx)(sinx)^2...and now i am stuck
i don't know of any IDs that could...
the way that i was conidering it: i take q1 and q3 to be + cuz they are acting in the right direction...then q3 is - cuz is it actin to the left
so since this isn't wokring hoe can i take this into account better
if i do that...how do i calcuate the electric field about 2 cuz the distance is 0 so according to eqaution for point electic charge...the field would be 0 too