# Search results

1. ### Kondo Lattice vs. Single Impurity Kondo System

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...
2. ### Falling bodies

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...
3. ### Au doped with Fe Spin Glass

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...
4. ### C/++/# Sovling schrodinger's equation in C++

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.
5. ### Determine the electric flux through each of the sides

right it is the first value...i don't have the notation for that in my keypad.... but the when i do that operation =2.37x10^5 N m^2/C which is wrong
6. ### Determine the electric flux through each of the sides

well the net charge inside is just 2.1x10^-6 C/e is that right?
7. ### Determine the electric flux through each of the sides

that means that the flux= the charge on the inside of the soild/ the constant e which is 8.85 x10^-12
8. ### Determine the electric flux through each of the sides

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
9. ### Determine the electric flux through each of the sides

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...
10. ### Partial fraction decompostion

thank you very much
11. ### Partial fraction decompostion

ohhh ok...i think i understand.. so if i have (2x+3)/(x+1)^2=(A/x+1)+(B/(x+1)^2) (*) then 2x+3=A(x+1)+B cuz i multiply both sides by (x+1)^2
12. ### Partial fraction decompostion

so should it be 2x+3/(x+1)(x+1)=(A/x+1)+(B/x+1) here..i just split up the (x+1)^2 this is right?
13. ### Partial fraction decompostion

what happened to to (x+1)^2 term?
14. ### Partial fraction decompostion

(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...
15. ### Charges on the vertex of a triangle

ummm...so do i have to find in they x and y compnents and all that sutff?
16. ### Charges on the vertex of a triangle

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...
17. ### Electric field and ball of charge

thank you so much. i really appriciate it.
18. ### Electric field and ball of charge

so the force due to the electric field is F=Eq so instead of E=Tsine(16) i would have Eq=Tsin(16)??
19. ### Electric field and ball of charge

Fx=0=E-Tsin(16)+the charge of the partical where is the Electric field coming into play agian....i don't understand
20. ### Electric field and ball of charge

so it would be E=Tsin(16)+Eq?
21. ### Electric field and ball of charge

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...
22. ### Some-what fancy trig inegral

oohhh ok...let me try it
23. ### Some-what fancy trig inegral

now i am getting the integral of tanx-sinxcosx i know how to integrate sinxcosx but not tanx
24. ### Some-what fancy trig inegral

ok let me try that
25. ### Some-what fancy trig inegral

right...so sinx/cosx would be tanx and that would it it (sinx)^2(tanx).....right?
26. ### Some-what fancy trig inegral

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...
27. ### Electric charge and total force

oohhh ok. i will try again
28. ### Electric charge and total force

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
29. ### Electric charge and total force

that what i thought but in the formula i calls that i multiply q2 and q3 which is (-)(+)= - so that is why i am confused what is wrong
30. ### Electric charge and total force

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