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/sci/ - Science & Math

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>> No.5671123 [View]
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5671123

>>5669580
Quantum computation.
Superposition=Awesomeness

>> No.5671118 [View]
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5671118

>>5671105
>nothing that can be tested using current or even foreseeable technology.
Lol dumbass you just defined what a theory is
>It's not a theory. A theory can be TESTED
I don't even...

And what's up with the string theory...bad example of what you're trying to say. Einstein's general relativity is the best example. 100 year old physics and yet we can't take it for granted since is just theory. So dont you be talking about string(which by the way contrary to the popular belief is a better theory than anything that's out there right now)

>> No.5671107 [View]

>>5671098
now that I take a better look myself I can see on the cover the word "eigenvalues"....
That's kinda the opposite of dummies.
This book (cba to search it up) is probably an introduction to the cuantic world once you completed the classical part of physics and have mastered calculus III and algebra up to the tensors.
Also never ever study from books named: "sth for dummies" They are made by stupid authors who think they are so funny and cool with their stupid wanna be "simple explanations" ...there's no such thing as simple explanation...there's only math bros.

>> No.5671088 [View]

>>5670941
There's no such thing as quantum physics for dummies; you need decent college math to work your way to understanding quantum mechanics. Also who the fuck suggested you to start studying quantum physics without having basic knowledge of classical mechanics/ classical electrodyamics and electricity and magnetism

>> No.5671073 [View]
File: 170 KB, 960x854, 04c040005fe211dfac7c000b2f3ed30f.jpg [View same] [iqdb] [saucenao] [google]
5671073

>>5668683
no reply? so you managed to understand?

>> No.5671071 [View]
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5671071

>>5671043
lol I formulated an answer without the need of radius before he even trolled there

>> No.5671066 [View]
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5671066

>>5671022
A=4*pi*r^2 <- A=0.1 m^2

V=(A*r)/3
or V=[A^(3/2) * 4^(-1/2) * (pi)^(-1/2) ]/3 if you cant even calculate your radius

>> No.5671021 [View]
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5671021

>>5670971
I'm not sure that he asked for a way to solve that problem; I pretty much think he asked how to get from that kind of form he got to the one showed in answers. That formula wasnt a overkill. Is the simplest yet more general way of dealing with that kind of radicals. Nuff said.

And about solving that problem of his....yee that cosinus formula lead him to a nasty result; with a little bit of focus and use of geometry he could have solved it way easier without "memorizing" formulas although that cosinus formula is also easily derivable.

>> No.5670962 [View]

>>5670945
get on your level of what? Highschool kid logic?

>> No.5670926 [View]

>>5670922
implying you can't derive them

>> No.5670908 [View]
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5670908

>>5670801
you got sqrt(8-4*sqrt3)
i dont hav the terminology so dnt laugh at my explanation: introduce the "4" under the sqrt so you get
sqrt(8-sqrt48) than apply the formula i uploADED

>> No.5670816 [View]
File: 101 KB, 624x224, 1362014952079.gif [View same] [iqdb] [saucenao] [google]
5670816

>>5668683
dF=(dq*Q)/(4*pi*E0*r^2)

Q<- the particle
dq<- very small piece of the arc
pi<- 3.14
E0<- vacuum permitivity
r<- radius of your arc

let L=dq/dl be the liniar density of your arc

thus dq=dl*L;

rewrite F=(Q*dl*L)/(4*pi*E0*r^2)
One of the components of F will be 0; Fx=0 cause of the simmmetry
Fy=F*sin(c)
so you now have

Fy=[(Q*dl*L)/(4*pi*E0*r^2)]*sin(c)

lenght of a tiny piece of the arc dl=r*dc
Fy=(Q*r*dc*L)/(4*pi*E0*r^2)*sin(c)
Integrate over your angle : from 0 to bla bla
The only part you have to integrate is sin(c)*dc the rest of the terms are constant
You should end up with sth
{(Q*r*L)/(4*pi*E0*r^2)}*(-cos(0)+cos(whatever angle you have))
where L=the total amount of charge on the arc / the lenght of the arc

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