The electron is taken to be a point particle. As a point charge, it has a coulomb potential that emanates radially from it's position... so yes, it will attract any positively charged particle equally from all sides (so long as we are considering positive charges of equal charge)
you may...
are you an undergraduate? what level of physics have you been through?
if you are an undergraduate, i would recommend two sources probably over all else:
1) MIT Open Courseware Video Lectures for Mechanics and Electricity & Magnetism by Professor Walter Lewin...
just to add:
nicksauce and hallsofivy are indeed correct, and the dirac delta function can often times not be treated the same way as a function (but instead as a distribution). It only makes physical sense when integrated over, and is often used in physics to simplify problems that have the...
i think you could answer it either way, though I don't think either of those answers is particularly "tidy". why is it you want to use this notation? it seems somewhat superfluous to me for the type of question it seems to be answering.
i think the most appropriate answer would depend on...
this is also a wonderful java applet that shows the effect quite well:
http://www.falstad.com/fourier/
i remember when i was first learning about Fourier series... took me a while to be comfortable with the idea of it all
so I'm going to go ahead and take a guess that your friend was referring to a Fourier series.
as arildno has pointed out, this is a sum of an infinite number of periodic functions, not just two. (ie, it would be an infinite sum of sines, each with a different period and amplitude)
you can...
ok so that step was done using the chain rule.
basacally: d/dt f(x) = f'(x)*(dx/dt)
I am on my phone on a train right now so I hope that explanation will be sufficient. in our case, f(x)=GM/x^2
So you are very close, and the math in your previous post looks correct, but you are making it a bit more complicated than it needs to be.
Indeed, in order to solve the differential equation
\frac{\mathrm{d}^2x}{\mathrm{d}t^2} = -\frac{GM}{x^2}
you will need a bit of help with...
is that true? what about this:
http://scitation.aip.org/getabs/servlet/GetabsServlet?prog=normal&id=PRBMDO000072000007075342000001&idtype=cvips&gifs=yes
Precisely.
The equation which allows you to do that is Newton's Second Law, ie: F=ma.
You will want to treat it as a one dimensional problem (set r=x, a=x'').
then you can find an expression (in terms of x,x') for jerk.
if i understand you correctly, what you are describing is not a physically possible situation... the Earth cannot just disappear... an electric charge cannot just go away... it takes energy to do so (allowing for conservation of energy when the potential energy goes away)
umm... you can think of the fan as providing a fictitious potential, but what is really going on is the electrical energy of the fan is moving the fan blades... which is then moving the air.
the air is pushing the ball and providing a force opposite the momentum of the ball.
this is a...
so let's think about it this way:
if we were to do this:
{U^*} = \int_{{x_1}}^{{x_2}} {(\rho gy\sqrt {1 + y{\single-quote^2}} } + \lambda l)dx = \int_{{x_1}}^{{x_2}} {(\rho gy\sqrt {1 + y{single-quote^2}} } + \lambda \int_{{x_1}}^{{x_2}} {\sqrt {1 + y{single-quote^2}} dx} )dx
then, all we...
well, the typical situation (where your coordinates are somewhat normal (ie, can be related somehow to the cartesian coordinate system in a time independent fashion) then the hamiltonian is the energy of the system.
ie, simple harmonic oscillator:
L=T-U= 1/2 m x'^2 - 1/2 k x^2
where m is the...
if \frac{\partial L}{\partial t}=0 then the hamiltonian is a conserved quantity. So yes. If the lagrangian doesn't explicitly depend on time, H is conserved.