I think yo need to know the function of the angle in time, lets say #(t), and then simply integrate two times over each aceleration to have x and y in function of time.
For example, if #=90º, then you have a_x=w, a_y=p-g, making some integration....
v_x=v_x0+w*t, x=x_0+v_x0*t+0'5*w*t^2...
Yes, that was the original idea. But I thought it would be interesting too the fact of solve the "time for equillibrium".
I did, the surface I took for absortion was Sa=pi·Re^2 (effective section, like a disk). And the surface for emision Se=4·pi·Re^2.
I tought that knowing the...
Oh, NOW I know what do you say, but still I'm not sure if the increase of intensity (and decrease of the shadow) will happen if the experiment is made in more optimal conditions. Maybe in this case is due to some sort of incident angle of the light. But I've not ideas.
No, I'm saying that the "sticked shadow" is created by the small space between the two objects, but that this shadow is more "gray" than the other two, and when you have light coming from an extense focus, then if that light grows in intensity, the interference becomes less visible due the light...
I think that the matter of "approach the slit to the screen" didn't work because the waves had less "space" to interact.
Diffraction
In the second image of that link, we can imagine what happens if one approach the screen to the double slit, enought close there is not interference.
In...
I've been thinking, and I've found that the original problem is much easier.
If we have that:
Pabs=Fs·pi·Re^2·(1-A) where A is the Albedo Bond, and 1-A=absortibity
Pabs=Ls·Re^2·(1-A)/(4·d^2)
Pem=4·pi·sigma·Re^2·Te^4
In equilibrium: Pabs=Pem so...
Like I said before, the gravitational field is conservative. That means that no matter the path, only the initial and final point for the Ep. You have the Ep of every point. Now, the same thing that you made for B you can do for other points. EkC=EpA-EpC, and so on. Think that EpA is the Et, so...
Exactly, now that you know that EkB=1'5·10^7 an that Ek=m·v^2/2, you can write:
1'5·10^7=m·v^2/2. Is an equation where you know m (4·10^4), so simply solve v in that equation.
For the other points, you have that the Ek is the difference between Et=EpA and the Ep in that point (same that...
Well, you have there. If the EpA-EpB=1'5·10^7 J, then that is the potential Energy that became kinetic, so you have that...
EpA-EpB=EkB=m(vB)^2/2
For a point x:
EpA-Epx=Ekx=m(vx)^2/2
Homework Statement
Trying to answer the question of "what would happen if Sirius take the place of the Sun", I've begin to try to calculate the average temperature of Earth in function of time for any incident flux (and therefore, the average temperature today), knowing the real average...
Now that you know the total energy, you can apply that to other point (for example, B).
Knowing that Ep=mgh, use the height of B to calculate Ep, then you have Et (done by yourself), soy you can have v with some algebra.
Knowing that gravitational field is conservative, the potential energy is:
Eg=mgh.
So if the heights of the points are given relative to the gorund, then the potential energy in every point is the wrote above.
Knowing that the total energy is:
Et=Ek+Eg
In the other side, I have...