Recent content by m.e.t.a.

Whoops, I don't want to spread misinformation. What part(s) in particular did I get wrong there?

Hi, saipathudut. There isn't much I can add, but if you want to learn more, you might find the aether theory interesting. This is the long-obsolete theory that the Universe is pervaded by a medium (the aether) which we can think of as a kind of stationary scaffold against which all velocities...

Hi, welcome to PF! You make a good observation, it had never occurred to me before. I think you are right in presuming that an electron, after excitation, will sometimes immediately (or after a short delay) undergo de-excitation via the reverse process, i.e. between the same two atomic states...

P.S. I apologise for the people who evidently got up on the wrong side of bed this morning.

Hi skipper. I have no technical knowledge of what constitutes a "typical" UV absorption spectrum for eyeglasses, but it could be possible that, in addition to being strongly absorbent to UV, your glasses are also weakly absorbent to near-UV (i.e. deep violet). As K^2 mentioned, the eye's...

To expand on espen180's answer: For a single atom or molecule in vacuum (far from any other atoms or electromagnetic fields) the energy levels available to its electrons are indeed few and highly discrete. However, if you bring more atoms into create a crystalline lattice, new electron energy...

The metal body of the aircraft has low emissivity, but high thermal conductivity. Perhaps heat moves from the metal body into the high-emissivity paint via conduction, and is then efficiently dumped into the atmosphere via a combination of conduction and radiation.

In addition to focusing/concentrating the gas flame (if this does indeed occur), I would think that the thermal conductivity of the gauze also plays a role. The gauze, being made of many tiny filaments, absorbs heat more efficiently than the flat bottom of a saucepan. Additionally, the gauze is...

I agree with PhanthomJay—it is the velocity ratio of the two ends of the see-saw that is important. In post #2 I hypothesised that the stationary ball will be launched at no greater than twice the speed of the end of the see-saw on which it sits. For a central fulcrum setup, the heavy falling...

This equation from http://en.wikipedia.org/wiki/Drag_(physics)#Velocity_of_a_falling_object"should provide a good estimate for a sphere falling through air under Earth gravity: v(t) = \sqrt {\frac{{2mg}}{{\rho A{C_d}}}} \tanh \left( {t\sqrt {\frac{{g\rho {C_d}A}}{{2m}}} } \right)

For an object in freefall, I would think that the number of degrees of rotational freedom is 1. I.e., it doesn't matter what the object's angular velocity is about the x, y and z axes—these perpendicular angular velocities always resolve into a single angular velocity about a single axis of...

Before you can start calculating a solution, you must set a location for the fulcrum. For example, if the see-saw's fulcrum is put at the halfway (10 m) point, then I believe it is impossible for the heavy ball to launch the light ball at a speed any greater than approx. 2 × 1,290 m/s = 2,580...

Going back to Newton's law of gravitation: {\vec F_{12}} = - \frac{{G{m_1}{m_2}}}{{{{\left| {{{\vec r}_{12}}} \right|}^2}}}{\hat r_{12}} This equation, which provides a highly accurate description of gravity, states that the gravitational forces, {\vec F_{12}} and {\vec F_{21}}, experienced...

If I understand your original question correctly, then HallsofIvy has stated the equation necessary to find the resultant acceleration. It might help to define "resultant acceleration", a_{\rm{res}}, as \ddot r, where r is the distance between the two bodies. So a_{\rm{res}} is not the...

Thank you, you have been a great help. Your explanation that the pairing energy is a somewhat ad-hoc theory is very helpful to me. This gives me something to say. Beforehand, I was resigned to merely refer to this mysterious thing, "pairing", with no explanation of its physical meaning, or lack...