swansont

Keplarian elitist?

You'd need a change in surface area (which may change the drag coefficient) or viscosity. Parachutes are an example of the former, the object hitting water is an (extreme) example of the latter. A less extreme example would be freefall from a very high altitude - terminal velocity 20 miles up should be faster than terminal velocity 1 mile up.

Didn't have them, perhaps? The development discussion wasn't public until the 80's, IIRC. That, plus the political fallout.

Isn't that the daisycutter, rather than the moab?

3 m/s2 * (1 km/1000m) * (3600 s/hr)2

On average, neutrons get about 2 MeV each. The bulk of the released energy goes into KE of the fission fragments.

Right.

That's the pronblem when you don't understand the process. Carbon dating only works if the object is taking in atmospheric carbon, because C-14 is made in the atmosphere. Organisms that take in non-atmospheric carbon will yield inaccurate dates. Which raises the question of why some idiot would bother to carbon-date a seal?

[math] A=\lambda N[/math] Pop that into the equation for activity and you get # of atoms as a function of time. You can also use N=N0(1/2)n, where n is the number of half-lives that have passed. You should be able to convince yourself that the two equations are equivalent.

About 200 MeV.

Which part? The problem with measuring c, is that if I am in a different frame of reference (i.e. moving with respect to you), we won't agree on the distance you are using. Thus our time measurements won't agree. The way around that is to just pick a reference frame to use. So that's what we do.

IIRC A common tactic is separatation of variables. Assume Vx = f(x), Vy = f(y), Vz = f(z), such that V=VxVyVz Then the solution is fairly straightforward. If you can't separate the variables it's a tougher problem, but many physical solutions behave this way.

Since the Q value of the reaction is 4.871 MeV, that's 28.8 mW (more accurate number than my prior back-of-the envelope calculation that I gave before), so 1 g will not do much for you, as I pointed out earlier in this thread.

[math]A = A_0 e^-^\lambda^t[/math] [math]\lambda[/math] = ln(2)/t1/2 A is activity (decays per unit time) Have at it...

While c is constant, the distance it travels is frame-dependent (length contraction) and that makes the time also frame-dependent. What you end up doing is just picking a reference frame artificially, and let everyone do the conversion. AFAIK we choose the ECI frame (Earth-center inertial) and arbitrarily set all GR corrections to zero on the geoid.

Conservation of energy is a consequence of the laws of physics not changing over time, and vice-versa. Can't have one without the other. So if energy nonconservation has happened, then the laws of physics changed.

a. it doesn't answer the question (how do you do it) and b. it's wrong. (3 m/s = 10.8 km/hr)

Hyper Physics Concepts is a good place for basic stuff. So is Eric Weisstein's World of Physics The recoil issue is just conservation of momentum. A neutron with some KE gets captured and there will be some KE left over in the resulting nucleus because momentum is conserved. This may be large compared to the ionization energy of the atom, in which case you ionize the atom - this causes more damage to surrounding cells.

If you want more info, Google on "freezing point depression" (and/or "boiling point elevation" since it's a related phenomenon)