Recent content by Alewhey

Haha that youtube video is definitely faked somehow. The gravitational constant is far far far too weak to have such an effect.

I'm fairly certain there is no algebraic method. If there was some algebraic way to get x as a function of {3,4,5} then we could replace {3,4,5} with (say) {a,b,c} and have an analytical solution for the general equation. But for a,b,c,x integral, we have Fermat's Last Theorem which clearly does...

Individual photons do NOT have mass (or more precisely, their invarient mass is zero). However they do carry energy and momentum.

One example is nuclear fusion when creating elements heavier than iron. I'm not sure of the precise process (no doubt it is very complicated) but basically two lighter atom smash into each other at immense energies and fuse together, putting some of the energy into mass. This process only occurs...

I don't even know where to begin with this. Its quite clever and totally batty. Shouldn't the Sun and the Earth repel one another by your logic? (Both having an excess positive charge). I should point out that nonstandard physics is disallowed in this forum as it is primarily a teaching...

Essentially yes, but it might be better to think of it as saying that mass *is* energy - that is, mass is one manifestation of energy, just like heat or light, and can in principle be turned into other types of energy. This is clearer if we define c=1 (yes, we can do that!), leaving simply E = m.

I'm not sure if you think that question is somehow related to your previous ones, but anyway: photons carry momentum and as such can also decelerate matter traveling through space. But there is no 'intrinsic' resistance to motion in free space. For such resistance to exist we would have to...

You certainly can calculate it. Since all H2/O2 molecules are identical, the same amount of energy will be released each time and so we can just quote a standard value from a textbook. I presume the reaction you have in mind is 2*H2 + O2 --> 2*H2O which according to...

Well it looks like the question is about a classical potential. If a particle has an energy E and is moving in a potential V, then we have Kinetic Energy = (E - V) = \frac{1}{2}mv^{2} Then solve for v: v = \sqrt{\frac{2(E-V)}{m}}

You have the right equations and the right idea - E fields can be summed linearly - however, Q1 has -ve charge and so will be attracted to the +ve plate, and also to (+vely charged) Q2. So the forces are in opposite directions and one should subtract them to find the net force.

Well OK, so forget the hydrogen molecule and just consider a single hydrogen ion. The s-orbital, for example, has an infinite extent (form exp[-r/a]). Yes, the probability of finding far from the atom is almost zero, but that is irrelevant; the point is that the PE in this distant region is...

Right, R gives the equation of a line in the x-y plane, with direction (-17.3,-20). So a vector perpendicular to the line could be L = m(20,-17.3) With m a parameter which can take any value (similarly to t). We then find the points (x,y) where the two lines cross: x = 20m = 10...

For part a), you have simply got the components confused. Sin() gives the 'west' component not 'south', and vice versa. Try drawing the triangle to see your mistake (sin = opp/hyp, etc). For part b), now that you have the relative velocity vector you can set up t-dependent vector R giving the...

So you are saying that in some bound orbital, there is a point at which the potential is almost zero (e.g. a point equidistant between two H+ ions). But that since the energy of the orbital is a negative constant, this appears to imply that the KE is negative, since total E is simply the sum of...

I get the same. What is the book's answer?