Recent content by gamesguru

Superconductivity and superfluidity open many doors. I agree I haven't seen anything else substantiated...it's all ********. I can probably dig up some of the crazier claims I've come across if you want. PM me. There might also be technologies which allow us to extract energy from the CMB...

The theory of Hawking Radiation which we have considered above in this topic is that the radiation is due to the emission of particles when particle-antiparticle pairs appear at the boundary of the event horizon and the particle protrudes slightly from the event horizon. According to this...

I fail to grasp why the evaporation rate of matter would be consistently greater than the evaporation rate of antimatter by a fixed amount. In a simplified view, both would evaporate at equal rates since their probabilities would be equal. Obviously this is not the case, and there is an obvious...

^ This is just a theory. Why do we need to assume particle-antiparticles come in pairs? Why can't we assume it's just regular blackbody radiation outside the event horizon?

Perpetual motion under the Maxwell Equations? Doubt it. Energy must have a source. There are some situations where we think quantum systems have a very slight chance of generating free energy...but it's a difficult inquiry and it doesn't seem promising. Under classical physics, it's a big fat...

If you carry out the same calculations I did in the first post I made in this thread, but use vectors, you get this result (you can do it yourself, it's very easy, esp. since I already did it): \vec{F}=m_o \vec{v}\frac{d\gamma}{dt}+m_o\gamma\frac{d\vec{v}}{dt} and using my above post...

Use the fact that v\frac{dv}{dt}=\vec{v}\cdot \vec{a}

As long as you work with fairly heavy weights, the role of air resistance will be negligible and so the weight will not matter, only the height will determine the impact velocity. If you use my above formula, you will find (in feet): velocity=\sqrt{64.34*height}. You can use google now. To...

If the products that you are dropping are relatively aerodynamic you can use a very simple formula: v=\sqrt{2ad}, where v is the velocity when the object hits the ground (ft/s) a is the acceleration due to gravity (32.17 ft/s2) d is the height you dropped it from (ft).

Not many people have this book. Maybe you should take a screen shot and upload it? From what you're saying, I don't see why you would need a Taylor series for the energy of an oscillating mass. It is already quadratic with the displacement (for a spring-mass system at least).

If there are no non-conservative forces acting on the car, then maintaining a constant velocity won't take any energy. Taking drag into account is relatively simple actually. Use the work-energy theorem W=\Delta E and the definition of work W=\int_a^b \vec F \cdot \vec{dx}. The drag equation...

It's not possible to solve, the reason is that inside the sine function there is the function n itself and the independent variable t. There is no way to separate them and the only way to solve this is using taylor polynomials with an approximation. The first degree approximation is only valid...

Yea, to get the power of the blade, just use the formula P=Fv and realize that in this case, v=r\omega so, P=F_dv=\frac{2}{3}C_d\rho h \omega^3 r^4. Easy as that, no integration involved. Do not integrate unless both sides have a differential. In situations where you need to carve something up...

Here's your error: I must admit, it was quite a creative way to approach the problem, perhaps a little more complex than the way I would have chosen. Here's what I would have done: Imagine there is just a single blade (1/4th of what there is in reality, we can quadruple it in the end)...

His lectures aren't as mathematically rigorous as they should be and he doesn't really expand on any concepts. It's the only set of quantum mechanics lectures on the web though, so beggers can't be choosers.