PeroK's latest activity

One thing, because I think that is your main issue - voltage is not a property of electrical current per se (e.g. flow of electrons in...

I'd prefer to say it includes an electric field, but "has voltage" will do, yes. Do note @weirdoguy's comment in #9, though. Although...

It's not very helpful to think of light as made of photons. There are two theories of light. First, there is classical electromagnetism...

Light waves have E and B components. The oscillating E-field component is what accelerates electrons in antennas and other metal...

Yes, in principle you can use the electric component of light to push a current on a circuit. Light oscillates too fast for us to do...

We can also work backwards: $$a^3 \times a^{-1} = a^2$$$$a^2 \times a^{-1} = a^1 = a$$$$a \times a^{-1} = 1 (= a^0)$$$$1 \times a^{-1} =...

Exactly. So you want "multiply by no ##a##s" to be "multiply by 1".

I guess the rationale is this: Start with ##0##, add ##a## you get ##a##. Add ##a## again you get ##2a##. In this case you start with...

Why not define ##a^0 = 0##? And take it from there. For example: The usual index rule for all integers ##m, n## is: $$x^mx^n = x^{m +...

You can find the pdf online, e.g. https://webhomes.maths.ed.ac.uk/~v1ranick/papers/niven.pdf

Yes, this is correct for the given problem. I hadn't understood it properly when I posted originally.

I would rather use “Suppose there are ## n ## possibilities and those (precisely) ## n-1 ## must happen.” Therefore, if the USA wins...

I think you can finish it off by looking at the imaginary part and using the rational root theorem. This would show that either ##a =...

Here's my thinking. We need an angle ##\theta = q\pi## (in radians), where ##q## and both ##\cos \theta## and ##\sin \theta## are...

Closed loops of virtual low-mass magnetic monopoles would modify the QED prediction of ##g-2## for the electron and muon. @Roberto...