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90

It does! However it doesn't change the frequency enough to matter. An FM transmission is not a precise frequency. Instead it spans a range of about 100 or 200kHz depending on which country you are in. So your FM radio actually accepts a range of frequencies either side of the central frequency. Let's suppose you're travelling at the maximum speed permitted ...


27

To further add to John Rennie's answer: you don't even need autotuning for a frequency drift of the magnitude John calculates (10Hz): all FM receivers I've ever dealt with (I qualified as an electrical engineer in 1985 and worked a few short years in communications before returning to study) demodulated with a phase locked loop detector, whose job it is to ...


18

John and Rod already pointed out that the expected frequency shift from driving a car is "small"; I would like to expand a little more on the way FM works. FM = Frequency Modulation. The carrier (nominal center frequency) is being modulated - that is, in order to convey the audio content, the frequency is actually moving around deliberately in order to get ...


14

You probably are under the misconception that heat travels only via molecular interactions. (i.e, heat transfer by conduction, which needs a medium of sorts). Heat also transfers by radiation, which the sun is an enormous source of. Electromagnetic radiation does not need a 'medium' to travel through. All types of electromagnetic radiation carry energy, ...


4

The heat 'comes' as electromagnetic radiation, that is light. Light from the sun is electromagnetic radiation, that is a wave having energy and momentum or a very big amount of quantum particles, photons, that have energy and momentum. The interaction of this electromagnetic interaction is what heats the earth. Hope this helps.


4

No. Light travels at the speed of ... light, when measured locally in inertial reference frames. And the relationship between wavelength and frequency is $\lambda = c/f$. As the universe expands, the wavelength of the cosmic microwave background photons is "stretched" and thus their frequency must decrease by the same factor of $(1 + z)$, where $z$ is the ...


3

If you or one of your friends has transition lenses for their glasses, then you can test the UV blocking ability with those. The transition to darker shades in these lenses is initiated by UV-light. So hold your sunglasses over some transition lenses and see if they start to turn darker. You can try this with any photochromatic material really, but ...


2

No, the photons do not travel in a helix, they travel in a straight line but with a phase delay that is dependent on position. Looking across the beam's wavefront there is a phase delay that is dependant on the polar angle $\theta$ around the beam axis. If we take a simple helical mode's complex amplitude as $\zeta(r,\theta,z) = u(r,z) e^{-ikz} e^{il ...


2

CMB hasn't a frequency but a typical black body frequencies x radiances distribution. With the Planck law, the curve distribution gives the temperature of the radiation. I found this image in a previous question Relationship between temperature and wavelength? As you can see, each temperature has a typical curve. Yes, red/blueshift affect the ...


1

Remember that $\vec{E}=E_{0}e^{-i\vec{k}\cdot\vec{r}}\hat{x}$, so that, just by identifying the variables $\vec{k}\cdot\vec{r}=k_{x}x + k_{y}y + k_{z}z$ ; from that you should be able to get the wave vector $\vec{k}$.


1

Both \begin{align} \tilde{\bf E}&=\hat{x} E_0 \exp(-i(\pm\frac{\sqrt{2}}{\mathrm{m}} (\hat{x}\frac{\sqrt{2}}{2}+\hat{y}\frac{\sqrt{2}}{2})\cdot(x,y,z)-\omega t))\\ &=\hat{x}E_0 \exp(-i(\vec{k}_\pm\cdot \vec{r}-\omega t)) \end{align} are part of a valid electrodynamic field travelling in opposite directions. (1/m is just the unit 1/meters) with ...


1

A Gaussian beam has a width that changes with distance because of diffraction, which is an effect that takes place in any wave phenomenon. It has a pretty similar description to the Heisenberg uncertainty principle in QM if you're familiar with that. Namely, as the position in the $x$ and $y$ directions (with the optical axis pointing in the $z$ direction) ...


1

Although glass is an amorphous material, it behaves surprisingly similar to crystalline materials in some respects. In this case, you can imagine glass to be a semiconductor with a large bandgap, at least large enough to be beyond the visible wavelengths. Therefore, all visible light passes through, which makes glass transparent. Obviously, there will be ...


1

Classically EM radiation is just a wave (not a particle) and the double slit experiment is only the result of the well known interference proprieties of waves.In quantum mechanics we've got the wave-particle duality, and so the light could be seen as composed by particles, photons. So the first thing to point out is that mixing the two approaches (talking of ...


1

DC current is organized as following: positive potential applied to one end of the wire, negative potential applied to the other. Electrons move from one end to another with some speed. If you have one electron in vacuum and electric field from A to B, then there will be force acting upon that electron due to $F=eE$. Movement should happen along line ...


1

I had the same question once, and scoured the Internet for advice. All I got was conflicting information, much of it from "experts". I ended up getting a signal strength app for my smart phone and one for my laptop and experimenting. In my house, with my router, I found no detectable difference between horizontal and vertical.


1

Without seeing your code, it is hard to know where you went wrong. I did write a few lines of code myself to see what might be going on - and found that the values converge quite nicely as you approach higher temperatures (for the lowest temperatures of 500 K and 1000 K, significant amounts of energy do indeed extend beyond 10,000 nm). updated There is a ...



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