apolo399

@apolo399@lemmy.world

• ⁨Comment⁩ on ⁨Space Alert⁩ ⁨⁨4⁩ ⁨months⁩ ago⁩:

  Commenting to check later.

• ⁨Comment⁩ on ⁨ochem⁩ ⁨⁨4⁩ ⁨months⁩ ago⁩:

  Carbohydrates are the ones with (H_2 O)_n

• ⁨Comment⁩ on ⁨[deleted]⁩ ⁨⁨7⁩ ⁨months⁩ ago⁩:

  It is unlikely that Mussolini ever made this statement because it contradicts most of the other writing he did on the subject of corporatism and corporations.

  From your own link.

• ⁨Comment⁩ on ⁨Google Pulls the Plug: The End of Third-Party Cookies and What it Means | TWiT.TV⁩ ⁨⁨8⁩ ⁨months⁩ ago⁩:

  Both of you should look up AdGuard. It’s the only adblocker I use and it works system-wide.

• ⁨Comment⁩ on ⁨The colour of the Sun is white⁩ ⁨⁨1⁩ ⁨year⁩ ago⁩:

  So we can see the where this weirdness comes from when we look at the energy for a photon, E=hf=hc/λ

  When we integrate we sort of slice the function in fixed intervals, what i called above df and dλ. So let’s see what is the difference in energy when our frequency interval is, for example, 1000 Hz, and use a concrete example with 100 Hz and 1100 Hz. Then ΔE = E(1100 Hz) - E(100 Hz) = h·(1100 Hz - 100 Hz) = h·(1000 Hz) = 6.626×10^-31 joules. You can check that this difference in energy will be the same if we had used any other frequencies as long as they had been 1000 Hz apart.

  Now let’s do the same with a fixed interval in wavelength. We’ll use 1000 nm and start at 100 nm. Then ΔE = E(100 nm) - E(1100 nm) = hc·(1/(100 nm)-1/(1100 nm)) = 1.806×10^-18 joules. This energy corresponds to a frequency interval of 2.725×10^15 hertz. Now let’s do one more step. ΔE = E(1100 nm) - E(2100 nm) = 8.599×10^-20 joules, which corresponds to a frequency interval of 1.298×10^14 hertz.

  So the energy emitted in a fixed frequency interval is not comparable to the energy emitted in a wavelength interval. To account for this the very function that is being integrated has to be different, as in the end what’s relevant is the result of the integral: the total energy radiated. This result has to be the same independent of the variable we use to integrate. That’s why the peaks in frequency are different to those in wavelength: the peaks depend on the function, and the functions aren’t the same.

• ⁨Comment⁩ on ⁨The colour of the Sun is white⁩ ⁨⁨1⁩ ⁨year⁩ ago⁩:

  www.scielo.br/j/rbef/a/mYqvM4Qc3KLmmfFRqMbCzhB/?l…

  This is something that bothered me when I was in undergrad but now I’ve come to understand. The article above goes through the math of computing different Wien peaks for different representations of the spectral energy density.

  In short, the Wien peaks are different because what the density function measures in a given parametrization is different. In frequency space the function measures the energy radiated in a small interval [f, f + df] while in wavelength space it measure the energy radiated in an interval [λ, λ + dλ]. The function in these spaces will be different to account for the different amounts of energy radiated in these intervals, and as such the peaks are different too.

  (I typed this on a phone kinda rushed so I could clarify it if you’d like)

• ⁨Comment⁩ on ⁨They are watching⁩ ⁨⁨1⁩ ⁨year⁩ ago⁩:

  Idiotic joke