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swansont

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Everything posted by swansont

  1. The circumstances surrounding this and Hiroshima are hardly comparable.
  2. Klaus Fuchs. The Rosenbergs. And others. Also, your premise that Japan was beaten does not match the facts. They did not acknowledge it. They rejected the Potsdam terms. They did not surrender, even after the first bomb was dropped. Did they do their own Manhattan project equivalent, or did they use pilfered results? One thing about research is the time, money and effort you spend finding out things that don’t work. Subsequent efforts don’t have to expend resources chasing these down.
  3. Yes. I don’t think your post has a sold basis in fact.
  4. Assuming it would be economically feasible to retrieve these metals, I think the question this raises is what would happen if you could suddenly e.g. double the availability of these rare metals. IOW, are there efforts that are supply-constrained? If the new availability drove prices down, it’s possible that you’d come up with new products that aren’t currently viable owing to cost.
  5. The blocked light forms a cone (see the eclipse link for an example) s = r*theta s is the size of the block. theta is the angular size. r is the distance to the screen blocking the light.
  6. There are situations where this isn’t true, though it’s not truly thermal equilibrium - you can cool a sample of a dilute gas with lasers, with the axes being independent. One- and two-dimensional cooling can take place. The atoms equilibrate with the laser light. The photons don’t interact with each other, so the velocity profile in each dimension can be different. The atoms are dilute and move very slowly so they scatter much less often with each other than with the photons.
  7. In thermal equilibrium you can’t decouple these effects. You can’t create a situation where you have the translational motion without the non-translational motion.
  8. Have you ever noticed that e.g. a tree or house some distance away can block your view of things further away?
  9. If you want to block the whole star from all of earth, you need something the size of the earth. If you just want to block one person from seeing it, you need something roughly the size of someone’s face, though diffraction will mess with this. Arago’s spot means light will still be there. https://en.m.wikipedia.org/wiki/Arago_spot
  10. The size of the “light beam” is approximately the size of the source; even bigger, since over 4 LY you can’t ignore the divergence. Alpha Proxima is bigger than a few mm. The beam illuminates the whole earth, i.e. you could observe the star from any point on the surface unless earth itself is in the way. Diffraction means you need something bigger. Diffraction, and that the sun is bigger than the moon, and that the moon is some distance away. The light is not parallel. If you were closer to the moon, the dark spot would be bigger.
  11. This makes no sense. An angle, by definition, assumes a vertex, which is a point. The light from Proxima Centauri hitting the earth can be treated as parallel over such a short distance as the diameter of the earth. The angular size of the star is not the same as the divergence of light from it.
  12. Coxy123 banned as a sockpuppet of splodge and JustJoe
  13. These itemized topics look like math or computer questions.
  14. Yes you can copy and paste links, as long as they comply with the rules https://www.scienceforums.net/guidelines/ Rule 2.7 in particular, especially this part - “Links, pictures and videos in posts should be relevant to the discussion, and members should be able to participate in the discussion without clicking any links or watching any videos”
  15. There are 7-day programmable timers. You’d only have to reset it once a week. Or do e.g. a 42 hour cycle I think there are 14-day timers, too
  16. The magic number is 18. That’s how many years it’s been since someone had last posted.
  17. AIkonoklazt has been banned for repeatedly arguing in bad faith and re-introducing closed topics
  18. ! Moderator Note From rule 2.7: Advertising and spam is prohibited. We don't mind if you put a link to your noncommercial site (e.g. a blog) in your signature and/or profile, but don't go around making threads to advertise it. Links, pictures and videos in posts should be relevant to the discussion, and members should be able to participate in the discussion without clicking any links or watching any videos. Videos and pictures should be accompanied by enough text to set the tone for the discussion, and should not be posted alone IOW, spamming us with videos is not a winning strategy
  19. I think if there were scientific evidence it would be documented better than in a youtube video (which, BTW, needs to comply with rule 2.7, found in the “guidelines” tab; a video is not a substitute for substantive discussion and documentation. Asking people to watch a 25-min video rather than you putting the effort in to explain the situation is not going to fly)
  20. We don't break out the energy stored in rotational modes as a separate term when doing an energy balance Q = mc∆T If temperature is only the translational motion, you'd need an additional term to account for the energy in the other degrees of freedom, but you don't do this, or need to, because you already accounted for it in the heat capacity. Equipartition of energy means that you can't separate the translational from the rotational modes in terms of energy. If the temperature changes, all of the modes gain or lose energy, and the temperature is proportional to this. The distinction shows up in calculations of the internal energy U = aNkT = aPV Seems to me there's no unambiguous argument in either direction, because the equipartition force the energy to be shared between the modes. I see nothing in literature making the distinction that it's only translational KE in systems with additional degrees of freedom. Nothing shows up in the equations, and IMO it's confusing to make that distinction when it doesn't show up in or matter to the calculation.
  21. I think I understand what you mean, but IMO this isn’t the way to express it. Of course the other degrees of freedom contribute to the temperature; it’s right there in the equation. As you said, each degree of freedom has an energy of ½kT. If the other degrees don’t contribute, why does their energy depend on T? If you place an object with 3 degrees of freedom in contact with one with 5, they will still equilibrate at the same temperature. It’s just there are more modes in which you store the thermal energy in the second object. If it’s not temperature, how is the energy accounted for in any of the thermodynamic potentials?
  22. That's the electrostatic interaction, which involves virtual photons. That's typically not referred to as EM radiation, which consists of real photons. The vibrational modes of a solid can be described in terms of phonons (not photons); there are non-radiative ways of changing those states. https://en.wikipedia.org/wiki/Phonon
  23. The translational modes, which are the vibrations mentioned elsewhere
  24. The original topic was about a solid, so the ideal gas law and kinetic theory is moot. However, the kinetic theory shows that it’s the KE of the atoms that matters; these atoms collide elastically with other atoms - these are the analogue of the vibrations in a solid. There is no “internal EM energy” You can’t have motion of atoms, having KE, and have a bunch of EM radiation interior to the system, and have the theory work. kT is directly related to mv^2 (with a constant related to degrees of freedom) If there’s energy stored as EM radiation, there’s less KE, but that means a lower temperature.
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