Skip to content

exchemist

Senior Members
  • Joined

  • Last visited

Everything posted by exchemist

  1. I didn't watch it all. The guy's accent is so strong it's a real challenge to understand him. What do you think he is proposing, then? A rule of thumb is it's the amps that kill you not the volts. A battery sounds dangerous to me as it can store a lot of charge, i.e. can supply a lot of amps in a short space of time. A van der Graaf generator on the other hand supplies a high voltage but only stores a tiny amount of charge.
  2. At the risk of being thought racist, I would NOT trust a video from the Indian subcontinent showing somebody putting a screwdriver into an electric socket. When I was in Dubai we were constantly stopping people from that part of the world from doing dangerous things with electricity. Their safety culture around it seemed non-existent. You can safely send sparks between people if you have a van de Graaf generator, but don’t even think of doing anything involving mains electricity.
  3. Many Englishmen get through a litre or more of tea each day, of course. But not in one go.
  4. exchemist replied to MigL's topic in Science News
    And in fact at that temperature it makes bugger all difference whether you use C or K.
  5. What you calculate is the integral. Have you looked at the example I gave you a link for? That shows you how you do the calculation.
  6. I find the graphical representation of this the most helpful in understanding it. If you plot f(x) against x as a curve, f(x)dx - i.e. f(x) times the infinitesimal length dx along the x axis - is an infinitesimally thin vertical strip of area under the curve at the value x. If you add up a series of such strips you get a block of area under the curve. That is what integration does: it gives you the area under the curve representing the function, between two points on the x axis. The integral sign, ∫ , indicates a sum of these infinitely thin strips, adding up to a finite value for the area. Here is one example: https://www.bbc.co.uk/bitesize/guides/zq3ggk7/revision/1 This basic idea spins off into a host of applications, all over natural science and engineering.
  7. Wouldn’t such a drive mean the laws of physics can change with time, though? In which case Noether’s theorem would not apply any more.
  8. No it isn't. N14 can react with a neutron to generate C14 plus a proton. That is how C14 is generated in the atmosphere from the effect of cosmic rays. So your bomb is going to convert nitrogen into carbon, which will probably eventually form CO2 with some of the oxygen in the air. So it seems to me that - aside from all the other issues with this preposterous idea - your bomb is going to increase the CO2 in the atmosphere rather than decrease it.
  9. Evidently not, at least if their freedom of motion is suitably damped. But it seems a lot depends on the speed of rotation, i.e. the vibration frequency: at some speeds it indeed makes things worse but the idea seems to be that at design speed of operation it tends to cancel the out of balance forces. It is evidently highly mathematical and it looks to me as if one would need to be a rotating machinery engineering specialist to understand and explain it. For instance, look at this paper: https://asat.journals.ekb.eg/article_22757_440eb67a81fd7bbd66129baee93f72cb.pdf I'm afraid I don't have the background to be able to follow this. The principle seems to have been discovered by a chap called Thearle in 1932. Perhaps if you look him up you may be able to find a simpler explanation of it.
  10. In the course of this thread we have I think established that a neutron flux will not convert C12 to C14 to any significant extent. But it will indeed be produced from 14N and apparently 17O. So what one would get is a net increase in the carbon dioxide in the atmosphere, created from N and O. Brilliant, eh?
  11. That's because French dinners are social occasions. They don't eat that much, generally. In fact, one benefit of that style is that there is plenty of time for one's system to react and send a signal when you have had enough. (My late wife was French, so I have some experience of family gatherings.) I understand that one factor in obesity can be the bad habit of wolfing down food very quickly, as it takes time for the body to react to the amount of food consumed and send a "full" signal. So if one eats fast, one runs the risk of overeating. A slow succession of small courses, in which ones chooses how much of each to take, is far better from that point of view. I doubt that anyone in France (apart, perhaps from Obelix, who is no role model) would drink a whole litre of milk in a few minutes. It would be thought rather barbarous, I suspect. The French are a lot slimmer than the Americans - or the British.
  12. I did not know what these were until I read your post but, having looked at a few references, e.g. this one: https://www.sciencedirect.com/science/article/abs/pii/S0094114X15001937. my understanding is the balls move so as to take up positions at which their mass tends to bring the axis of the total moment of inertia closer to the axis of rotation. They thus don't just damp the out of balance forces, but actually reduce them by shifting the CoG. Secondly, by doing this they reduce the lateral accelerations (i.e. vibration normal to the axis of rotation), not changes in angular velocity. At least, that would be my reading of how they work.
  13. exchemist replied to skulldude's topic in Physics
    I don't know but I imagine it is that such an engine might be rather impractical for doing useful amounts of work. A Stirling cycle has pseudo-isothermal heat addition and removal, but then there's the thermodynamically complex issue of the heat regenerator.
  14. OK but that would just heat the air up, surely?
  15. Except it would not, as has been explained to you. Very little, if any C14 would be produced and the subsequent decay into N14 would take thousands of years. "Kinetically charged" is meaningless, by the way. What do you mean by kinetic charge, and what is it that you think would be "kinetically charged"?
  16. Yes, I'm very rusty on this but I think you raise quite a subtle point, actually. Setting aside the electrochemistry angle, which gets in the way of the essential issue slightly, this is a general feature of the equation for Gibbs free energy, as a reaction proceeds: ΔG = ΔG⁰ + RT lnQ, Q being the reaction quotient, i.e. [products]/[reactants]. At the start, Q=0, so ln Q is undefined. Note however that this does not predict infinite Gibbs free energy or anything bonkers like that, as it is delta G i.e. dG/dξ, (using ξ to denote the reaction coordinate). So what this is saying is that the gradient of G, when graphed vs. extent of reaction, tends to infinity at the extremes. There is a discussion of this here: https://chemistry.stackexchange.com/questions/115634/calculating-Δg-at-the-extremes-of-reaction-extent/115701?noredirect=1#comment218308_115701 Returning to electrochemistry, since ΔG = -zFE, you do indeed seem to get a theoretically infinite E, at the theoretical extremes in which the activities of products or reactants are mathematically zero. In practice there are a number of catches to that, however, since firstly such extremes are never quite realised in practice and secondly in practice kinetic effects are important, e.g. in your pure water example you have virtually no ionisation to allow the electrochemical reaction to get going. But I had to think quite hard about this and I may have got it a bit wrong after all these years, so I'd quite welcome comments from others.
  17. That does not address my query, which was about how you imagine you would generate C14, as opposed to what it decays into, which is indeed well-known. I can't find anything to suggest that irradiating C12 with neutrons will produce C14, as you seem to think it may. Where do you get that idea from?
  18. Yes and there are antiquarks. An antiproton is composed of 3 antiquarks, for instance. So whether you consider hadrons or the quarks that make them up, it comes to the same thing: they can annihilate with the appropriate antiparticle.
  19. I presume your idea is to generate C14, which decays to N14 by β-emission. I'm not an expert on this but I'm not sure you can readily convert C12 to C14 by neutron bombardment. Have you checked whether there is a pathway for that? (C13 is stable, I gather, so even if you could produce that it would not help.)
  20. I do not believe anyone can identify you merely from the username “therammo”. Nothing else in this list of yours would appear to be applicable.
  21. Read the link. It suggests an association.
  22. Blue light from IT devices is said to be bad at bed time: . https://www.health.harvard.edu/staying-healthy/blue-light-has-a-dark-side. but I don't think anyone says it is generally harmful. Some forms of blue light from LEDs contain a UV component. There are concerns that this can be harmful: https://journals.lww.com/health-physics/Fulltext/2020/05000/Light_Emitting_Diodes__LEDS___Implications_for.6.aspx But I'm not sure whether definitive conclusions have been reached as to whether the risk in commercially available LEDs is significant or not. Maybe someone here will know.
  23. There is also the opioid crisis in the US, which I understand results in suicides. https://www.nimh.nih.gov/about/director/messages/2019/suicide-deaths-are-a-major-component-of-the-opioid-crisis-that-must-be-addressed But I can't account for the seasonality. On looking into it briefly, I am unable to substantiate what I had previously understood about a suicide peak in winter in N Europe. It may be that that is just a popular myth.
  24. I’ve heard the same, for N Europe, I think. But that graph looks as if it may be for the US. Possibly the pressures are seasonally different there.

Important Information

We have placed cookies on your device to help make this website better. You can adjust your cookie settings, otherwise we'll assume you're okay to continue.

Account

Navigation

Search

Search

Configure browser push notifications

Chrome (Android)
  1. Tap the lock icon next to the address bar.
  2. Tap Permissions → Notifications.
  3. Adjust your preference.
Chrome (Desktop)
  1. Click the padlock icon in the address bar.
  2. Select Site settings.
  3. Find Notifications and adjust your preference.