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The best answers address the question directly, and back up facts with wikilinks and links to sources. Do not edit others' comments and do not give any medical or legal advice.
I've read from various non-authoritative sources online that given the choice between a digital volume control and a potentiometer to adjust the volume of sound, I should use the latter (as long as the former isn't set to cause clipping). This makes sense for minimizing quantization noise, since it keeps all the possible digital levels available; but how much difference does it make if other sources of noise are accounted for? For example, will a potentiometer always reduce the DAC's noise floor as much as it reduces the signal? NeonMerlin02:01, 21 November 2022 (UTC)[reply]
Digital volume control does not diminish the number of digital levels. For a simple example, assume the levels of the source are 0, 1, 2, 3 and 4. Then perhaps on the max control setting you get 0 V, 0.50 V, 1.00 V, 1.50 V and 2.00 V. One notch lower may give you 0 V, 0.45 V, 0.90 V, 1.35 V and 1.80 V. So there are still five levels. There is no control setting that will turn input level 4 into output 1.90 V, but the jump from 2.00 V to 1.80 V corresponds to 0.9 dBV, which is barely noticeable. --Lambiam10:20, 21 November 2022 (UTC)[reply]
A potentiometer alone doesn't affect the Signal-to-noise ratio of an analog signal and has no useful effect on a digital signal. A potentiometer at the input to a DAC is useful to avoid overloading the DAC but any further reduction of the input signal leaves some conversion levels unused, resulting in poorer quantisation noise than the DAC with its given number of resolution bits is capable. A Digital potentiometer is a digitally controlled circuit that may be used to scale an analog signal, or it may be a multiplying function incorporated in a DAC. The Signal-to-(quantising plus random)noise ratio established at the analog to digital conversion cannot be improved but it may be worsened if the final digital output lacks extra resolution bits to encompass data MSB when gain >1 (amplification) is applied and to retain data LSB when gain <1 (attenuation) is applied. Philvoids (talk) 20:15, 21 November 2022 (UTC)[reply]
Is the question about controlling the volume at the input to a digital recorder, or at the output? My response above assumed the latter. --Lambiam19:16, 22 November 2022 (UTC)[reply]
There's a number of factors which can make potentiometers start getting a bit crackly especially if they've got a high resistance or are old. And if they are a wiper running over a coil you're digitising anyway :-) NadVolum (talk) 23:17, 21 November 2022 (UTC)[reply]
Potentiometers with logarithmic resistance tracks suited to audio volume control are susceptible to scratchy wear. A design advice is to add a capacitor to block direct current through the slider and its tiny contact point on the track. Philvoids (talk) 12:21, 22 November 2022 (UTC)[reply]
measuring oxygen level in air
Short videolonger similar one The short (30 sec) video shows the experiment, longer one has explanation. The idea is that burning a candle inside a closed jar creates vacuum in the jar, as O2 is consumed. The vacuum is measured by water lifted into the jar. About 20% of the jar's volume is filled with water, so that means 20% of the air is gone. The part that is gone is the oxygen, and the part remaining is nitrogen and other insert gases.
There would be some carbon dioxide. The burning wax make water vapour and carbon dioxide, and later may yield carbon monoxide. The steam will mostly condense into the water anyway. Carbon dioxide will also dissolve a bit in the water. You might get a truer result if you burn something that has a solid combustion product, eg steel wool. Graeme Bartlett (talk) 06:22, 21 November 2022 (UTC)[reply]
Thanks. I did find some pages about using steel wool, but it apparently takes a long time, like days. You wait for it to slowly oxidize, at least on the pages I found. I don't need a very precise measurement so if the CO2 and CO don't affect it too much, I will try to find a way to do the candle method. The complication is that I'm trying to test the output of an O2 generator rather than atmospheric air, so I have to get the generator output into the jar and then light the candle, but I think I can find some hacky way to do that. 2601:648:8200:990:0:0:0:4775 (talk) 06:30, 21 November 2022 (UTC)[reply]
A small spark will ignite steel wool, so if you include a spark plug (as part of a powered circuit you can close from the outside), you can set the reaction off while the jar is closed off from surrounding air. --Lambiam10:29, 21 November 2022 (UTC)[reply]
Hmm you mean a car spark plug with an ignition coil? Seems way overboard. I might order some nichrome heating wire which could maybe also work. I had figured to use it to ignite a match tip, and use the match to light a candle. The idea is to just do a crude estimate in my kitchen, not anything lab grade. The electric ignition is because I'll have to immerse the jar in water, then collect oxygen in it from a plastic tube, then light the candle. Another way might be to use a flex-type barbecue lighter but idk how well that would handle being partly immersed in water. I'll keep thinking about this. Thanks. 2601:648:8200:990:0:0:0:4775 (talk) 10:48, 21 November 2022 (UTC)[reply]
I've no idea about your background, so this is simply a friendly warning. Please be aware that normal, slow, combustion can accelerate to very fast, hot and intense burning in the presence of pure oxygen. You need to ensure that your containers can tolerate the sudden increase in temperature and, depending upon your setup, pressure. As an absolute minimum wear eye protection in case of glass shattering. There's a video here which is designed for medics showing how oxygen makes fires far hotter and fiercer. Martin of Sheffield (talk) 12:05, 21 November 2022 (UTC)[reply]
If you use steel wool as the fuel, you do not need a candle, just enough steel wool. Apparently you can use steel wool directly as the heating element as well.[1] --Lambiam18:40, 21 November 2022 (UTC)[reply]
Lambiam, I think I would have trouble igniting the steel wool with a spark since I don't see how to avoid immersing it in water while collecting the oxygen. Jayron32's suggestion wet steel wool sitting for a week might be doable, though I'd have to do a jar with ordinary air to compare. Actually just waiting 1 day or so might be enough to see a difference in the oxidation speeds. Martin of Sheffield, good point about safety glasses. 2601:648:8200:990:0:0:0:4775 (talk) 05:54, 22 November 2022 (UTC)[reply]
It is possible to ignite steel wool by passing electricity through it. You don't need a spark plug and thousands of volts. Just have a pair of wires and have a few amps passing through the wool will heat it up till it burns. Graeme Bartlett (talk) 10:20, 22 November 2022 (UTC)[reply]
If you follow the link in my reply above, you'll see a method not involving a spark. It may not work well, though, with wet steel wool. --Lambiam10:24, 22 November 2022 (UTC)[reply]
The "about 20% rise of the water due to about 21% original oxygen content of air" is a great example of a logical fallacy (and one that can easily be explained alternately and even disproved), though I can't remember the correct term for this specific mistake. The easy forgotten detail is that the air around a burning candle is hot whereas the jar is cool. Cooling the air near the candle causes a decrease in pressure and therefore water is pulled up into the jar. Here's a good detailed explanation of that, a way to prove it, and yet another interesting idea related to it: [2]. DMacks (talk) 18:15, 21 November 2022 (UTC)[reply]
The air in the container is initially in thermal equilibrium with the outside air. The burning will heat the air, so the pressure rises and will push the water down until the pressure is equalized. On the other hand, the volume of the hot air decreases because of oxidation removing O2, so the water will rise. It is not immediately clear which of the two counteracting effects will be the stronger, but that is not relevant. In the end, the air will cool down to thermal equilibrium again, and only the effect of the volume reduction remains. --Lambiam18:52, 21 November 2022 (UTC)[reply]
I have to agree with DMacks. The crude way that the candle experiment is usually performed would mean that temperature differences (and hence pressure differences) likely play a large role in the result. Even just lowering a jar over a lit candle is going to trap hot air, which will lead to volume changes when it cools. In addition, burning wax will (mostly) create carbon dioxide and water, but the portion that is carbon dioxide will still continue to occupy space in the jar, so even complete consumption of the oxygen would not be expected to reduce the volume by 20%. Dragons flight (talk) 09:22, 22 November 2022 (UTC)[reply]
My interpretation of the envisaged experiment was that by some still-to-be-determined method (like nichrome wire igniting a match lighting the candle) the candle was going to be lit after the container is sealed off. --Lambiam10:33, 22 November 2022 (UTC)[reply]
If I do the thing with the nichrome wire, it will also be easy to check whether there is vacuum created by just heating the air by electrical resistance, with no candle. Yes the idea of igniting the match with a nichrome wire is to be able to light the candle after the jar is sealed. It seems like a pain and I will keep thinking about alternatives. 2601:648:8200:990:0:0:0:497F (talk) 06:44, 23 November 2022 (UTC)[reply]
I've done the activity many times, but with wet steel wool over the course of about a week or so. As the steel wool rusts, it will react away the oxygen from the air. Burning is faster, but not strictly necessary. I just set the apparatus up on the desk at the front of my classroom, and let it react slowly. The water will rise inside the test tube about 1/5 of the way and then stop rising. It's usually good enough for a quick demonstration for school children. See here for an example of how to do it. --Jayron3219:02, 21 November 2022 (UTC)[reply]
a. Wiedemann-Franz law is an empirical law indicating the linear relation between electrical & thermal conductivities. What's the underlying microscopic-quantun explanation ? After all, thermal electronic motions are random in all directions, but electrical conductance is directional (opposing the electrical field).
a. The W–F law applies to metals, in which some electrons are not bound to atoms but can move around, the higher the electrical conductivity, the more freely. These electrons carry not only a charge, giving rise to an electrical current, but also thermal energy, giving rise to a flow of heat. Consider that the motion of the particles of a gas is also random in all directions. What gets transferred as heat are not the particles themselves, in this case electrons, but their kinetic energy.
b. In a sufficiently hot material, hot to the point of incandescence, the atoms and therefore the electrons and protons are violently jostled around, with velocities that can be modelled as a three-dimensional continuous Gaussian distribution. Since they are charged, their motion produces electromagnetic radiation, which in this situation is called thermal radiation.
a. Your comments here repeat known facts, already included & implied in the question. Hence, I'll pinpoint & emphasize it. 1st, let me make a comment on your last phrases: what is transferred isn't just Ek, since the process of collisions between electrons & (metallic) ions on the line of the electric field is a very slow one (drift velocity). So the picture of conductance is more complicated - the energy is transferred like an EM wave. So it's not the mere charge motion. Moreover, this phenomenon depends on the existance of phonons etc. These are the background of my question. To conclude the focus of the 1st question, why electric & thermal are linearly dependent ?
A relevant remark: pay attention to fact that thermal energy transfer is slow, while the electric one is (much) fast(er).
b. Essentially, your answer here is based on the classical model of accelerating or decelerating charges, while in the question the QM picure was addressed - typical metallic energy band structure of the electrons.
c. Following the above discussion, how these two models (classical & QM) correspond to each other ?
@בנצי: Your harsh dismissal of Lambiam’s answer was unwarranted. You did link to the Wiedemann-Franz law, but you called it an "empirical law" and asked for a "microscopic-quantun [sic] explanation". It seemed reasonable to infer that you had not read or understood that article fully, as the section Wiedemann–Franz law#Derivation (the first section past the lead, with a fairly explicit name) explains that the linear relationship follows easily if you assume both transport phenomena follow the Drude model. Lambiam’s level of explanation was perfectly reasonable in light of these assumptions.
Sharpness & focusing aren't harshness, and your comments don't contribute to the discussion. The original question was very clear, and I don't see why you find saying "empirical" a problem. Well, it's not a fundamental law but an empirical one, and was quoted to indicate the linear relation between the two phenomena. I was sorry to find & read this response of yours, let alone postulating on a "perfectly reasonable" with no substantiation on physics grounds, and apologize for having a professional standpoint, and even critical. Raising a question should not lead to inferiority, let alone comments regarding the extent of civil. My intentions were completely honest & genuine, trying to get better insight/s into the mentioned topic. The answers led to none.
Maybe "poor communication" or maybe not. Many thoughts & efforts were made to phrase the questions clearly, but they seem being diminuated, which is insulting enough, let alone taking the position of a judge, in the personal comments like "wisdom", "squarely" blaming me, and the "civil". Thank you very much indeed.
Eventually, I'll find out about what were asked elsewhere, and maybe add it here for the benefit & completion of the discussion, for all. בנצי (talk) 11:56, 23 November 2022 (UTC)[reply]
I have no doubt that you have genuine intentions to learn, and I think you were not uncivil enough to warrant sanctions.
You seem to think that your post is entirely clear, and that Lambiam’s answer was useless. Fine, here’s a simple solution: let’s wait and see if someone else answers. If you don’t get any further replies, it must be because nobody on the RefDesk is qualified in those topics. In that case, make sure to post it on https://physics.stackexchange.com/ and give us a link so we can read the answers and be enlightened. TigraanClick here for my talk page ("private" contact)13:00, 23 November 2022 (UTC)[reply]
I found neither harsh nor uncivil exressions in the OP's response to the first responder.
WF law has two aspects: the empirical one, and the theoretical one the OP asks their question about. Please note that this law really has the empirical aspect, in the sense that this law can be experimentally verified (at least in the specific cases checked out in the experiments), as opposed to some physical arguments which (hypothetically) can only be refuted yet not be verified (like many well known astrophysical theories, and even many mathematical hypotheses, e.g. "this random infinite sequence of digits contains no zeros", and the like, while WF law is different because it can also be experimentally verified). Anyway, the OP only claimed that this law, having its empirical aspect, must (probably) also have its quantum theoretical background, so what is it?
I won't be surprised if it turns out that the OP was one of the main contributors to our article about WF law. Anyway, I couldn't find out how your comment about "knowledge exceeds wisdom" could help.
As far as I could understand, the OP's first question is clear: An electric flow is measured as a vector, while heat is measured as a scalar, so how can this vector and this scalar be linearly related to each other, from a quantum theoretical viewpoint? How does Drude model explain this linear connection?
A last comment, but to the OP this time: Maybe you should have introduced yourself, before you asked your question on this reference desk, beacuse the previous editor could see no userpage of yours. סמי20 (talk) 18:39, 23 November 2022 (UTC)[reply]
I am not convinced by the remark that "thermal energy transfer is slow, while the electric one is (much) fast(er)" is an essential fact because a practical measurement of either must be affected by a finite thermal or electrical capacitance respectively. Philvoids (talk) 20:02, 23 November 2022 (UTC)[reply]
"Not convinced" in what ? what claim is not convincing ? Hold an iron bar in its end, while the other end is put in flame. You must be knowing you don't have to remove your hand too immediately. Hoever, making a call to you on the phone intercontinentally for example, or sending this reply via Wiki interface, require ~microsec. בנצי (talk) 11:23, 24 November 2022 (UTC)[reply]
The iron bar has an isobaric heat capacity of 25.09 J⋅mol−1⋅K−1. That has to be taken into account in your estimate of "time to say 'ouch' ". Philvoids (talk) 14:33, 24 November 2022 (UTC)[reply]
Yes, electric flow is measured as a vector, while heat is measured as a scalar. Conductivity, however, whether electrical or thermal, is neither. It measures the effect of applying a difference (voltage or temperature). One might just as well offer the objection that charge is measured as a scalar, while heat flow is measured as a vector. It is equally irrelevant. --Lambiam20:04, 23 November 2022 (UTC)[reply]
The current density is definitely in essence a vector quantity.
In many useful applications, electricity can be assumed to flow along a single axis (because the length of a wire is much greater than its diameter). Then one can reason in of scalar properties (current, resistance), use Ohm’s law in scalar form, etc. Similarly, when looking at a one-dimensional heat transfer problem, one can use scalar values: for instance, if a thin insulating layer is placed on a large section of wall, the (vector) heat flux is mostly perpendicular to the surface, and one can reason as if temperature / flux varied only along that axis.
However, when looking at a problem where multi-dimensional effects are important, one needs to consider the vector formulation of current density, electric field etc.
Oh, I didn't write it applies to special cases. Of course, it's a physical rule, so it must be universal (regarding metals), so it's supposed to apply to all cases. Maybe I had to add more words to my previous response, for making it clearer. I've only indicated that WF law (like every physical law) is experimentally verified at least in the specific cases checked out in the experiments. Like every physical law. Actually, no physical law can be experimentally verified for all cases, because the set of all cases is infinite. For exmaple, if I measure the mass of one electron, the result is valid for this electron only. For this result to be valid for a second electorn, one has to repeat the experiment for the second electron, and so forth. This principle is true for all physical laws, including WF law, and I'm sure you agree. סמי20 (talk) 16:59, 24 November 2022 (UTC)[reply]
Oh, I didn't write current is a vector. I've only indicated it's measured (e.g. by it's density) as a vector. Anyway, I had to be more precise, because you were asking about the electrical conductance, i.e. the ratio of current to voltage. סמי20 (talk) 16:59, 24 November 2022 (UTC)[reply]
I'm sure if Tigraan had seen your userpage, they wouldn't have thought what they still think about you. Actually, you are more than a username: You are an individual, a person, a human being, and as such you may own several usernames on Wikipedia, so I still claim I won't be surprised if it turns out you are one of the main contributors to our article about WF law (not necessarily under your current username). סמי20 (talk) 16:59, 24 November 2022 (UTC)[reply]
I won't be surprised if it turns out that the OP was one of the main contributors to our article about WF law You know you can check that, right? (The answer is no.)
A last comment, but to the OP [...] Nobody is under any obligation to create a userpage or introduce themselves to edit Wikipedia, and the RefDesk is no exception. I do not modify the answers I give based on the identity of the person who asks (except maybe if the question is clearly susceptible of two interpretations, and I can guess which interpretation to take based on that, but that’s rare). If anyone is provably giving poor-quality answers to people they do not know or like, I support the usual escalation mechanism towards sanctions (friendly warning on talk page, if it fails stern warning, if it fails admin actions).
The OP’s command of English is poor (presumably because it is not their native language); there is no shame in that, but it does impede communication. There might also be a cultural barrier thing, where a perfectly acceptable sentence in one language becomes too harsh when translated in another language. Speaking for myself, I find Your comments here repeat known facts... pay attention to fact [sic] that... to be much harsher than what one would reasonably expect to find in a Q-and-A session.
Finally, I will elaborate on my "knowledge exceeds wisdom" line. I suspect that OP believes they have understood many things which in fact they have not mastered, moved on to much more complex topics, and as a result of that asked a poorly-formulated question (the abovementioned English difficulties compounding it). For example, suppose that someone asks why humans still exist, when our article Cretaceous–Paleogene extinction event says that no tetrapods weighing more than 25 kilograms (55 pounds) survived. Answering that question requires to make assumptions about what misconceptions the asker might have (for instance, do they believe humans coexisted with dinosaurs?); it might require counter-questions to find out where the misunderstanding is.
If you look at the original question, it is hard to tell if the OP would be best served by a basic explanation of how electrons conduct electricity, a hand-waving explanation of Drude model basics, or a technical discussion of the limitations of Drude and alternative models; or an analysis of experiment vs. models for the WF law; or something entirely different. Their following comments do not help much, either. TigraanClick here for my talk page ("private" contact)14:45, 24 November 2022 (UTC)[reply]
November 23
Truth of claims
All Indian school books continuously tell school kids that Jagadish Chandra Bose first proved that life exists in plants
It is rather unclear what the statement means. Therefore it is not possible to say whether any scientific community accepts this. You might as well say that early Indian astronomers proved that the Moon exists in the sky. The second link states: "Before the advent of the 20th century, science did not acknowledge the vitality of trees and plants." Bose is credited with showing this botanical vitality. But any farmer, since prehistoric times, was aware of the vitality of plants. If by "vitality" the concept of the theory of vitalism is meant, then you should be aware of the fact that in today's scientific consensus this is a bogus theory, and "proving" it does not earn one credit. Long before Bose, Erasmus Darwin, a grandfather of Charles Darwin, argued in his book Zoonomia that vegetable life and animal life sprang from a common source and were not fundamentally different. What is certain is that Jagadish Chandra Bose was the first to propose the theory that plants have an internal signalling system, as laid down in his book The Nervous Mechanism of Plants. While the significance of the theory is acknowledged, I think that the consensus is that his experiments did not establish his precise claims, in particular the existence of signalling pathways in plants conveying electrical signals carrying messages. Instead, plants use phytohormones, which are chemical compounds. --Lambiam07:50, 23 November 2022 (UTC)[reply]
Regarding the radio wave claim... For some reason, many people want to claim the "inventor of the radio" for their country. (See Talk:Guglielmo Marconi and the archives thereof.) My reading of our (detailed) article invention of radio is that:
The early theoretical work on radio waves in general was mostly Hertz/Maxwell
Many prototypes of detection apparatus were created in the 1890s, by many researchers. Yes, that includes Bose, but also others with roughly equal contribution (Branly, Lodge, Popov).
The first working long-range communication devices were definitely Marconi’s.
The Year 2038 Problem can manifest itself at any time when dates after 2038-01-19 03:14:07 are considered; you don't actually have to wait until then. OR, but I am already aware of an instance where a company (or perhaps their legal department) requested a 20-year software license key - and it failed as expired. catslash (talk) 21:45, 23 November 2022 (UTC)[reply]
@Catslash: My friend experienced a somewhat similar problem in late 1990's: at some point the results of periodical work-related medical examinations of his employees started to be automatically invalidated by the HR system due to expiration, as a result of 'wrapping' a year last-two-digits representation from the range 2000 .. 2002 into 1900 .. 1902. :) --CiaPan (talk) 00:26, 24 November 2022 (UTC)[reply]
The Year 2038 problem affects systems which store time as seconds since 1970 in a signed 32-bit integer. This format originated in Unix systems, although it's possible that some parts of some older Windows systems also do this. But a 64-bit Windows 11 system almost certainly doesn't store time in a 32-bit variable anywhere in the operating system. The only problem would be if you are running a particular application which stores time this way. CodeTalker (talk) 23:49, 23 November 2022 (UTC)[reply]
If you wish, it is possible to do the experiment now merely by setting your computer's clock to a few minutes before the timestamp you are worried about, switching off its automatic timing updates, if it has them, and seeing what happens. I know that this was how some IT folk tested whether their systems were going to hit the Year 2000 problem and they would need to do something about it. Mike Turnbull (talk) 18:13, 24 November 2022 (UTC)[reply]
Suppose I wanted to pump water at a rate of 5m3/s to a height of 5m. How much energy would that require in kW? If I doubled the height to 10m, would it require double or quadruple the energy? Mjroots (talk) 07:56, 24 November 2022 (UTC)[reply]
From my very rusty physics:
5m3 weighs 5,000 kg, so exerts a force downwards of 5,000g N. To lift this water 5m would therefore take 25,000g J. To do it in one second (ie 5m3/s to a height of 5m) requires 25,000g W, that is 25g kW or roughly 250 kW. All assuming no losses anywhere in the system. To lift to twice the height would indeed require twice the input. Martin of Sheffield (talk) 08:44, 24 November 2022 (UTC)[reply]
You will have to ensure that your pumps can deliver the higher pressure, 10m of water is near enough 1 bar. As the pressure increases the efficiency of the pump (particularly for non-displacement types) will reduce. With any luck a hydraulic engineer will happen along soon. Martin of Sheffield (talk) 14:15, 24 November 2022 (UTC)[reply]
5 m3/s is quite a lot of water, so this is not a small pump. In the image I put on the right (just uploaded from my collection) you can see two pumps that can handle 4 m3/s over 5 m each, powered by 450 kW electric motors, so that's the size of pump you're talking about. PiusImpavidus (talk) 10:10, 25 November 2022 (UTC)[reply]
An amusing thought, but our Dutch neighbors are fairly competent at water handling. It's probably just intended to deal with the small amounts of water arising from normal operations, some of it perhaps condensation, or when the janitor mops the floor. Any substantial volumes would probably flow to larger drains elsewhere out of the picture. {The poster formerly known as 87.81.230.195} 90.217.47.60 (talk) 16:18, 25 November 2022 (UTC)[reply]
When there's a leak, best to shut down the pump quickly. It's above polder water level, so no more water will come through. It may be below river level, but there's a check valve on the outside. This pumping station, in operation since 1933, drains about 160 km2, about half of it in Germany (hence the name), so a single pump can handle a precipitation rate of around 15 mm/week. There are 4 pumps, giving some redundancy. PiusImpavidus (talk) 18:41, 25 November 2022 (UTC)[reply]
Is this true considering the Pump Lawshead is proportional to the square of shaft speed and power is proportional to the cube of shaft speed for centrifugal pumps? If looking at the power applied to the pump shaft, wouldn't doubling the head require 23/2 or ~2.83 times the power? fiveby(zero) 17:23, 25 November 2022 (UTC)[reply]
Not in practice. I would guess that gases get into the mirror from the side and tarnish the silver. If you do an internet search on tarnished mirror you will see plenty of examples. I took a photo of one of our mirrors with tarnish. See the thumbnail image. --Phil Holmes (talk) 12:08, 25 November 2022 (UTC)[reply]
Basically, yes. The glass is a transparent support and seal, the actual mirror is the bright untarnished metal that reflects the light. Just putting the bulk metal behind glass though will not produce a good mirror, there may be surface imperfections and without an adequate seal air will get in and tarnish the reflective surface. Fifty years ago it used to be (maybe still is) a standard school chemistry experiment to make up silvering solution and apply it to a microscope slide. You ended up with a fragile mirror. Industrially the backing is then built up to protect the microscopically thin metal layer from scratches and wear. Martin of Sheffield (talk) 13:56, 25 November 2022 (UTC)[reply]
No. The sandwich consists of glass, foil and something to keep the foil pressed against the glass, such as another sheet of glass, MDF or anything reasonably flat. There is no glue. Read the articles quoted and when you've understood them, then come back here. Martin of Sheffield (talk) 15:14, 25 November 2022 (UTC)[reply]
No, there is no spoon glue. Unless you lay the whole thing flat, you need something to clamp the layers together though, which could be just your hands. The only thing that is important is to keep the aluminium smooth, without wrinkles. For most metals, any object of the material with a smooth polished surface works as a mirror; no glass is needed. --Lambiam19:24, 25 November 2022 (UTC)[reply]
Note, OP, that you have introduced an artificial situation. Actual mirrors are created by the methods described in the two articles linked by Martin of Sheffield above, where the deposited reflective material bonds to the glass at an atomic level (creating a very even and reflective surface). It was you who introduced the ad hoc comparison of placing a piece of pre-existent aluminium behind a sheet of glass: while this corresponds to the broad set-up, it is not how useful mirrors are actually made.
If you actually want to make a mirror at home, use chemicals as described: the process is not difficult – amateur astronomers have been silvering their own telescope mirrors for centuries. In that instance the silver is deposited on the front surface of the glass, not the rear, so it tarnishes quite quickly and has to be regularly re-done. Nowadays, however, most amateurs send their mirrors to professional companies who use vacuum deposition of Aluminium, which tarnishes much more slowly – this is likely beyond the capabilities of home hobbyists. {The poster formerly known as 87.81.230.195} 176.249.29.80 (talk) 08:31, 28 November 2022 (UTC)[reply]
November 25
If an incandescent and LED light bulb both are the same watts...?
The formula relates the wattagedenoted by the letter to the voltage and the current Ohm's law, relates the voltage to the current and resistance We can rewrite both equations by solving for The first gives us the second one It follows that Solving for establishes that So the resistance of an element is determined by the combination of the voltage and its wattage. --Lambiam19:06, 25 November 2022 (UTC)[reply]
Okay, I just Google'd "are LED light bulbs ohmic?" and the 1st response was "LEDs are not Ohmic materials, meaning the relationship between current and voltage is not linear." So back to my older question on how do LED light bulbs work with the same voltage as incandescent bulbs. 67.165.185.178 (talk) 01:46, 26 November 2022 (UTC).[reply]
the efficiency of the LED is about 4-5 times that of the incandescent (light watts/electrical watts), and they undoubtedly have some electronics in the base of them, presumably some sort of rectifier and a voltage downconverter.. Greglocock (talk) 07:18, 26 November 2022 (UTC)[reply]
These are diverse and complicated circuits; broadly, they convert and limit the power from the main supply (often using a switched-mode power supply).
Many vendors place a confusing "wattage equivalent" to help users compare brightness against an old-fashioned incandescent bulb. They often also print the actual power consumed, and often also specify the brightness in units of lumens.
It turns out that a retail lightbulb is quite complicated, in this century; equally, we have more different types of lightbulb with diverse inside-bits.
One manufacturer that I like is Philips - specifically because they provide detailed engineering information about their bulbs - like these LED bulbs with datasheets. It just turns out that retail purchases across their product line-up can be difficult - you might have to order wholesale or through a specialty electrician!
Neither incandescent nor LED lamps are "ohmic" if that means having constant resistance R so that Ohm's law quoted by Lambian exactly applies. An incandescent lamp has a positive temperature coefficient of resistance. The high inrush current to a cold lamp (e.g. R = 40 ohms) causes the tungsten filament to heat and the current to reduce in milliseconds to a stable illumination value e.g. I = 0.86 RMS amp, V = 120 RMS VAC, R = 140 ohms, P = 103 watt. A constant-voltage mains supply is suitable for an individual lamp. In equipment where many vacuum tube (valve) filaments are connected in series, direct connection to a mains supply can cause uneven overheating during switch-on so this is avoided by adding a negative-temperature-coefficient thermistor to the series circuit. LEDs have a steeply rising exponential current/voltage characteristic so LED lamps that contain them include a driver circuit that converts the alternating current from the mains to an effectively constant-current through the LED(s) - often this is achieved by switching the LED(s) on/off with variable duty cycle at an imperceptible high frequency. Philvoids (talk) 15:46, 26 November 2022 (UTC)[reply]
If you don't require R to be constant (whether as a function of temperature or as a function of voltage), Ohm's law applies just fine; simply define resistance by the formula R = I/V. Regardless, I = P/V, so two appliances with the same wattage for a given voltage draw equal amounts of current. --Lambiam11:19, 27 November 2022 (UTC)[reply]
That definition exercise seems to confirm the OP's opening words, that the currents in amps of the lamps are equal (while forgetting that the published "watt rating" of LED lamps is not objective). It is more constructive here to define luminous efficacies for tungsten incandescent bulbs 15 lumens/watt and for LED technology 75-110 lumens/watt. Explanation of the LED lamp driver technology can begin with the article Switched-mode power supply. Philvoids (talk) 18:06, 27 November 2022 (UTC)[reply]
November 27
When UV breaks bonds, are majority time they are just rebonded?
For O2, it takes <241 nm of radiation to break O=O bond, that 241 nm is in the UV-C radiation. For triple bonds like N2, that requires <127 nm. Now I'm looking at plants, and C-C bond, which <343 nm of radiation can break it. That is in the UV-A range. But you don't see plants or animals being broken down by sunlight. My guess is, are these bonds instantly rebonded after a split? So if sunlight breaks an O2 bond, do the 2 singlet oxygen just rebond right back? Then is this also done in longer carbon chains like plants and animals? Maybe for complex molecules like plant/animal structures, there is rearrangements when C-C bonds are broke like 1% of the time. 67.165.185.178 (talk) 10:30, 27 November 2022 (UTC).[reply]
Really interesting question, and animals and plants do indeed break down under UV light! The reason why they appear to not is that they have a myriad of natural systems in place to block UV damage in the first place, and more systems to repair UV damage after it occurs. For example, in plants their cuticles serve to absorb UV light in the B/A range via vacuole-bound flavinoids and phenolic compounds in cell walls or cutin. Also, in most organisms DNA is susceptible to damage caused by UV irradiation, a common form being pyrimidine dimers. These are restored by dedicated protein repair mechanisms.
However, I am no physical chemist so I can't speak to what percentage of UV-induced homolytic cleavage of O2, N2, C-C bonds recombine under natural conditions. I'm sure it occurs some of the time, but not often enough for living organisms to rely on! Synpath (talk) 21:01, 28 November 2022 (UTC)[reply]
What are leaves look like on the molecular level, are they like polymers? -C-C- bonds, or? And yes, just because <343 nm can break a C-C bond, is the maximum largest nm that can. But that could mean that <330 and <320 are still very unlikely to break a C-C bond, so, I never got into the details of that. And would it not have to be at a very precise location? Sunlight is 53-55% IR, but I never found the demographics for what % is visible light and UV. 67.165.185.178 (talk) 00:19, 29 November 2022 (UTC).[reply]
What generates what distribution?
Intuitively it's clear, why throwing a handful of dice and adding the result, would output normal distribution. Combinations like 6-1, 4-3, 5-2 will populate the middle, around an average of 3.5, and extreme combinations like 6-6-... or 1-1-... will be less common, and populate the left and right tail.
Purists would say that real world examples may sometimes resemble certain mathematical functions. All that is required is that the area under the function must be 1. Any attempts to assign a hypothesis or principle to a distribution or pattern of distributions is to be viewed through the lens of statistical hypothesis testing. Abductive (reasoning)21:00, 27 November 2022 (UTC)[reply]
You missed out 'Many' in your quote. It isn't proper quoting when you leave out words like that so the 'quote' says something quite different. Power law distributions typically arise when there is a multiplicative effect in what is being measured. For instance one might expect something like it in how many people access websites or how much money people have. A popular website is talked about more and gets more resources and can develop. A rich person is more easily able to pay for better investment and tax advice and get better rates on loans. You don't get such an effect with for intance the heights of people which is better modelled with the normal distribution. NadVolum (talk) 21:31, 27 November 2022 (UTC)[reply]
Additionally, a power law fitting these observed "natural" distributions is merely an empirical fact without a generic underlying theoretical explanation. The fit is often mainly adequate for the tail only and even then far from perfect, saved only by the bell paucity of high-valued observations. A log-normal distribution may work as well or better. In fact, given an empirically obtained distribution of a random variable assuming values in the set of (theoretically unbounded) positive real values when seeking a fit with a model distribution it is often wise to seek a fit of the empirical distribution of instead. --Lambiam14:03, 28 November 2022 (UTC)[reply]
Yesterday, my mom took a garbage bag of rusty tack strips out, and she got hurt. But it was just a very small scratch that didn't bleed at all and didn't have a lot of dirt in it. Should my mom get a tetanus shot or no? And the last time she got a tetanus shot was 14 years ago. 67.215.28.226 (talk) 22:50, 27 November 2022 (UTC)[reply]
I have a lot of questions about my dad's leg ulcers. Firstly, why did his very severe case of COVID-19 itself cause these ulcers. Secondly, why was he so malnourished? I know he was in agony, but I didn't know one could get so malnourished because of it. Thirdly, why does his leg leak more when he drinks water than other times. Fourthly and lastly, why wouldn't he let the care-giver do anything to it while Jo and Cleo did? Sadly, he passed away on October 30th of this year. And several months ago, he went to a doctor. They gave him an antibiotic, a pee pill, and some ibuprofen. But he wasn't allowed to take the pee pill or the antibiotic because his creatine in his kidneys was a little high, which isn't surprising because he took all those ibuprofens and Tylenols. My mom wasn't even allowed to touch his legs because they were so painful. She checked his oxygen and heart rate. His oxygen on the oximeter was 97%. And his resting heart rate was 110 bpm, which isn't surprising because of the pain. Hours later, his heart rate was 92 bpm. And his blood pressure was 116/76. 67.215.28.226 (talk) 22:15, 28 November 2022 (UTC)[reply]
[Edit Conflict] The following notice is prominently displayed at the top of the Reference desk's Main Page:
"Note: Legal or medical advice is prohibited. Further information is available at Wikipedia's legal and medical disclaimers. See also the medical advice guidelines."
Please follow all three of those links and read the contents carefully. You may then understand why Wikipedia (a) is unable to answer such questions, because it is not in a position to know the answers, and (b) is not allowed to answer such questions, even if it could.
The only people who could answer such questions are the medical staff who attended your dad, or another doctor who could understand and interpret his medical notes, if they were supplied. Many countries in the world have Healthcare systems in which might be possible for either thing to happen at no or little cost. I understand that in your country (I've checked what that is), the Healthcare industry is organised in a way that means this would either not be possible or would be very expensive.
We sympathise with you, but we cannot help in any way I can think of. If I'm wrong, I'm sure another editor will give any advice they can. {The poster formerly known as 87.81.230.195} 176.249.29.80 (talk) 23:12, 28 November 2022 (UTC)[reply]
In the past, bananas would be overripe in two seconds of leaving them there. But my mom and I did a science project. We peeled the banana's half and then left the other half unpeeled. If was supposed to show that the peel kept it safe. But it didn't overripe. I would've done so in the past. But, nowadays, they don't overripe as easily anymore. Did they change the chemicals of a banana or something? 67.215.28.226 (talk) 00:18, 29 November 2022 (UTC)[reply]
