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July 21

Microplastics detachment

Why, even though plastics degrade slowly, over hundreds to thousands of years, microplastic particles detach from its parent body (often bottles), contaminating the surrounding area, but glass bottles, for example, seemingly don't have such property? 212.180.235.46 (talk) 10:58, 21 July 2024 (UTC)

Because plastics degrade and glass doesn't. See Polymer degradation. Alansplodge (talk) 11:08, 21 July 2024 (UTC)

Glass in the sea does degrade too. As it washes up and down on a beach, the surface becomes rough. Graeme Bartlett (talk) 12:20, 22 July 2024 (UTC)

Also, most glass bottles are made from materials that, when worn down into particles, are virtually identical to natural sands, so do not contaminate the marine environment. See also Sea glass (which I have always known as drift glass). {The poster formerly known as 87.81.230.195} 94.2.67.235 (talk) 00:27, 23 July 2024 (UTC)

Colors. (The Color Painter)

I'm not sure where to start. This work defines 12 basic colors. Some are named, and I tried to match them up to sRGB equivalents...

However, I'm no expert on colors.

The original is written in the 1880's, and the author was connected with commercial printers in the US.
It seems reasonable to assume the basic colors were originally intended to match up with available basic pigments available at the time of publication.

So what are the likely 12 original colors (ideally based on pigments that would be generally available to a printer in the US) as modern (and sRGB equivlants?)

The page on Wikisource: s:Page:The color printer (1892).djvu/23 , someone on commons tried to tweak the colors in s:File:The color printer (1892) - Basic tones.svg ShakespeareFan00 (talk) 16:15, 21 July 2024 (UTC)

The author writes, "These colors were adopted because the writer believes that a greater variety of mixed colors can be produced from this selection than from any other containing the same number; besides, these colors are not only the most useful, but also, the most common, and best known among printers." So they were obviously widely available as printing inks, although I doubt they were thought of as "basic" pigments. We have a List of inorganic pigments; I expect most of these 12 are among them. Even if identified, it may not be obvious how to place them in sRGB colour space.  --Lambiam 19:23, 21 July 2024 (UTC)

I wouldn't sadly know where to start looking for information as to common printing inks in the 1880's, but from initial reading around, I'm finding that the articles on pigments don't necessarily have sample colors on them. ShakespeareFan00 (talk) 17:10, 22 July 2024 (UTC)

You probably know this already, but 'The Printer's Manual: An Illustrated History: Classical and Unusual Texts on Printing from the Seventeenth, Eighteenth, and Nineteenth Centuries' by David Pankow says that Earhart's 12 colors came from "twelve stock inks". Maybe they had the same names that he used...not that that helps you get closer to your objective in any way whatsoever. Sean.hoyland (talk) 17:23, 22 July 2024 (UTC)

Wasn't there a museum that kept a collection of old pigments / inks? ShakespeareFan00 (talk) 19:46, 24 July 2024 (UTC)

The Forbes Pigment Collection at the Harvard Art Museums.[1] Their online CAMEO database, of which the Forbes Pigment Database is a section, runs on MediaWiki. DMacks (talk) 23:53, 25 July 2024 (UTC)

Thanks. So it's a case of working out what the stock inks were. In the scan some of the color samples like Vermillion and Rose Lake seem to be different from what I was finding with those names. And whilst the work says it used zinc white, using an approximation I found for that as an HTML triplet, gave colors that were too desturated compared to whats in the scan. Some degree of pigment fading or color change is not unepexected in a more than 100 year old work, but it would be nice to try and figure out collectively what the originals might have been :) ShakespeareFan00 (talk) 19:45, 24 July 2024 (UTC)

Vermilion is actually not one specific colour but a colour family. The pigments are very likely still available today – if not as commercial printing inks, then as artists' paints. There is no guarantee, though, that two colours with the same name from different producers are identical. Also, there is no guarantee that the colours you see on your screen are the same as those on the physical pages of the book. The average RGB triple for the scan of the colour labeled BLUE is (63, 125, 161), or  #3f7da1 , which is not very blue, more like  steel blue , so the scanned B values are perhaps systematically too low.  --Lambiam 00:12, 25 July 2024 (UTC)

July 24

The challenge of space surgery

What are the main challenges of performing surgery in outer space? Have there been any actual attempts to do so either on animals or on humans (other than in fairly controversial Russian disaster movies), and if so, were they successful? 2601:646:8082:BA0:55D1:7827:9FF8:5400 (talk) 08:32, 24 July 2024 (UTC)

The challenges in a zero-gravity setting are manifold. Surgery commonly induces some bleeding. In Earth-bound surgery, gravity keeps the blood from floating away. Free-floating blood droplets will present an unacceptable risk, so foolproof procedures need to be developed to prevent the blood or other bodily fluids from escaping. Free-floating surgical instruments are also not acceptable, but tethering them is awkward. AFAIK no surgery has been attempted under zero-gravity conditions, whether experimental or out of necessity.  --Lambiam 10:40, 24 July 2024 (UTC)

What, not even cutting off a rat's tail, as claimed in the above-mentioned Russian movie? 2601:646:8082:BA0:55D1:7827:9FF8:5400 (talk) 11:11, 24 July 2024 (UTC)

(Zero) Gravity

Before the Russkies went and actually filmed a movie in space, how did moviemakers film zero-gravity scenes for older movies set in space (such as You Only Live Twice, Apollo 13, Moonraker, Deep Impact, Armageddon, Space Odyssey 2001 (what, no article?!), Mission to Mars and Gravity)? Did they film the scenes in the pool, or did they use the Vomit Comet, or what? 2601:646:8082:BA0:55D1:7827:9FF8:5400 (talk) 09:03, 24 July 2024 (UTC)

The article is at 2001: A Space Odyssey.  --Lambiam 10:15, 24 July 2024 (UTC)

Space Odyssey 2001 was directed by Sham Lee Qbrick. Clarityfiend (talk) 11:02, 25 July 2024 (UTC)

For some techniques, see 2001: A Space Odyssey § Zero-gravity effects and Technologies in 2001: A Space Odyssey. Other films have used CGI; see e.g. Gravity (2013 film) § Visual effects.  --Lambiam 10:25, 24 July 2024 (UTC)

The weightless scenes in Apollo 13 were filmed in short sections on a Reduced-gravity aircraft (see Apollo 13 (film)#Filming). Deor (talk) 13:47, 24 July 2024 (UTC)

Yes, the Vomit Comet as noted above. ←Baseball Bugs ^{What's up, Doc?} carrots→ 18:48, 24 July 2024 (UTC)

July 25

Vicarization

Greetings! I was reading the https://en.wikipedia.org/wiki/Fauna_of_Puerto_Rico article and came across the above word, it is blue-linked to "Speciation" but I cannot find that word on the speciation article. I couldn't find a definition in several dictionaries i checked either...Do you know what this word means? could it be made more clear in the article?140.147.160.225 (talk) 18:47, 25 July 2024 (UTC)

The link should be to Allopatric speciation. To my ignorant mind it appears that the correct word is "vicariance", not "vicarization". --Wrongfilter (talk) 19:06, 25 July 2024 (UTC)

Yes. VICARIANCE n. (biology) The separation of a group of organisms by a geographic barrier, resulting in differentiation of the original group into new varieties or species [2] Philvoids (talk) 12:07, 26 July 2024 (UTC)

thanks so much! I've updated the article 140.147.160.225 (talk) 18:42, 29 July 2024 (UTC)

Fels-Naptha

At Talk:Fels-Naptha I raised the issue that the article makes that unsourced claim that "Fels-Naptha once contained naphtha, a skin and eye irritant", but I was unable to find any source showing that Fels-Naptha (note the single "h") soap used to have Naphtha (note the two "h"s) in it, when it was removed, or why it was removed. This may be an urban myth. Can anyone find a source for those claims? --Guy Macon Alternate Account (talk) 19:52, 25 July 2024 (UTC)

Following up links in the references of that article and of Naphtha leads to several sources that state naphtha was dropped from the ingredients, at least one saying this was due to fears it might be carcinogenic. I don't know if any of those sources count as 'Reliable', and none give their sources. This seems unsurprising given the degree to which naphtha was and is used in many products and processes.

'Naptha' seems merely to be a common variant spelling, particularly in product names, presumably because the string -phth- is uncommon in English and likely to be mispronounced and misread by non-chemists. {The poster formerly known as 87.81.230.195} 94.2.67.235 (talk) 20:49, 28 July 2024 (UTC)

July 26

Absorption of matter, without changing the absorber's restmass. Possible?

When a stationary system absorbs a new stationary body, the system's restmass increases by the new body's restmass.

Something similar happens when an electron absorbs light. See Wikisource: The free electrons absorb some of the ultraviolet energy that initially set them free and form an ionized layer.

However, when a stationary body collides with a moving body, the stationary body gains kinetic energy, without changing this body's restmass.

Can a moving body be absorbed by an absorber (like in the first case), but with the absorber's restmass remaining the same as before (like in the second case)? HOTmag (talk) 12:35, 26 July 2024 (UTC)

A free electron cannot absorb light. Ruslik_Zero 13:24, 26 July 2024 (UTC)

Photons caught by a trap increases its mass. As Ruslik_Zero points out, photons interacting with free electrons are not absorbed. Instead they interact elastically, but inelastic collisions are always additive increasing the invariant mass of the absorber. Modocc (talk) 14:26, 26 July 2024 (UTC)

Must the answer to the question in the title be negative?

As for free electrons absorbing light, I've struck it out, but is the book "Electronics Technician" wrong? It's quoted in Wikisource: [3]: The free electrons absorb some of the ultraviolet energy that initially set them free and form an ionized layer.

HOTmag (talk) 14:46, 26 July 2024 (UTC)

Not wrong. The electrons were bonded so the photons' energies broke them without changing their intrinsic rest masses. Modocc (talk) 15:38, 26 July 2024 (UTC)

Ok, so I've just added back the first comment about free electrons absorbing light (I've also added your clarification). Anyway, I'm still curious to know the answer to the question in the title. HOTmag (talk) 16:27, 26 July 2024 (UTC)

In Compton scattering, a free electron gains energy and momentum from a photon, but it does not "absorb" it. --Wrongfilter (talk) 16:52, 26 July 2024 (UTC)

Right, matter absorbs radiation per Quantum electrodynamics. Its rest mass increases unless the energy gets scattered elsewhere. Modocc (talk) 18:06, 26 July 2024 (UTC)

The electron's mass does not increase. --Wrongfilter (talk) 18:26, 26 July 2024 (UTC)

With Compton scattering, it cannot absorb the photon. Modocc (talk) 18:29, 26 July 2024 (UTC)

Who said it does? I'm out. --Wrongfilter (talk) 18:31, 26 July 2024 (UTC)

Sorry. I certainly did not nor did I say the electron's mass increased! And when I realized I simply repeated what you said I was going to fix that. Modocc (talk) 18:35, 26 July 2024 (UTC)

Right, the mass of the bound masses increases and with respect to electrons only their energy increases. Modocc (talk) 18:38, 26 July 2024 (UTC)

HOTmag, as I was trying to say, bound matter's rest mass increases unless the energy absorbed by it gets emitted again. Modocc (talk) 18:49, 26 July 2024 (UTC)

In general, the energy of waves are absorbed: See Absorption (acoustics) and Absorption (electromagnetic radiation). Also, all matter is thought to be comprised of matter-waves. Perhaps that helps. Modocc (talk) 19:50, 26 July 2024 (UTC)

To sum up, electrons' masses are intrinsic and elemental, but the bound rest masses of objects are not. Both absorbed light and matter can increase the latter's (bound rest masses) mass as you observed, but neither can increase the former's mass (the electron's). Thus the answer is no, absorbers do not absorb matter without increasing their rest masses unless they release it, like you noted with particle annihilation, if only because their masses are not as elemental as their constituents... Modocc (talk) 00:24, 27 July 2024 (UTC)

Let me be more clear, now without mentioning photons:

When a stationary system absorbs a new stationary body, the system's restmass increases by the new body's restmass.

However, when a stationary body collides with a moving body, the stationary body gains kinetic energy, without changing this body's restmass.

Can matter be absorbed by an absorber (like in the first case), but with the absorber's restmass remaining the same as before (like in the second case)? HOTmag (talk) 19:30, 27 July 2024 (UTC)

The first case, absorption, always adds rest mass (the second case changes the object's KE, but not its rest mass like within particle accelerators). Modocc (talk) 20:11, 27 July 2024 (UTC)

I didn't ask about the first case, i.e. about a stationary system absorbing a stationary body so that the whole system's restsmass increases, nor about the second case in which the restmass doesn't change.

I wonder, why there can be no third case, i.e, a case in which a system (whether a stationary one or a moving one) absorbs a moving body, so that the system's kinetic energy increases but the whole system's restsmass doesn't change. Is there any reasoning or explanation behind this fact of absence of such a third case? HOTmag (talk) 21:56, 27 July 2024 (UTC)

When an object is at rest its KE is zero, but conservation of energy requires that every object's total energy to be the sum of its parts. We call it rest mass and absorption(s) increases it. Modocc (talk) 22:36, 27 July 2024 (UTC)

I think the conservation of energy is not sufficient for the full explanation: Without the conservation of momentum, one can still argue, that before the absorptoin, the absorber was at rest - hence carried no kinetic energy, while the other body about to be absorbed carried some kinetic energy. After the absorption, the whole system remained surprisingly with the same mass as before, but gained the absorbed body's kinetic energy. What's wrong with that? The wrong thing is my neglecting the conservation of momentum. HOTmag (talk) 00:58, 28 July 2024 (UTC)

The absorbed body adds, at a minimum, a mass-energy equal to KE/c² to the absorber which gains its KE. In addition, for the n-body system, its rest mass and total energy is conserved and unchanged whether they are far apart or bonded together, or internalized and perhaps superimposed. To calculate their combined rest mass one simply adds up their energies in its center-of-momentum frame. In this reference frame the momentum vanishes and their total energy is therefore its rest mass. Modocc (talk) 03:25, 28 July 2024 (UTC)

As to your first sentence about adding a mass $E_{K}/c^{2}$ to the absorber: Please notice, that without the conservation of momentum, one can still argue that before the absorption, the body about to be absorbed carried a total energy $E$ that included - both an internal energy $E_{0}$ - and a kinetic energy $E_{K}$ equivalent to a mass of the size $E_{K}/c^{2}$ you've mentioned. After the absorption, the whole system remained surprisingly with the same mass as before, whereas the absorbed body's internal energy $E_{0}$ was not added to the absorber's internal energy as an addition of the size $E_{0}$, but rather the absorbed body's total energy $E$ was added to the absorber's kinetic energy as an addition of the size $E$. What's wrong with that, without assuming the consevation of momentum, which may actually be not conserved (as you can see in my following thread)?

As to your last claim that "for the n-body system, its rest mass and total energy is conserved and unchanged whether they are far apart or bonded together": AFAIK, what's conserved is the mass-energy as a whole, but the mass alone doesn't have to be conserved: Check: an electron-positron pair, becoming energy alone, without conserving the mass alone (unless one attributes mass to photons, which is a controversial and debatable possibility). HOTmag (talk) 07:11, 28 July 2024 (UTC)

Energy contributes to the rest mass. For example, the gluons' energy within the proton contributes to its overall rest mass. It's a widely accepted concept. Modocc (talk) 12:35, 28 July 2024 (UTC)

Not always. Check: an electron-positron pair, becoming energy alone, without contributing any mass to the light emitted (unless one attributes mass to photons, which is a controversial and debatable possibility). HOTmag (talk) 12:38, 28 July 2024 (UTC)

The rest mass of the 2-body system is conserved. Modocc (talk) 12:54, 28 July 2024 (UTC)

Before the electron and the positron annihilated each other and became energy, the system's rest mass was positive, but after they annihilated each other, the system became light carrying no restmass. HOTmag (talk) 13:24, 28 July 2024 (UTC)

The light carries only energy and momentum yes, but there is a center-of-momentum reference frame in which the particle-waves' momentum vanishes, but their energy, their 2-body rest mass, does not. Modocc (talk) 13:44, 28 July 2024 (UTC)

Yes, their energy does not vanish, but their 2-body restmass does vanish, once they annihilate each other and become light, which actually carries no restmass, so the energy they carried before they annihilated each other does not contribute any mass to the light emitted. HOTmag (talk) 17:19, 28 July 2024 (UTC)

Place the event in an opaque container. Both the container's total energy and momentum are unaffected because its rest mass includes the photons. Modocc (talk) 19:55, 28 July 2024 (UTC)

Yes, when a photon is absorbed it contributes to the absorber's restmass.

However, when an electron-positron pair becomes light in the free space, the pair's restmass vanishes.

This proves that restmass alone, in the free space (rather than in an opaque container), doesn't have to be conserved.

Indeed, restmass is conserved if one assumes both the conservation of energy and the conservation of momentum, but if one only assumes the conservation of energy without assuming the conservation of momentum, then one can still argue, that although any massive body about to be absorbed carries a total energy $E$ that includes - both a kinetic energy $E_{K}$ - and an internal energy $E_{0}$ equivalent to restmass, still after the absorption, the absorber remains surprisingly with the same restmass as before, whereas the absorbed body's internal energy $E_{0}$ is not added to the absorber's internal energy as an addition of the size $E_{0}$, but rather the absorbed body's total energy $E$ is added to the absorber's kinetic energy as an addition of the size $E$. What's wrong with that, without assuming the consevation of momentum, which may actually be not conserved (as you can see in my following thread)? HOTmag (talk) 22:35, 28 July 2024 (UTC)

The container need not contain free space and even if it does as a system its total energy still does not change. Think also of the atmospheric cloud. It consists of enormous numbers of massive and massless particles moving at various velocities, but it is nearly stationary to you and the clouds' total energy that can be calculated is called rest mass. Modocc (talk) 23:00, 28 July 2024 (UTC)

Yes, the container, as well as the atmospheric cloud, are systems each of which conserves the restmass. However, in my previous response I didn't talk about any container, nor about any atmospheric cloud. I only said, that "when an electron-positron pair becomes light, [not in a container nor in an atmospheric cloud, but rather] in the free space, then the pair's restmass vanishes". This proves that restmass alone, in the free space (rather than in a container or in an atmospheric cloud), doesn't have to be conserved. Then I added the crucial last paragraph in my previous response. HOTmag (talk) 06:45, 29 July 2024 (UTC)

We've had the two-photon thing before. --Wrongfilter (talk) 09:32, 29 July 2024 (UTC)

Yes, I remember, but this time I'm talking with Moddoc about an electron-positron pair, which is another issue. HOTmag (talk) 10:54, 29 July 2024 (UTC)

How many photons do you think come out of an electron-positron annihilation event? --Wrongfilter (talk) 11:13, 29 July 2024 (UTC)

2. HOTmag (talk) 12:54, 29 July 2024 (UTC)

July 27

Grooming behavior

In videos of grooming monkeys put all the trash they clean into own mouth, as if eating it, rather than throwing out (same happens when a pet monkey is grooming a human). Why do they do that and isn't harmful for their health? 212.180.235.46 (talk) 10:00, 27 July 2024 (UTC)

How would they know any different? ←Baseball Bugs ^{What's up, Doc?} carrots→ 11:41, 27 July 2024 (UTC)

If they find lice, these are delicious and nutritious. Also yummy and more important are flakes of salt (dried-up sweat). Dandruff is harmless. If it was generally harmful to their health, evolution would have weeded out this specific behaviour long time ago.  --Lambiam 12:32, 27 July 2024 (UTC)

I guess when you are a social animal, putting parasites you have found back on the ground probably isn't a good idea. This is, however, what my dog does with ticks. But the ticks are apparently not delicious, even for fire ants. Sean.hoyland (talk) 05:23, 28 July 2024 (UTC)

Also, small things sometimes (often?) have some antibacterial and antifungal things in their biological toolkits to help them stay alive. Maybe eating them can be beneficial. Sean.hoyland (talk) 05:36, 28 July 2024 (UTC)

What's the opposite of "sticky"?

When a given object tends to get attached to close objects, it may be called "sticky". What about the opposite phenomenon? i.e. How should a given object be called, when it's "reluctant" to get attached to close objects? Even if all objects are at rest, so the adjective "elastic" is not sufficient for describing the opposite of "sticky". HOTmag (talk) 20:20, 27 July 2024 (UTC)

Repulsive, eg. the wall repels objects and the suffix -phobic is used in chemistry, eg. hydrophobic. Modocc (talk) 20:33, 27 July 2024 (UTC)

Thank you. HOTmag (talk) 22:01, 27 July 2024 (UTC)

There is non-stick, which describes objects made of (or coated with) a material with a very low coefficient of friction. Mikenorton (talk) 20:39, 27 July 2024 (UTC)

Thank you. HOTmag (talk) 22:02, 27 July 2024 (UTC)

Hoban (1975) used the term "anti-sticky". [4]  Card Zero  (talk) 03:42, 29 July 2024 (UTC)

"anti-sticky" is good. I guess you could write "not sticky" as "$\neg$ sticky", a downside being that it looks like a stick being described as stick-like. Sean.hoyland (talk) 07:06, 29 July 2024 (UTC)

Thank you. HOTmag (talk) 13:11, 29 July 2024 (UTC)

Teflon? Teflon-like? Huldra (talk) 21:59, 29 July 2024 (UTC)

Also nonadhesive. Modocc (talk) 11:12, 30 July 2024 (UTC)

Slippery Doug butler (talk) 11:24, 30 July 2024 (UTC)

July 28

What is the geological composition of Lascaux?

I am an artist currently working on a project themed around prehistory and I'd like to know about the geological composition of Lascaux (as well as other early human settlements and pre-human geography in general), so I can better represent it. Additionally, is there any sort of database I can use for these purposes? Aedenuniverse (talk) 01:20, 28 July 2024 (UTC)

See karst. Not sure that helps much. Sean.hoyland (talk) 06:05, 28 July 2024 (UTC)

Specifically, it is reported as being calcarenite from the upper Coniacian. Sean.hoyland (talk) 06:10, 28 July 2024 (UTC)

The function of Lascaux is a topic of academic debate, but it was not a settlement. It, and similar caves, are not suitable for serving as dwellings. Human groups of that time lived in tents or the open air, and may occasionally have found shelter in much shallower caves.  --Lambiam 12:02, 28 July 2024 (UTC)

Conserving the kinetic energy, without conserving the momentum. Possible?

There being no external forces, inelastic collisions conserve the whole system's momentum, without conserving the whole system's kinetic energy.

What about the opposite physical process? I.E. is there any physical process, e.g. with external forces, which conserves the whole system's kinetic energy, without conserving the whole system's momentum? HOTmag (talk) 07:27, 28 July 2024 (UTC)

Resolved

I've just thought about it:

The whole system is: a single elastic body.

The physical process is as follows: the single elastic body collides with an elastic wall, being the external force.

Result: the single elastic body is pushed back to the oppposite direction, with the same speed as before the elastic collision took place.

Therefore: the elastic body's kinetic energy is conserved, yet the elastic body's momentum is not. QED. HOTmag (talk) 10:05, 28 July 2024 (UTC)

Nah, you've ignored newton 3. QED Greglocock (talk) 23:47, 28 July 2024 (UTC)

Why do you think I ignored it? I think I didn't. HOTmag (talk) 06:47, 29 July 2024 (UTC)

My second thought is to underline 3 words in my response below. Ignoring what is happening by calling one involved object "the whole system" is disingenuous. Philvoids (talk) 18:21, 29 July 2024 (UTC)

The decision of what the "whole system" is, depends on our choice. Here is a typical example: If two perfectly elastic bodies collide with each other, we can choose, whether to call the two-body system: "the whole system", in which case no external force is involved - hence the momentum is conserved in our "whole system" chosen, or to call one colliding body alone "the whole system", in which case the other body exerts an external force on our "whole system" chosen - hence the momentum is not conserved in our "whole system" chosen. This is how all of physics works, with no exceptions. HOTmag (talk) 20:39, 29 July 2024 (UTC)

The wall will gain some momentum. In a multi body collision the centre of gravity of the system remains at constant velocity (N1) in the absence of external forces. You are 100% wrong, and are just making up random explanations to cover up your lack of understanding. I'll be ignoring you from now on. Greglocock (talk) 23:54, 29 July 2024 (UTC)

I'm referring to a very specific reference frame, which is the wall's reference frame, in which every observer referring to the elastic body as "the whole system" attributes no change to the wall's momentum.

I'm bad at analyzing personal comments, so I'm letting the users decide who is right and who is wrong.

HOTmag (talk) 09:18, 30 July 2024 (UTC)

If you're analysing a collision, you can't just pick one of the things that's colliding and call it the whole system. If it was the whole system, there would be nothing else for it to collide with. "The whole system" is not a reference frame. It's the whole set of objects, by definition; you can't pick and choose which objects it includes. AlmostReadytoFly (talk) 09:57, 30 July 2024 (UTC)

A side note: I've never claimed that "the whole system" is a reference frame.

As to your main response: Can you give any example of a two-body system, for which our referring to one of them as the "whole system" may contradict the laws of physics? AFAIK, there is no example of this kind. Recommendation: before you try to think about such an example, take another look at the example I've already given in my previous-previous response (i.e. the one beginning with "The decision"). HOTmag (talk) 10:35, 30 July 2024 (UTC)

If you have a two-body system, the two bodies are the system. If you say you have a one body system, then say that that body collides with something else (e.g. a wall), you're contradicting yourself. AlmostReadytoFly (talk) 11:30, 30 July 2024 (UTC)

Maybe the expression "the whole system" confuses you. I can replace it by the expression "the sub-system chosen", in which case no contradiction follows, even according to your attitude.

One contradicts oneself when one says "x" and then says "not x". My analysis is not the case, even when the expression "whole system" is used, but I'm changing it for you, to avoid confusion. HOTmag (talk) 12:32, 30 July 2024 (UTC)

"When I use a word," Humpty Dumpty said in rather a scornful tone, "it means just what I choose it to mean——neither more nor less."

If you just want a physical process where a body keeps its kinetic energy but not its momentum, consider a body in a circular orbit. AlmostReadytoFly (talk) 13:00, 30 July 2024 (UTC)

Humpty Dumpty didn't care if the way he spoke could confuse others, but I do care, and that's why I changed the expression "whole-system" to "sub-system".

Yes, also the body you suggest can be chosen as the sub-system. But also the body I've suggested can. HOTmag (talk) 17:56, 30 July 2024 (UTC)

Your Question: is there any physical process, e.g. with external forces, which conserves ... kinetic energy, without conserving ... momentum?

Answer: Yes, a circular orbit. AlmostReadytoFly (talk) 18:24, 30 July 2024 (UTC)

I've already approved your answer, in my previous response, so I wonder why you had to repeat the same answer. I only added that also my answer (that preceded yours) was correct. HOTmag (talk) 18:59, 30 July 2024 (UTC)

(ec)

All types of collision (inelastic or elastic) conserve momentum. Total kinetic energy would be conserved (meaning no release of sound or heat) only in an impractical perfectly elastic collision. To identify a process that conserves a system's kinetic energy but may be affected by external forces, one needs firstly to clarify whether the system shall qualify as an Isolated system where thermodynamic laws apply. The Second law of thermodynamics observes that the entropy of isolated systems left to spontaneous evolution cannot decrease, as they always tend toward a state of thermodynamic equilibrium where the entropy is highest at the given internal energy. Philvoids (talk) 10:10, 28 July 2024 (UTC)

July 29

Access to reference 9 in Petrichor

Garg, Anu (2007). The Dord, the Diglot, and an Avocado Or Two: The Hidden Lives and Strange Origins of Words. Penguin. p. 399. ISBN 9780452288614.

Do you have access to the page 399? I would like to find out what it says about the two researchers, Isabel Bear and Dick Thomas. (In ruwiki the same source is cited and it is written that Thomas was from UK, whereas the enwiki says they are both from Australia).

Thank you in advance for your help. Gryllida (talk, e-mail) 00:41, 29 July 2024 (UTC)

The ref says, "In 1964, two Australian researchers, I.J. Bear and R. G. Thomas..." but does not provide any further biographical informaion about either one. Our article links to enwiki articles about each of them: Isabel Joy Bear and Richard Grenfell Thomas. Bear's article makes a strong claim for her being Australian, even though she did work for a few years in the UK. Thoman's article does not have any hint of any national connection other than Australia. DMacks (talk) 01:02, 29 July 2024 (UTC)

Thanks, Bear's article says "In the 1950s Bear moved to the United Kingdom, where she worked at the Harwell Science and Innovation Campus. She moved to the University of Birmingham, where she worked as a postdoctoral researcher in the department of metallurgy. Whilst working in Birmingham Bear became interested in solid-state chemistry. Bear joined the Council for Scientific and Industrial Research (CSIRO) in 1953, [...]" -- does it mean she apparently worked in the UK between 1950 and 1953?

The Nature paper was in 1964. Gryllida (talk, e-mail) 04:05, 29 July 2024 (UTC)

Yes, Reference [3] after the first sentence in your quote says explicitly "

In the UK (1950-53)

During three years in the UK she was employed first as an Experimental Scientist in the Metallurgy Division of the Atomic Energy Research Establishment at Harwell, and later as a Research Assistant in the Metallurgy Department of Birmingham University." AlmostReadytoFly (talk) 15:45, 29 July 2024 (UTC)

The fastest Internet speed during rain?

So ChatGPT says the fastest Internet speed during rain is fiber topics, then cabled Internet, then mobile wireless Internet. I just want a 2nd opinion if anyone agrees or disagrees? Specifically, ChatGPT said:

• Most Affected: Mobile wireless internet is the most affected by rainy weather due to signal attenuation.
• Moderately Affected: Cabled internet can be affected if the infrastructure is old or damaged, but it is generally more resilient than mobile wireless internet.
• Least Affected: Fiber optic internet is the least affected by rain due to its well-protected, light-based transmission system.

Thanks. 66.99.15.162 (talk) 19:36, 29 July 2024 (UTC).

Fibre optic cable is always the fastest. The bot is right that wireless technology suffers from rain by signal attenuation, which, provided the protocol is designed to make use of good conditions, can slow down internet speed in rain. Some attenuation isn't too bad, as it reduces interference between nearby cellphone towers, without significantly affecting signal strength on short distances. Once the transmitter reaches maximum power, more rain reduces possible speed.

Wired technology is generally unaffected by rain, unless there's so much rain that it enters the cabinets housing routers etc. or causes landslides, ripping the cables apart. Maybe ChatGPT thinks (to the extend that machines can think) that copper cable networks tend to be older than fibre optic networks and therefore more susceptible to such water intrusion.

Copper networks are typically faster than wireless for the same reason as why speaking tubes are better than providing everybody with a megaphone: the more people shout over the same medium, the more confusion. PiusImpavidus (talk) 09:47, 30 July 2024 (UTC)

The speed of mobile wireless Internet access depends more on the generation of broadband cellular network technology deployed locally (2G, 3G, 4G, 5G) than on the weather conditions. If there is no cellular coverage, the only solution is satellite Internet access, which can stream at a high rate but has a high latency. In all cases (wireless, cable, fiber) the bandwidth may depend on the contract with the provider – often one can opt for a subscription with a higher rate at a higher cost. And in all cases the actual latency and streaming rate may be much lower than the promised one.  --Lambiam 09:57, 30 July 2024 (UTC)

Historic (pre-1800s) Wildfires in California

I read that historically, about 2-4 million hectares would burn a year in California, https://www.propublica.org/article/they-know-how-to-prevent-megafires-why-wont-anybody-listen and https://www.sciencedirect.com/science/article/abs/pii/S0378112707004379. A claim made in the sources was that smoke was a feature of the landscape, rather than an oddity as it is now. If the "extreme" modern season tends to be around 2 million ha with moderate air quality impact, what would be some rough estimates for the average pm2.5 levels across pre-1800s California in late summertime? Takedalullaby (talk) 21:01, 29 July 2024 (UTC)

Sounds like The Burning City by Larry Niven and Jerry Pournelle. They claim that the Los Angeles valley was originally called Iyáangẚ, "the valley of smoke". by the Tongva. Abductive (reasoning) 20:17, 30 July 2024 (UTC)

July 30

Rare quasi-cancer disease in children

Friends recently received a diagnosis for their toddler, and they told me what the doctor had called it, but now I can't remember its name. I know that it has a Wikipedia article, because I read the article when they first told me.

All I remember of the little guy's symptoms is that he frequently has joint pain at unexpected times. The doctor explained to them that the disease causes lesions in random places (including in the bones, if I remember rightly), and because these bulges occur in places where they shouldn't be, some interfere with ordinary movement and cause pain. The disease is treated with chemotherapy, and apparently there's some debate among the experts over whether it should be classified as a kind of cancer. I think the doctor gave a reasonably good prognosis for the disease with treatment and a dreadful prognosis without treatment. I don't remember if the Wikipedia article mentioned if the cause is known, or if it is, what causes it. It's not so rare that the exact number of diagnoses is known, but it's classified as rare (at least here in Australia) because it occurs only once per several thousand individuals. I've looked through Category:Syndromes with musculoskeletal abnormalities without finding it.

While this desk doesn't provide medical advice, remember that I'm not asking for diagnosis: I'm basically starting with a diagnosis and trying to work out the name. This is similar to the strep-infection question from Wikipedia:Kainaw's criterion. Nyttend (talk) 19:27, 30 July 2024 (UTC)

Histiocytosis? Ruslik_Zero 20:39, 30 July 2024 (UTC)

Ah, it's Langerhans cell histiocytosis. Now I remember my confusion when talking with the friends, since initially I thought they meant it was something pancreatic. Thanks for the pointer! Nyttend (talk) 21:35, 30 July 2024 (UTC)

July 31

Is there any simple necessary sufficient condition that conserves kinetic energy, without mentioning kinetic energy?

Note that the absence of external net force is a simple necessary sufficient condition that conserves momentum, without mentioning momentum (just as the absence of external torque is a simple necessary sufficient condition that conserves angular momentum, without mentioning angular momentum).

However, the absence of external net force is not a necessary condition (that conserves kinetic energy), because a given body's kinetic energy can be conserved even when external net force are exerted on that body, e.g. when it's in a circular orbit (in which case the space must have more than one dimension), or when the body elastically collides with a wall sharing a reference frame with an observer who measures the body's kinetic energy (in which case the space is allowed to have a single dimension).

The absence of external net force is not a sufficient condition (that conserves kinetic energy) either, as can be shown when two bodies inelastically collide with each other: The two body system's kinetic energy is not conserved [when the whole system is not seen at rest], although no external net force is exerted on that two body system.

HOTmag (talk) 07:11, 31 July 2024 (UTC)

In a system of two equal bodies circling each other around a common centre of mass, the net momentum of the system is constant, yet there are forces at play.  --Lambiam 09:18, 31 July 2024 (UTC)

The forces you are talking about, are internal ones, each of which is exerted on only a part of the two body system, on which no external force is exerted. Any way, for the sake of clarity, I've just added the word "external" to my first post. HOTmag (talk) 09:37, 31 July 2024 (UTC)

.

.

Resolved

I've just thought about it, assuming that restmass does not change:

The conservation of velocity is a simple necessary sufficient condition that conserves momentum, without mentioning momentum.

The conservation of speed (i.e. of the absolute value of velocity) is a simple necessary sufficient condition that conserves kinetic energy, without mentioning kinetic energy. HOTmag (talk) 10:49, 31 July 2024 (UTC)

You write ‘The absence of external forces is not a sufficient condition ...’ This is debatable. I say that the kinetic energy of the constituents of the system doesn't necessarily contribute to the kinetic energy of the entire system. The energy associated with the movement of the inelastically colliding bodies relative to their common centre of mass is counted as internal energy of the system, not as kinetic energy. Take for example a bottle of warm gas in a circular orbit around Saturn. After a while, the gas cools down. The kinetic energy of the gas molecules decreases, but the kinetic energy of the bottle of gas remains the same; the thermal energy decreases. PiusImpavidus (talk) 18:08, 31 July 2024 (UTC)

Had the absence of external forces been a sufficient condition that conserves kinetic energy, then under that condition - the kinetic energy would've been conserved in all cases - hence in all reference frames and not only when the whole system is seen at rest. Hence, the absence of external forces can't be a sufficient condition that conserves kinetic energy, because when two bodies inelastically collide with each other while the kinetic energy is measured relative to any point that doesn't see the whole system at rest - then the system's kinetic energy does change after the collision - even though no external force is exerted on the system (because the system's momentum does not change). Any way, for the sake of clarity, I've just added this clarification to my paragraph you've quoted. HOTmag (talk) 20:54, 31 July 2024 (UTC)

We have two bodies, inelastically colliding. The kinetic energy of each of the bodies isn't conserved and external forces act on each of them. Namely, the force exerted by the other body. On the other hand, the kinetic energy of the system of two inelastically colliding bodies is conserved, as no external force acts on the system.

Consider the combined kinetic energy of two bodies with masses $m_{1}$ and $m_{2}$ and velocities ${\vec {v}}_{1}$ and ${\vec {v}}_{2}$:

$K={\frac {1}{2}}m_{1}{\vec {v}}_{1}^{2}+{\frac {1}{2}}m_{2}{\vec {v}}_{2}^{2}$

The velocity of the centre of mass is

${\vec {v}}_{cm}={\frac {m_{1}{\vec {v}}_{1}+m_{2}{\vec {v}}_{2}}{m_{1}+m_{2}}}$

Subtracting ${\vec {v}}_{cm}$ from ${\vec {v}}_{1}$ and ${\vec {v}}_{2}$ and rearranging some terms gives the combined kinetic energy of the two bodies in the centre-of-mass frame as

$K_{cm}={\frac {m_{1}m_{2}}{2\left(m_{1}+m_{2}\right)}}\left({\vec {v}}_{1}^{2}+{\vec {v}}_{2}^{2}-2{\vec {v}}_{1}\cdot {\vec {v}}_{2}\right)$

Now suppose we put the two masses into a black box and calculate the kinetic energy of the system in the box:

$K_{s}={\frac {m_{1}+m_{2}}{2}}{\vec {v}}_{cm}^{2}={\frac {1}{2\left(m_{1}+m_{2}\right)}}\left(m_{1}^{2}{\vec {v}}_{1}^{2}+m_{2}^{2}{\vec {v}}_{2}^{2}+2m_{1}m_{2}{\vec {v}}_{1}\cdot {\vec {v}}_{2}\right)$

Now add $K_{s}$ and $K_{cm}$ together and notice how the dot products cancel. Simple calculation shows that

$K=K_{s}+K_{cm}$

That was the maths, now the physics.

The two bodies each have their own kinetic energy. However, when we package the two bodies into a system, the system as a whole has a kinetic energy less than the kinetic energies of the constituent parts. The rest of the kinetic energy of the constituent parts isn't kinetic energy of the system, but internal energy of the system. And this internal energy is equal to the kinetic energy of the constituent parts in the centre-of-mass frame.

So, in your system of two inelastically colliding bodies, the kinetic energy of the bodies is indeed not conserved, but there's an external force acting on each of them. The kinetic energy of the system of two bodies (on which no external force acts) is conserved; it's the internal energy that decreases. PiusImpavidus (talk) 12:14, 1 August 2024 (UTC)

I agree to all of your maths, as well as to all of your claims about physics, except two sentences: "the kinetic energy of the system of two inelastically colliding bodies is conserved, as no external force acts on the system....The kinetic energy of the system of two bodies (on which no external force acts) is conserved".

The correct fact that "no external force acts on the system", is not sufficient for justifying your claim that the system's kinetic energy is conserved. As I've already pointed out in my previous response, the system's kinetic energy is only conserved when the whole system is seen at rest. For more details, see our article inelastic collision: "inelastic collisions do not conserve kinetic energy" [i.e. not in all reference frames].

Btw, for simplicity, let's assume that ${\vec {v}}={\vec {v}}_{1}=-{\vec {v}}_{2}$ and that $m=m_{1}=m_{2}$ (hence $K=K_{cm},$ and $K_{s}=0).$ HOTmag (talk) 14:23, 1 August 2024 (UTC)

The kinetic energy of the system $K_{s}$ only depends on the total mass of the system and $v_{cm}$. In the absence of external forces, neither of those change, so $K_{s}$ is conserved, in every reference frame. What the bodies do to each other is irrelevant. $K_{cm}$ and $K$, the difference between which is constant, are not conserved in an inelastic collision and the loss of energy is the same in every reference frame, including the centre-of-mass frame.

The nice thing I tried to show you is that the sum of the kinetic energies of the bodies can be broken into two parts, being the kinetic energy of the system of two bodies and the internal energy of the system, which can simply be added together. When you calculate the kinetic energy of a rock, you don't include the kinetic energies of all its vibrating atoms, right? Because that's the thermal energy of the rock. It's exactly the same here. PiusImpavidus (talk) 08:07, 2 August 2024 (UTC)

Thank you ever so much for your clarifications. So, I'm striking out the wrong thing in my first post (See above). HOTmag (talk) 11:12, 2 August 2024 (UTC)

Consider again a system of two equal bodies orbiting around a common centre of mass. Choose the coordinate system such that the common centre of mass of the two-body system is at rest. If one body's momentum equals $\mathbf {p}$ at some instant of time, that of the other at the same instant of time equals $-\mathbf {p} ,$ so the momentum of the system is $\mathbf {0} .$ Now apply external forces rotation-symmetrically to the objects so as to reverse their motion, making them circle again around their common centre of mass – which has not budged – but in the opposite sense. The symmetry guarantees that the momentum of the system remains $\mathbf {0}$ at all times, so the absence of external forces is not a necessary condition for conserving momentum.  --Lambiam 20:06, 31 July 2024 (UTC)

Your case does not involve an external force but rather involves an external torque. Any way, for the sake of clarity, I've just added this clarification to the first paragraph of my first post. HOTmag (talk) 20:54, 31 July 2024 (UTC)

...but equal and opposite applied forces like compression conserves momentum too. In other words, absence of external forces is a sufficient condition but it is not necessary. Modocc (talk) 21:35, 31 July 2024 (UTC)

By "equal", I guess you mean "having the same absolute value".

Anyway, when the external forces are equal [in their absolute value] and opposite, then the sum of those external forces is zero. Hence, saying that the external forces are equal [in their absolute value] and opposite, is like saying that there are no external forces. Hence, if you want to prove that the absence of external forces is not a necessary condition that conserves momentum, you will have to give an example in which the sum of the external forces is not zero. HOTmag (talk) 06:47, 1 August 2024 (UTC)

The fact that external forces can sum to zero does not imply they are absent... For example, without material internal resisting forces submarines and neutron stars implode. Modocc (talk) 08:18, 1 August 2024 (UTC)

All depends on what we mean by "absence of external forces". By "absence of external forces" I intend to include also all cases of external forces summing to zero. HOTmag (talk) 08:41, 1 August 2024 (UTC)

If you mean "absence of external net force" then why not write that instead? Also if F=0 then dP/dt=0 per Newton's second law, see Force. Modocc (talk) 12:09, 1 August 2024 (UTC)

Re. your first sentence: Ok, I'm adopting your current suggestion (See above in my first post).

Re. your second sentence: Of course, but what was wrong in my saying (ibid.) that "the absence of external [net] force is a simple necessary sufficient condition that conserves momentum, without mentioning momentum"? HOTmag (talk) 14:23, 1 August 2024 (UTC)

Nothing wrong, so I deleted the second part. Then in hindsight I deleted the first part because in many contexts it's understood, but not always. Then you restored the entire comment. Sigh. Modocc (talk) 15:02, 1 August 2024 (UTC)

Probably I had started responding to your response before you deleted it? Anyway, surprisingly, I didn't get any warning of "edit conflict" when I responded to your deleted response. HOTmag (talk) 15:56, 1 August 2024 (UTC)

Also, it is best practice to strike the original phrase(s) when modifying them so Lambiam's and my comments retain context. Modocc (talk) 15:54, 1 August 2024 (UTC)

Agree. Next time... HOTmag (talk) 15:57, 1 August 2024 (UTC)

${\vec {F}}\cdot {\vec {v}}=0$ PiusImpavidus (talk) 12:50, 1 August 2024 (UTC)

In my first post I gave two counter-examples:

1. On the one hand, when the body elastically collides with a wall sharing a reference frame with an observer who measures the body's kinetic energy, then the body's kinetic energy is conserved even though this case does not satisfy your condition. Hence it's not a necessary condition.

2. On the other hand, when two bodies inelastically collide with each other, the two body system's kinetic energy is not conserved [when the whole system is not seen at rest], even though this case does satisfy your condition. Hence it's not a sufficient condition. HOTmag (talk) 14:23, 1 August 2024 (UTC)

Both counter-examples fail:

1. While bouncing, ${\vec {v}}$ changes such that it is on average either 0 or parallel to the wall, whilst ${\vec {F}}$, the normal force, is perpendicular to the wall. Therefore, their dot product is zero, so the condition is satisfied.
2. I explained that one above. You have to make a distinction between the sum of the kinetic energies of the two bodies and the kinetic energy of the system of the two bodies. The former decreases in every reference frame, but the latter is conserved in every reference frame, so this doesn't prove that the condition is insufficient.

PiusImpavidus (talk) 08:39, 2 August 2024 (UTC)

So by ${\vec {v}}$ you meant the average velocity. I couldn't understand it before you made it clear. Instead of you, I would write: ${\vec {F}}\cdot ({\vec {v_{2}}}+{\vec {v_{1}}})=0.$

Anyways, your necessary sufficient condition could also be expressed verbally: "a zero external net force - or a zero average velocity - in every coordinate", right? HOTmag (talk) 11:12, 2 August 2024 (UTC)

Is there any simple necessary sufficient condition, that conserves a given system's total energy, without mentioning energy/mass?

Just as absence of external force is a simple necessary sufficient condition that conserves momentum, without mentioning momentum. HOTmag (talk) 10:57, 31 July 2024 (UTC)

The net work done by the system is equal to the net heat received. PiusImpavidus (talk) 17:45, 31 July 2024 (UTC)

It's not a necessary condition: Consider a harmonic oscillator, in which the total energy is conserved, even though the (changing) net work done by the system is not equal to the (zero) net heat received. HOTmag (talk) 21:09, 31 July 2024 (UTC)

What work? An isolated harmonic oscillator does no work on its surroundings. The parts forming the isolated harmonic oscillator do work on each other and their energies oscillate, but the whole system does no work and has constant energy.

Like in the above discussion, you don't keep proper track of what's in your system and what's not. PiusImpavidus (talk) 12:25, 1 August 2024 (UTC)

Sorry for not clarifying myself. Anyway, I meant the following:

When a gravity pendulum is influenced by gravitation, the pendulum's total energy is conserved, even though the (changing) net work done by the gravitation on the pendulum is not equal to the (zero) net heat received by the pendulum's movement. HOTmag (talk) 14:23, 1 August 2024 (UTC)

There are two ways to look at this:

1. When the bob of the pendulum climbs, its kinetic energy is converted into potential energy, but still present in the bob. The total energy of the bob doesn't change and no work is done. When the bob descends, the potential energy is converted back into kinetic energy, but again no work is done. This is mathematically the easy way, but not entirely correct.
2. When the bob of the pendulum climbs, it performs work on the gravitational field. The bob loses energy, the field gains energy. When the bob descends, the field performs work on the bob. The field looses energy, the bob gains it. The energy of the bob isn't conserved, but an isolated bob without a gravitational field is no harmonic oscillator. The system of bob+string+gravitational field is a harmonic oscillator and energy in that is conserved. Unfortunately, this approach is mathematically very hard.

In physics, we store energy in fields all the time. PiusImpavidus (talk) 08:22, 2 August 2024 (UTC)

no work is done. What? Don't you agree that the work is equivalent to the change in kinetic energy? HOTmag (talk) 11:18, 2 August 2024 (UTC)

In physics, Work is the product of a force and a lasting, permanent displacement. Philvoids (talk) 18:43, 2 August 2024 (UTC) Underlining added for clarity.

If the force is denoted by ${\vec {F}}$ and the displacenent is denoted by ${\vec {s}},$ then the work is defined as $W=\int {\vec {F}}\cdot d{\vec {s}},$ and it's equivalent to the change in the kinetic energy, isn't it? HOTmag (talk) 21:02, 3 August 2024 (UTC)

Physicists carefully define Energy a measurable property that is transferred in interactions and Mass the intrinsic quantity of matter in a body because these are fundamental to both our study and understanding of many Systems i.e. functional groupings of elements that act according to a set of rules. Posing yet another challenge to define a physical system "without mentioning those words" is just inviting responders into a handicapped word play that may provide the engaging debate that the OP is seeking to expose their own ideas but that is a misuse of the reference desk that should not continue. Underlining added for clarity. Philvoids (talk) 10:03, 2 August 2024 (UTC)

I didn't ask about defining something (without using specific words). I only asked about a necessary sufficient condition for the conservation of (total) energy (without mentioning energy/mass), which is a definitely different thing!

Just as the absence of external net force is a necessary sufficient condition for the conservation of momentum, without mentioning momentum. This is a definitely legitimate question. HOTmag (talk) 11:27, 2 August 2024 (UTC)

August 1

Imane Khelif

I'm posting this question in the most neutral and, I hope, respectful way possible. I do not intend to offend anyone or start a fight. What I'd like to know is the actual biological situation of Imane Khelif. Our article is not very clear on the subject. I understand that it is a delicate topic so some of the details may not be public. Does she have X-Y, X-X or some pathological variation of chromosomes? Does she have male or female genitalia? Does she have a specific medical condition (I am aware of the concept of intersex but my understanding is that it could be the outcome of different medical conditions)? Thank you! 79.35.53.87 (talk) 14:30, 1 August 2024 (UTC)

It appears that no details have been shared officially with the public. The IBA has declined to divulge any specifics, but its chairman declared on March 23 to TASS that, based on the results of DNA tests, "it was proved they had XY chromosomes". That is all we know.^[5]  --Lambiam 15:50, 1 August 2024 (UTC)

I believe there are over 50 different 'situations' that fall under the umbrella title of Intersex. We have no access to evidence suggesting which, if any, of these Imane Khelif may fall under, but the OP (and others) might find the article of overall interest.

When categorization for sporting purposes is (some might say as a matter of practicality) binary, but the human species is not, quandries are bound to arise, alas. {The poster formerly known as 87.81.230.195} 94.1.169.77 (talk) 01:16, 2 August 2024 (UTC)

One may also consider the gender of Schrödinger´s Cat. --2001:871:6A:1B71:292B:D775:91F7:34E5 (talk) 16:49, 3 August 2024 (UTC)

Schrödinger´s Cat's gender is not regarded as an unresolved quantum Wave function so how shall that help? Philvoids (talk) 22:12, 3 August 2024 (UTC)

August 3

Heat capacity, sand battery, check calculation

Could someone please check my work? Sand has a specific heat capacity of around 800 kJ/kg·K. I calculate that if you put 10 000 kWh (10 MWh) into 900 kg (0.6 cubic metres) of it, you raise its temperature by 50 °C. This video is talking about a massive ~40 cubic metre sand battery but it's capacity is only ~10 MWh? Either I'm wrong or the video is wrong, right? https://www.youtube.com/watch?v=KVqHYNE2QwE&t=377s 80.46.251.32 (talk) 01:05, 3 August 2024 (UTC)

According to the article Thermal energy storage, there is a prototype 8 MWh sand battery, built in Finland in 2022. The source given on this system says it is a "steel container, which is 4 m wide and 7 m high, [and it] is filled with 100 tonnes of builder’s sand", and is heated to 500 °C. There is a picture of the silo. So it seems that your calculation must be wrong. Abductive (reasoning) 06:09, 3 August 2024 (UTC)

The specific heat capacity of quartz sand is more like 800 J/kg·K, lower by a factor of 1000. The specific heat capacity is temperature-dependent. At 500 °C it is substantially higher, about 1230 J/kg·K.^[6]  --Lambiam 10:47, 3 August 2024 (UTC)