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May 25Edit

Should I list every species in a genus?Edit

Many articles about genera list only a few species in it, even if there are dozens of species that are classified in that genus. Should I go ahead and list every single one, or should I only do a select few? Svenard (talk) 21:14, 25 May 2023 (UTC)Reply[reply]

If you have a reliable source or sources, go ahead and list them all. But if the number is ridiculous, eg >1000, it would be better to break it up into smaller subdivisions. Graeme Bartlett (talk) 03:30, 26 May 2023 (UTC)Reply[reply]
No, the solution is a spinoff article, List of Foogenus species, as can be seen in Category:Lists of plant species, Category:Lists of animal species, and Category:Lists of fungal species. Please consult with the appropriate WikiProject for guidance. Please do not break up genus articles in any other way without consensus. Abductive (reasoning) 05:06, 26 May 2023 (UTC)Reply[reply]
Ghost Light (Doctor Who) is rather disjointed, but thankfully Dr Who stopped the creature in it killing everything. It was cataloguing all the various species but the list was never complete because evolution kept making new ones. NadVolum (talk) 22:21, 26 May 2023 (UTC)Reply[reply]

Do particles in a linear particle accelerator lose speed/momentum for some reason?Edit

Just wondering if particles in a liner accelerator are observed to lose any speed/momentum after travelling through the last E field? Byron Forbes (talk) 23:45, 25 May 2023 (UTC)Reply[reply]

I think the short answer is no, but they could spread out due to repulsion. Otherwise if there is no electric or magnetic field, their inertia carries them forward at a constant velocity. But.. They could interact with a target or the end of the accelerator and lose energy. Hopefully there is enough shielding to stop anything dangerous. And once particles like neutrinos escape, they will lose energy due to leaving Earth's gravitational field, and then through the expansion of the universe. Graeme Bartlett (talk) 00:28, 27 May 2023 (UTC)Reply[reply]
Not really relevant but since you mentioned them - how can a fixed speed neutrino without a frequency, lose energy? Byron Forbes (talk) 17:09, 27 May 2023 (UTC)Reply[reply]
If the neutrino has mass, then it will lose energy by slowing down, and the speed is not fixed. But if it is massless, then it will still be able to lose energy by increasing its wavelength, or decreasing its frequency, as it is red shifted, but it would keep moving at the speed of light. Graeme Bartlett (talk) 04:30, 28 May 2023 (UTC)Reply[reply]

May 26Edit

The problem with gravity.Edit

Not sure if this is taken care of by GR, but I'll mention it anyway.

In the Newtonian universe, we have a G force acting perpendicularly to a planet orbiting a sun. But if this is so, then there is no force to decelerate the planet so that it can change direction whilst maintaining speed. So there is no way this force can be perfectly perpendicular - it would need to point in a direction a little backward from the motion of the planet. Byron Forbes (talk) 01:14, 26 May 2023 (UTC)Reply[reply]

None of this requires GR. You're forgetting velocity is nonzero and perpendicular to the radius (or else conflating velicity and acceleration). One way to think about it is that the acceleration changes the velocity, so the planet's future position is angled compared to previous. It doesn't just affect the position directly (pulling it inward). Or another way to think about it is that there is forward motion and sideward pull, so the net effect is motion "forward and a little sideward". Because "sideward" is towards the sun, the long-term effect that it keeps curving around. Or yet another way, one way it's explained to kids, is that the planet is falling towards the center but moving forward, so it keeps "missing" and instead goes around. DMacks (talk) 01:25, 26 May 2023 (UTC)Reply[reply]
Lets look at this at a point in time where the velocity of the planet is due north at the 3 o'clock position. Now look at time t' just a moment later. In order for both direction to change and speed be maintained, the velocity component in the Nth direction must decrease. But if the G force was/is always perpendicular, then how can this happen? Byron Forbes (talk) 02:58, 26 May 2023 (UTC)Reply[reply]
The velocity component in the West direction increases because the force of gravity is pulling due-west (from 3 o'clock towards the center). So the direction changes (the angle of the velocity but not its magnitude, so speed is constant). DMacks (talk) 03:10, 26 May 2023 (UTC)Reply[reply]
The speed remains constant because momentum is conserved. The westward acceleration due to the sun's gravity increases the velocity in the westward direction, so the velocity in the northward direction must decrease to keep the total speed unchanged. As a simple analogy, imagine a stick lying on the ground pointing northward. Its projection in the north-south direction is equal to its length, and its projection in the east-west direction is zero. If you move the stick so that it's pointing slightly northwest, the length of the stick remains unchanged. But its projection in the north-south direction is smaller that it was originally, and its projection in the east-west direction is larger. CodeTalker (talk) 04:36, 26 May 2023 (UTC)Reply[reply]
Try putting a weight at the end of a string and spinning it round. You can stop moving your hand and the weight will continue to go round in a circle until frictin stops it. The string can't exert a force except inwards towards your hand. NadVolum (talk) 08:40, 26 May 2023 (UTC)Reply[reply]
Have a look at orbit. Kepler (1571–1630) first realised that the orbits are not circular, they are ellipses. His laws included the "equal areas" observation (his second law) though his assumption that the sun was at exactly one focus of the ellipse was subsequently modified by Newton to use the combined centre of mass of the whole system. To return to the OP's question, gravity will always act towards the focus, and so for half the orbit it is ahead of the perpendicular to the planet's path, thereby accelerating it. For the other half it is behind the perpendicular as so retards the orbital speed. Martin of Sheffield (talk) 09:16, 26 May 2023 (UTC)Reply[reply]
True. But I think the OP's question is also germane to near circular orbits such as those of our geostationary satellites which orbit with near constant speed. In this case, the change in orbital speed is negligible and one can note that the OP's Nth component deceleration is balanced by a perpendicular Wst component of acceleration (as pointed out above by DMacks), assuming a uniform inward force exists... Modocc (talk) 15:41, 26 May 2023 (UTC)Reply[reply]
It was the OP's use of the phrase "there is no way this force can be perfectly perpendicular" that rang an alarm bell. It's correct, but not for the reason argued. Is it possible that the OP is thinking of simple vector diagrams which show neat little triangles and hasn't let the calculus go to the infinitesimal calculus? I'm neither a mathematician nor a teacher so having sown the thought I'll beat a hasty retreat! Martin of Sheffield (talk) 15:54, 26 May 2023 (UTC)Reply[reply]
When I was at NCSU I improved a circle drawing algorithm and noted that every linear Cartesian coordinate step in one direction corresponded precisely one-to-one with steps in the perpendicular direction. In theory, one can model an infinite number of such steps, but pragmatically it's best to take just a few steps (for a circle) or even just one or two on account of the fact that smaller component steps sum to larger ones even if these are more abstract than real because the circles' arc primarily passes near finite rectangular coordinates. With a spherical coordinate system centered on the central body, there is less ambiguity regarding a particle's trajectory for which we do actually observe perpendicular inward forces that constrain masses to circular motion, such as the weight and string mentioned by NadVolum. Net forces can be perfectly perpendicular, including gravitational forces, but orbits often deviate from that. Modocc (talk) 17:01, 26 May 2023 (UTC)Reply[reply]
Now that makes a lot of sense! Cheers! Byron Forbes (talk) 16:49, 27 May 2023 (UTC)Reply[reply]
No reason a string cannot have shear forces, and I would suggest it must have. Byron Forbes (talk) 16:48, 27 May 2023 (UTC)Reply[reply]
The string is only under tension and the linear force components are sinusoidal. At 3 o clock the magnitude of the Nth/Sth component of the tension crosses zero. Modocc (talk) 19:11, 27 May 2023 (UTC)Reply[reply]
You could try putting the weight with a string attached onto a table and then flick the end of the string sideways with your finger. It will be interesting to see how much shear force you can impart to the weight. You could try with a thread as well. NadVolum (talk) 20:07, 27 May 2023 (UTC)Reply[reply]
Mathematically orthogonality of the acceleration to the velocity if the absolute value of the latter is kept constant can be easily proved. Suppose that   is the velocity, then the acceleration is  . Now, if  , we have  , which proves the orthogonality. Ruslik_Zero 20:46, 26 May 2023 (UTC)Reply[reply]
dv/dt = 0? The speed isn't changing, but the Velocity (direction) is always changing. That is a vector quantity you have there. Byron Forbes (talk) 16:56, 27 May 2023 (UTC)Reply[reply]
The scalar product of the velocity and its derivative is zero, which means orthogonality. Ruslik_Zero 18:15, 27 May 2023 (UTC)Reply[reply]
I do not understand the argument why the force cannot be perpendicular. Applying Newton's laws of gravitation and of motion to the two-body problem gives us a differential equation whose solutions have trajectories that are conic sections. Given the right initial conditions the bodies are in a circular motion around the centre of gravity, in which case the gravitational force is always perpendicular to the direction of motion. If the trajectory is a non-circular ellipse, the force is perpendicular only at the apsides and otherwise alternatingly either slightly forwards or backwards – forwards on the planet's way to the perihelion, and backwards on its way to the aphelion.  --Lambiam 21:48, 26 May 2023 (UTC)Reply[reply]

Can Vitamin K in say Spinach or Broccoli be destroyedEdit

Does drying / cooking / freezing vegetables containing Vitamin K (such as Spinach or Broccoli) destroys the vitamin?

Thanks. 2A10:8012:17:CDC6:49DF:B235:F14:5969 (talk) 17:52, 26 May 2023 (UTC)Reply[reply]

This page at the University of Rochester notes that it is not particularly temperature sensitive, but that light exposure can damage vitamin K, and recommends avoiding long exposures to strong light. --Jayron32 18:20, 26 May 2023 (UTC)Reply[reply]

Journal auto-skimEdit

I'm looking to set up an auto search and email notify for recent scholarly articles on distributed cognitive architectures. How would I go about it? Google Scholar and Google alerts? Temerarius (talk) 23:18, 26 May 2023 (UTC)Reply[reply]

Google scholar will be much more selective for scholarly articles, so recommended. Try here after you are logged on to Google:scholar_alerts Graeme Bartlett (talk) 00:24, 27 May 2023 (UTC)Reply[reply]

Should a glass of red wine daily cause the human body to produce more endosteroids (natural steroid hormones from the adrenal gland)?Edit

Thanks. 2A10:8012:17:CDC6:E023:B45A:943A:B761 (talk) 02:03, 27 May 2023 (UTC)Reply[reply]

This would refer to the corticosteroids (cortisol, corticosterone, aldosterone and testosterone), steroids produced in the adrenal cortex. I did not find a relation with the consumption of red wine in the literature. There is an advice to avoid drinking alcohol while taking prednisone, a synthetic corticosteroid, because it can worsen some side effects.[1] Resveratrol can inhibit steroid metabolism.[2] Neither has a bearing on the natural production of corticosteroids.  --Lambiam 07:59, 27 May 2023 (UTC)Reply[reply]
For information: the Healthline site (ref 1) is not available in the UK or EU unless you enable trackers and cookies. If it falls foul of GDPR, readers outside of these areas may wish to exercise due prudence in trusting it. Martin of Sheffield (talk) 08:39, 27 May 2023 (UTC)Reply[reply]
Hello User:Lambiam. After reading your answer I wonder what are the "human-natural" endosteroids which science so far discovered.
In the article Steroid hormone, I didn't find a category such as Category:Human Produced Steroid Hormones. If there was such a category, than it should have included two sub categories, Category:Human Produced Corticosteroids and Category:Human Produced Sex Steroids, right? 2A10:8012:17:CDC6:78FE:7B18:6EF6:D13A (talk) 15:08, 27 May 2023 (UTC)Reply[reply]
We have categories Category:Corticosteroids and Category:Sex hormones. No one has chosen to organize our articles in categories Category:Natural steroid hormones and Category:Artificial steroid hormones. If we had such articles, the more specific ones would be category intersections.  --Lambiam 09:01, 28 May 2023 (UTC)Reply[reply]

May 27Edit

Are all types of whales endangered or close to being endangered?Edit

Are all types of whales endangered or close to being endangered? 2001:569:5026:8A00:C97:246D:109A:295E (talk) 05:45, 27 May 2023 (UTC)Reply[reply]

The Common minke whale, a close relative of the Blue whale, is assigned the "Least Concern (LC)" status on the IUCN Red List.[3] Our List of cetaceans, which also includes dolphins and porpoises generally not referred to as whales, reports a "threatened" status (CE, EN, VU) for 23 species, versus 10 NT and 50 LC.  --Lambiam 07:12, 27 May 2023 (UTC)Reply[reply]

May 28Edit

Scientists and Flat-EarthersEdit

Can someone be both a Flat-Earther and a scientist simultaneously? I'm not referring to someone who genuinely believes in a flat Earth and dismisses all evidence contrary to it, but rather someone who approaches it as a thought experiment. This would be akin to a physicist or mathematician speculating about a world existing in two dimensions. Astrobiology or Exolinguistics could easily slip into this category too. Aren't there more scientific endeavors about imaginary worlds like this? Bumptump (talk) 12:09, 28 May 2023 (UTC)Reply[reply]

No. Anyone, including a scientist, can enter into a thought experiment in which they assume the Flat Earth theory is true, and then see whether it stacks up in the face of scientific evidence. Guess what? It doesn't. So they can't remain a Flat Earther for any longer than the length of time it takes to test the theory. Alternatively, they could remain a Flat Earther but only by abandoning their adherence to scientific principles. It's one or the other, not both. -- Jack of Oz [pleasantries] 12:18, 28 May 2023 (UTC)Reply[reply]
I meant flat earther in the sense of a participant in an elaborate mental game. Like The Planiverse is not intended to be a description of a world or the testing of a theory. It's just a what-if scenario. Bumptump (talk) 12:42, 28 May 2023 (UTC)Reply[reply]
A cartographer? fiveby(zero) 17:19, 28 May 2023 (UTC)Reply[reply]
Scientists can enjoy science fiction just like anyone else. Shantavira|feed me 19:03, 28 May 2023 (UTC)Reply[reply]
I had a college roommate and fellow lab technician (and future scientist) who joined the Flat Earth Society for the fun of it. He also enjoyed the Journal of Irreproducible Results, and if memory serves me, he contributed to the latter. (talk) 20:06, 28 May 2023 (UTC)Reply[reply]

Fictional portrayal of diving into a methane oceanEdit

In the current season of the science-fiction MMORPG shooter Destiny 2, users are tasked with travelling to Titan and diving into the moon's liquid methane oceans to defeat enemies, sometimes as much as several kilometres deep. The setting is also depicted with alien marine life that in my opinion, is too similar to our own (fish, coral, sea anemones, bioluminescent life, etc). I would like to know if there is any merit to this portrayal, whether of Titan, the possibility of hosting marine life, or the very act of diving/swimming in a body of liquid methane.

I assume a person would instantly die of hypothermia regardless of how advanced or purportedly futuristic the protective gear they're wearing is, considering that to my knowledge, methane must be around -200 Celsius in order to be kept at the liquid state, nevermind the average surface temperature of Titan being just as extremely cold. The oceans would also have less density than the atmosphere of Titan, or liquid water for that matter, so there would be little-to-no buoyancy. (talk) 22:45, 28 May 2023 (UTC)Reply[reply]

The density of liquid methane is 423 g/L, compared to 1000 g/L for water, so there would be some buoyancy but less than that of water. For a human to float you'd need a buoyancy compensator or something similar. There's no theoretical reason that heated protective gear couldn't keep a human at a survivable temperature while submerged in liquid methane, although there would probably be a lot of vigorous boiling in the immediate area as some of the heat is dispersed into the environment. CodeTalker (talk) 23:49, 28 May 2023 (UTC)Reply[reply]
I would assume that the rapid evaporation induced by a protective suit's heat would lead to/cause the methane to explode? There's also the fact that the game's users are using firearms, directed-energy weaponry, and rocket launchers and such at the "seafloor"; the ignitions of which would also release tremendous amounts of heat into the surrounding area. (talk) 01:17, 29 May 2023 (UTC)Reply[reply]
No, to 'explode' it in the sense I think you mean would require there to be an oxidant (such as oxygen) present in substantial quantites to react violently with the methane. Simply introducing heat would merely heat up and perhaps boil a little of it, but an ocean of methane, which would be subject to conduction and convection, could absorb and disperse a lot of heat. {The poster formerly known as} (talk) 03:22, 29 May 2023 (UTC)Reply[reply]
Some issues with life in methane oceans: it is much colder and metabolism would be at a much slower rate to match; liquid methane dissolves much less variety of chemicals than water, so there may be a lack of minerals around to build solid coral. The ocean would have higher density than the atmosphere, and so not float up into the atmosphere. The densest gas is less dense than the least dense liquid, so liquid is always going to fall in any kind of gaseous atmosphere. Graeme Bartlett (talk) 07:17, 29 May 2023 (UTC)Reply[reply]

May 29Edit

On the beachEdit

I'm hoping someone with some knowledge of biology has a better idea than I do as to just what I've photographed here. - Jmabel | Talk 04:24, 29 May 2023 (UTC)Reply[reply]

Someone with better knowledge of the area's geography might not have to read the description of the images, which I repeat here, adding links: "Sandy floor of Dumas Bay at low tide, near but not within Dumas Bay Sanctuary, Federal Way, Washington, U.S." So the bay is an arm of Puget Sound. -- (talk) 04:55, 29 May 2023 (UTC)Reply[reply]
Possibly they are closed up sea anemones, though they are more common on rocks. When submerged tentacles would come out. Graeme Bartlett (talk) 07:09, 29 May 2023 (UTC)Reply[reply]
I hadn't thought of anemones. Plausible. Any other ideas? - Jmabel | Talk 02:40, 30 May 2023 (UTC)Reply[reply]
Given the location (Puget Sound), my thought would be geoducks. Did you dig down a bit to see what was in there? Lots of burrowing animals will also leave similar hole patterns. --Jayron32 14:29, 30 May 2023 (UTC)Reply[reply]
You're looking at a group of Aggregating Anemones (Anthopleura elegantissima) which are closed as they are exposed to air. Underwater they'll open and expose their tentactles a Graeme Bartlett mentioned (talk) 17:35, 31 May 2023 (UTC)Reply[reply]
Compare with this photo: Young sea anemones - Anthopleura elegantissima - Wikimedia Commons (talk) 17:37, 31 May 2023 (UTC)Reply[reply]

May 30Edit

Blackbody energy densityEdit

In blackbody radiation, how do you go from measuring radiation power per unit area to obtain energy density per unit volume? I am looking for this formula and one of the first publications that uses it, by Wilhelm Wien for example ? Because in the laws of Kirchhoff and Stephan-Boltzmann it is always a question of radiation power.

~~~~ Malypaet (talk) 07:36, 30 May 2023 (UTC)Reply[reply]

You can imagine the radiation crossing a surface at the speed of light into a new volume. In one second it will have traveled c meters. So to get what is in one cubic meter, divide that energy by c to get energy per unit volume.   However only half the energy is heading in the correct direction (half is going backwards) and also half is coming at different angles other than perpendicular, so you have to multiply by cos of the angle of incidence and integrate over possible angles. so then you have to multiply the answer by 4. (Another way is to consider the area of a sphere versus the area of the circle 4 πr2 versus πr2. Graeme Bartlett (talk) 10:04, 30 May 2023 (UTC)Reply[reply]
So when you write the word energy, as you multiple power by one second, you use the system unit W-s, right ?
~~~~ Malypaet (talk) 11:52, 30 May 2023 (UTC)Reply[reply]
Sure, but a joule is identical to a watt-second. --Jayron32 14:26, 30 May 2023 (UTC)Reply[reply]
Where can I find a reference for this formula? Malypaet (talk) 21:07, 30 May 2023 (UTC)Reply[reply]
see Stefan–Boltzmann law#Energy density for our article. Graeme Bartlett (talk) 01:28, 31 May 2023 (UTC)Reply[reply]
The problem with Wikipedia articles is that they present the Stephan-Boltzmann law with references from the future (Planck, Bose-Einstein...). But at least you gave me a reference on the source. I just have to translate the German! Malypaet (talk) 07:24, 31 May 2023 (UTC)Reply[reply]
I have not tried to discover the first publications as you ask. It appears that people first worked out what was happening in a spherical cavity at a particular wavelength, whereas you are interested in the result over all wavelengths. Though your question is more general than blackbody radiation, and applies to any stuff moving at the speed of light (light, neutrinos, gravitational waves). So there may be an answer for you from before Wien. Graeme Bartlett (talk) 12:22, 31 May 2023 (UTC)Reply[reply]

May 31Edit

Copernicus Open Access HubEdit

I was exploring the Copernicus Open Access Hub and was wondering how do you get the images and data from the satellites to usable images like the one's at image of the day. I was look at the IMG folder in the Granule but the files are jp2 and when converted to jpg are very dark.

PS: Here's a tutorial on accessing the images. PalauanLibertarian🗣️ 13:56, 31 May 2023 (UTC)Reply[reply]

The article JPEG 2000 shows comparison of .jpg and .jp2 files for the same source image and there are no obvious brightness differences. I downloaded a Copernicus image file that arrived as a .jpg with 24 bit/pixel "True color" depth and little or no compression, resulting in a large and relatively slow loading file. The free image editor IrfanView is useful for converting between .jp2 and .jpg. It can display a histogram of pixel intensities and it offers a range of brightness, contrast, tint, and gamma modifications. Philvoids (talk) 21:38, 31 May 2023 (UTC)Reply[reply]
@Philvoids How did you download a .jpg. I thought all the img's were jp2's. I tried downloading some of them and converting them to jpgs but they are just black and white. Example1 Example2. PalauanLibertarian🗣️ 01:19, 1 June 2023 (UTC)Reply[reply]

June 1Edit