Wikipedia:Reference desk/Archives/Science/2017 September 2

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September 2[edit]

Spider identification[edit]

Is there a way to identify this small spider I took some shots of? I live in southern California, and a small tan and black hairy spider has been living on my fig tree for at least a month, and I've seen it with a fly in its fangs on a couple of occasions when I caught it while it was feeding. The best shot of it is this one which has a ladybug on the same leaf for perspective of its size, and you can see a tan cross/plus sign on its abdomen. The front of it near the fangs is also a bluish green color. And you can see from this close up how hairy it is.-- 00:50, 2 September 2017 (UTC)[reply]

This is a Jumping spider, family Salticidae. Looks familiar but can't remember the species off the top of my head, will look it up. Dr Dima (talk) 04:02, 2 September 2017 (UTC)[reply]
I thought Phidippus californicus but it doesn't look quite right. It's almost certainly Phidippus, but it's up to the BugGuide to figure out what the species is. Sorry. --Dr Dima (talk) 04:21, 2 September 2017 (UTC)[reply]
Phidippus putnami looks pretty close, so this may be it. Still, you better ask the BugGuide people to be sure. Dr Dima (talk) 04:26, 2 September 2017 (UTC)[reply]

Chicken feet[edit]

Chicken feet does not contain muscles. So how does 100 grams of chicken feet contains 17 g protein— Preceding unsigned comment added by RabbitDog (talkcontribs) 03:42, 2 September 2017 (UTC)[reply]

"Type I collagen, is present in many forms of connective tissue, and makes up about 25% of the total protein content of the mammalian body". - Fibrous connective tissue. —PaleoNeonate – 03:56, 2 September 2017 (UTC)[reply]
Also, don't forget that some protein exits in bones. Aspro (talk) 11:28, 2 September 2017 (UTC)[reply]

aircraft airframe material[edit]

What is "LA"[1] in the context of aircraft airframe material? Mũeller (talk) 04:59, 2 September 2017 (UTC)[reply]

Adding small amounts of Lanthanum (I have seen 0.9% specified) improves the machinability of Titantium Ti-64 alloy. Adding 1% Lanthanum to A390 Aluminum alloy improves the ductility from 0.7 to 1.8% and increases the tensile strength from 100 to 150 MPa. I have seen these alloys referred to as "A390 La" and "Ti-64 La". But in this case, I think our article on AgustaWestland AW101 has the answer: "The fuselage structure is modular and comprises an aluminium-lithium alloy, designed to be both light and damage-resistant". --Guy Macon (talk) 07:45, 2 September 2017 (UTC)[reply]
If it would describe Aluminium-lithium alloy it would read Al-Li or alike. Certainly not "LA". The "LA" honeycomb part is very special tho and searching for that i found the company http://www.lacomposite.com/, which also is an official subcontractor (parts supplier, see SoA)for the company AgustaWestland which manufactures and sells these helicopters. Thus i bet the "LA" only means these parts are from lacomposite. --Kharon (talk) 20:10, 2 September 2017 (UTC)[reply]
Interesting. I have never heard of anyone making stringers out of honeycomb, but it could work. --Guy Macon (talk) 11:19, 3 September 2017 (UTC)[reply]

Martial arts question[edit]

How does breaking power correlate with knockout power? For example, if you can break 4 boards with one strike, how big of an attacker can you knock out with the same strike? How about if you can break 6? 2601:646:8E01:7E0B:4DFA:5B5D:E0B4:4FDF (talk) 10:10, 2 September 2017 (UTC)[reply]

Boards are not standardized so as we know how much force/power is needed to break them. But it also does not matter, since even extremely weak people have enough force to break some bones, which kind of knocks any opponent out. --123abcnewnoob (talk) 10:57, 2 September 2017 (UTC)[reply]
See Martial arts and Breaking (martial arts). An ability to break things does not correlate well with one-on-one Combat sport where there can be many more effective ways of disabling an opponent than trying to break his bones. See Self-defense#Unarmed. Traditional Japanese martial art schools place little, if any, emphasis on board-breaking. Blooteuth (talk) 12:41, 2 September 2017 (UTC)[reply]
Also people "playing on" with broken bones is the stuff of sporting legend. All the best: Rich Farmbrough, 13:38, 2 September 2017 (UTC).[reply]
Be aware that most board breaking is with the grain, boards will break fairly easily in this orientation, and are rarely used that way structurally.
A knockout punch - that renders one's opponent unconscious, may be more a combination of placement and timing than extreme force. All the best: Rich Farmbrough, 13:38, 2 September 2017 (UTC).[reply]
OK, in this case, let me break this question into parts: (1) Is there a correlation between a person's size and the amount of force needed to knock him/her out? (2) About how much force is needed to knock someone out? (3) How much force is needed to perform a power break on an unpegged stack of 4 standard 1-inch-thick boards, and how much more force is needed if the number of boards is increased to 6? 2601:646:8E01:7E0B:4E8:FDC:7D20:30AB (talk) 08:56, 3 September 2017 (UTC)[reply]
Oak or pine? balsa? Maple? Ash? Plywood? Particle board? Sagittarian Milky Way (talk) 09:30, 3 September 2017 (UTC)[reply]
Assume pine, because it's the most commonly used wood for this application. 2601:646:8E01:7E0B:4E8:FDC:7D20:30AB (talk) 12:19, 4 September 2017 (UTC)[reply]
1) No. Unfortunately, our relevant article at Chin (combat sports) is pretty useless. Size may play a small role, if only because a larger person has a larger head, but such differences are overwhelmed by technique, timing, etc. In combat sports it's generally accepted [citation needed] that it's extremely difficult to knock someone out who is expecting it and surprisingly easy to knock out someone who is not expecting it. Matt Deres (talk) 02:27, 4 September 2017 (UTC)[reply]

Problem solving abilities of scientifically trained/mathematically literate[edit]

If you have, for example, extensive training in some science or math, but over the years do not integrate it in your day to day job, would that still make your general problem solving abilities better than average joe's (without science/math training)? Is there some scientific research about how science/math educated people perform outside their direct field of expertise? --123abcnewnoob (talk) 10:55, 2 September 2017 (UTC)[reply]

Problem solving requires two distinct types of mental skill, analytical and creative. This article contrasts analytical thinking that predominates in solving closed problems with creative thinking characterized by fluency, flexibility and elaboration in using the imagination in solving open problems. Science/math education purportedly develops left-brain thinking, is more logical and analytical, and is predominantly verbal, possibly with neglect of right-brain thinking that is more holistic and is concerned with feelings and impressionistic relationships. Problem solving skills may be trained as a distinct discipline, see Problem solving. Blooteuth (talk)
Knowing the scientific method would help to solve problems. For example, say you notice you get sick after eating X. Some would just stop there, and assume that X is the cause. But a better way would be to keep a food journal, so you can look back at what you ate and correlate that to when you got sick. You might then find out that you often eat X and Y together, and get sick after. Then, eating them separately would help to determine which, if either, is the true cause. StuRat (talk) 13:09, 2 September 2017 (UTC)[reply]
We certainly have Life itself as "benchmark". Have you ever wondered why for example not even one of all our very famous Nobel Memorial Prize in Economic Sciences Laureates managed to make a fortune at the stockmarket or anywere else in financial biz for himself? Economic Science is mostly based on math btw. To the contrary surprisingly many of the richest people actually did not even finish their education, not to speak of becoming an respected scientist afterwards. So no, it does not look like a huge help. --Kharon (talk) 20:29, 2 September 2017 (UTC)[reply]
@Kharon: Economists do not get assessed by speculating in the stock market. Some are not even allowed to do that (all those who work for central banks, for example). And it's completely false that famous economic laureates have never become rich in the financial industry. Your answer has so many issues that it's funny. Hofhof (talk) 06:13, 3 September 2017 (UTC)[reply]

Chemistry questions.[edit]

1. Water has a higher density as a liquid than it is as a solid. What other substances are like that?

Some answers that I've heard are paraffin, bismuth, and hydrogen peroxide, but can't verify. On Wikipedia, paraffin is a solid that is what candles are made of, but I couldn't find the density as a liquid. Same with density of bismuth as a liquid. And can't find the density of hydrogen peroxide as a solid either.

2. When magnesium sulfate is put in water, water turn hot. When ammonium nitrate is put in water, water turns cold. What's the mechanism of how you can tell whether reaction will turn hot or cold? For ionic compounds at least. Is the answer also the same for covalent compounds?

3. What's a better drainer to use acid-based or base-based?

Acids dissolve greases, proteins, and carbohydrates.
Bases dissolves greases, proteins, and fats and oils.

Most drainers like Drano are base-based, I guess, because bases dissolve a wilder range of substances. Thanks. 12.130.157.65 (talk) 12:21, 2 September 2017 (UTC).[reply]

Didn't you answer your own question just now for no. 1? Regarding no. 2, whether ther overall reaction is endo- or exo-thermic depends on the bond-energies involved. Electron-deficient cations tend to readily undergo hydration during dissolution. Hydration typically releases energy. In your example, the magnesium cation is very much electron deficient, being able to accept four water molecules to form the tetraaquo complex, whereas the ammonium cation is electronically satieted, not being able to accept any water molecules. Ammonium can however, form hydrogen bonds with water, but that does not release energy in the same way as hydration does. Regarding no. 3, in my opinion, more often than not, a block will contain more fat/oil than other carbohydrates, so I would recommend a base-based drain cleaning agent for general use. Plasmic Physics (talk) 13:10, 2 September 2017 (UTC)[reply]
Well, all cations are electron-deficient. Okay, so what I'm absorbing here is that it has to do with whether the cation is a single atom cation or a polyatomic cation? Thanks. 12.130.157.65 (talk) 13:53, 2 September 2017 (UTC).[reply]
I'm using the term 'electron deficient' in the valence-theory sense, not in the general sense. In ammonium, nitrogen has a full octet, and so does bond to other electron rich species, it is just attracted to the anion. So, it has nothing to do with being monoatomic or polyatomic. Point in case: the polyatomic trichlorocobalt(3+) ion is electron deficient, and bonds to ammonia like the monoatomic magnesium(2+) ion bonds to water. Plasmic Physics (talk) 21:34, 2 September 2017 (UTC)[reply]
Regarding drains containing more fat/oil than carbs, but that is not necessarily a complete list, of what acids and bases dissolve. There could be more stuff that acids dissolve that bases do not, that I did not include, that could give acids an overall edge than bases. Shrug. 12.130.157.65 (talk) 14:14, 2 September 2017 (UTC).[reply]
Basic drain cleaners work better than acidic ones because drains get a lot more acidic solutions run down them under ordinary working conditions. Therefore the buildup will be resistant to acids, but not so much to bases. Abductive (reasoning) 18:30, 2 September 2017 (UTC)[reply]
Do acids damage metal pipes more? Sagittarian Milky Way (talk) 15:56, 3 September 2017 (UTC)[reply]
2) See endothermic and exothermic reactions.
3) Aren't greases essentially the same as fats and oils ? StuRat (talk) 13:13, 2 September 2017 (UTC)[reply]
"Wax expands considerably when it melts and this allows its use in wax element thermostats for industrial, domestic and, particularly, automobile purposes." Paraffin wax All the best: Rich Farmbrough, 13:47, 2 September 2017 (UTC).[reply]
For (1), this happens when some part of the structure of the solid is broken to allow denser packing as a liquid. Among the elements, this happens for Ga, Ge, Sb, and Bi, for instance. (Source: Greenwood and Earnshaw 2nd edition, p. 222.) Double sharp (talk) 15:43, 2 September 2017 (UTC)[reply]
For (1), Plutonium's weird like that, too. It contracts (becomes more dense) as it melts (transitions from solid to liquid state). but after it melts, liquid Pu expands with heating. loupgarous (talk) 12:57, 4 September 2017 (UTC)[reply]
Thank you! I knew I was forgetting something, but I forgot which one it was. Double sharp (talk) 14:19, 4 September 2017 (UTC)[reply]

Heliosphere Moved from Math Desk.[edit]

(Moved from Math Desk.)

I don't quite understand why the solar wind interacts with the interstellar medium at all. Aren't the particles such low density that they would just pass by each other without any effect on the other ? StuRat (talk) 02:00, 2 September 2017 (UTC)[reply]

The net effect of the charged particles supposedly constitutes a significant enough magnetic field to cause large-scale interactions even though direct collisions between particles may be "relatively" rare. See Interplanetary magnetic field. — Preceding unsigned comment added by 73.232.241.1 (talk) 02:35, 2 September 2017 (UTC)[reply]
Perhaps what's missing are the elements of time and distance. That is, that while they don't interact on a human scale, over the course of a billion years and one light year, they do ? StuRat (talk) 13:18, 2 September 2017 (UTC)[reply]
I guess it's a matter of scale. The density is low but it's not zero. Consequently the mean free path of particles is large, but not infinite. The mean free path sets the "thickness" of the transition or interaction layer. I'm not familiar with the conditions around the heliopause, but we might take the value for "extreme vacuum" from the table in article mean free path as indicative (for charged particles in a plasma, the interaction should be stronger, hence the mean free path shorter). Even increasing generously from 105 km to 106 km or more, that is still small compared to the radius of the heliopause, which is some 1010 km. That is a rather sharp transition. --Wrongfilter (talk) 14:13, 2 September 2017 (UTC)[reply]
No, at densities prevalent in the solar wind and in interstellar medium the collisions are so rare that they can be neglected (at least at the scales of about 100 au). So, the plasma is effectively collisionless. However, because charged particles can interact over long distances electromagnetically, plasma behaves more like a liquid than like a cloud of particles. You can look here. Ruslik_Zero 16:59, 2 September 2017 (UTC)[reply]
Thrown off by the previous question into thinking that this was WP:RDS? -- ToE 02:49, 2 September 2017 (UTC)[reply]
Lol, got so caught up with the whole alien discussion, I hadn't even noticed myself!73.232.241.1 (talk) 03:05, 2 September 2017 (UTC)[reply]
Yep, that's exactly what happened. StuRat (talk) 13:16, 2 September 2017 (UTC) [reply]

Why do the solar wind and interstellar medium remain charged ?[edit]

That is, if they contain positive and negative ions and particles, why don't they attract each other and form neutral particles, given billions of years ? StuRat (talk) 17:06, 2 September 2017 (UTC)[reply]

Take a look at Interstellar medium, in particular at the section on heating and cooling. To a first approximation, much of the ISM is very hot (and tenuous), so it forms a plasma. For a nucleus to catch an electron, not only do they need to meet, there needs to be a way for the pair to to get rid of the extra kinetic energy. In the very low pressure environment, there are not many ways to do that. --Stephan Schulz (talk) 18:18, 2 September 2017 (UTC)[reply]
Solar wind ions move hundreds (thousands?) of km/s (at least around Earth), they don't have billions of years. Sagittarian Milky Way (talk) 18:32, 2 September 2017 (UTC)[reply]
According to Heliosphere, they are emitted from the sun at 400 km/s, and at the termination shock slow to below the speed of sound, which is, under these circumstances, 100 km/s (And boy, Speed of sound really goes over my head ;-). So it takes them on the order of magnitude of one year to go from the sun to the heliopause, unless I dropped a zero or five somewhere... --Stephan Schulz (talk) 18:44, 2 September 2017 (UTC)[reply]
How about just considering solar wind particles with opposing charges, then. Presumably they are moving at much the same velocity, so that their relative velocity is low. Why don't they all "hook up" ? StuRat (talk) 01:31, 3 September 2017 (UTC)[reply]
They are still at high temperature, and at high temperatures, gases are mostly ionised. The kinetic energy of the electrons is high compared to the binding energy of the outer electrons, so chances for combination are low and chances for knocking off electrons are high. See Plasma (physics), in particular Plasma_(physics)#Temperatures. --Stephan Schulz (talk) 08:23, 3 September 2017 (UTC)[reply]
Speed of the solar wind is about 500 km/s, temperature is about 100 kK. This means that the thermal velocity of the protons is about 40 km/s, much less than the bulk motion, and that of the electrons about 1700 km/s, much more than the bulk motion. The protons mostly move with the flow, the electrons have a random motion on top of that, constrained by the magnetic field. Of course they can recombine and they will, emitting a photon, in some fraction of the cases when a proton and an electron meet. However, when one of the few neutral atoms collides with an electron or a proton, there's a large probability it will get ionised again.
Not mentioned by anyone yet is the ultraviolet radiation. The UV from the sun and other stars (and energetic non-stellar objects) is quite intense and responsible for keeping almost the entire universe ionised. It's the UV that keeps the ISM hot. PiusImpavidus (talk) 09:07, 3 September 2017 (UTC)[reply]

Why do the solar wind and interstellar medium repel each other ?[edit]

For this to happen, they would need to have the same electric charge, but aren't any two particles passing near each other just as likely to have opposing charges and attract each other ? StuRat (talk) 17:08, 2 September 2017 (UTC)[reply]

Well, your assumption about required charges is wrong - two baseballs that hit each other in there are quite able to bounce back, even if both are uncharged. But I think the misunderstanding goes deeper. Why should the solar wind and the ISM repel each other? The solar wind eventually merges into the interstellar medium. --Stephan Schulz (talk) 18:26, 2 September 2017 (UTC)[reply]
Our heliosphere article states: "The solar wind flows outward from the Sun until encountering the termination shock, where motion slows abruptly". That sure sounds like the two are repelling each other. If they simply passed through each other, why does the solar wind slow so much ? Could it be that the charged particles in the solar wind do merge with those in the interstellar medium, and take on the resultant of the two motions ? StuRat (talk) 19:08, 2 September 2017 (UTC)[reply]
What is "interstellar medium"? There is maybe a single atom per cubic mile in the space inbetween star systems. Want to calculate how many atoms you could send flying thu that cubic mile befor a single one might have a chance to come close to the one waiting there? --Kharon (talk) 20:43, 2 September 2017 (UTC)[reply]
Interstellar medium says densities don't go below 1 ion per 10 liters. Sagittarian Milky Way (talk) 20:49, 2 September 2017 (UTC)[reply]
An interstellar medium is a medium who took an astral plane beyond menopause and all the way to the heliopause. StuRat (talk) 20:58, 3 September 2017 (UTC) [reply]
They could claim it has 2 little unicorns per 10 liters. Its all assumptions. Maybe educated calculations, maybe bluntly made up to fit another assumption or theory. Noone and nothing was or is out there to measure the factual. Same with the Dark matter. They could claim its from the apples the unicorns drop. If they would stick together, agreeing to that, noone could prove they bullshit us all. --Kharon (talk) 21:06, 2 September 2017 (UTC)[reply]
And your atom per cubic mile was somehow based on better information? --Trovatore (talk) 21:19, 2 September 2017 (UTC)[reply]
The fundamental forces of individual particles generate interference patterns which form tensor fields which can most certainly cause relatively strong collisionless reactions. And geometrically speaking, it's fairly easy to work out the average density of particle distributions, so your argument that it is somehow "all assumptions" is a little weak. And just take a look at the densities that are directly measured here within the Earth's orbital radius[2], 150 million kilometers from the Sun. Particle speeds of over 500 kilometers per second and densities of at least 1 to 10 particles per cubic centimeter. So yes, the resulting force fields can be substantial.73.232.241.1 (talk) 23:32, 2 September 2017 (UTC)[reply]
I can't answer Stu's question, per se, but I can provide a timely xkcd. --Trovatore (talk) 21:33, 2 September 2017 (UTC) [reply]
Also, the force of attraction must exceed the relative momentum difference between the particles in order for them to interact directly to any appreciable degree. Consider a hypothetical situation where only two particles exist in the universe, separated by billions of light-years of distance. If they are initially moving away from eachother at a separation velocity above some threshold, then the shared gravitation between them is overcome and they will remain separated until the gravitational force finally has time to catch up (and in this case, it eventually will, because gravity is an acceleration (quadratic) while velocity is linear).73.232.241.1 (talk) 22:58, 2 September 2017 (UTC)[reply]
I don't think that the interstellar medium should be able to repel particles with a given charge inward from all directions. This isn't one of Maxwell's equations per se, but a charge on all sides should cancel out somehow. (shell theorem? There must be something more general) But it should be possible for opposite charges to stop moving charged particles - I mean, proton meets electron, they settle down, make beautiful babies together. Cathode rays travel in vacuum, but not in air - charged particles can interact with neutral atoms. As for the specifics, there's some general prose here but I can't say I really understand the situation as yet. Many astronomical objects, Sun included, are tremendously hot and are surrounded by big bubbles of hot outward moving gas, whose pressure gets weaker as it expands, but why that slows so abruptly to form a termination shock is another question. Wnt (talk) 01:17, 3 September 2017 (UTC)[reply]
The termination shock is just an acoustic shock, where speed suddenly goes down and pressure up. Compare it to a jet engine in a supersonic flow. Right ahead of the engine inlet, the air flows in at supersonic speed. In the inlet you get a shock, where the flow suddenly slows to subsonic, whilst the pressure increases. (Actually, jet engines use a number of oblique shocks.) It's not much different with the termination shock of the solar wind. Given the Mach number of the solar wind, there is a distance from the sun where pressures and densities are just right to form the shock, and that's where the shock forms. The only difference is that the solar wind is a magnetised plasma, so part of the pressure is gas pressure, part is magnetic pressure. PiusImpavidus (talk) 09:49, 3 September 2017 (UTC)[reply]
See https://en.wikipedia.org/wiki/Heliosphere#Termination_shock. According to this lecture the termination shock is expected to be a fast mode magnetohydrodynamic shock wave. A "reverse shock" directed toward the sun. Oh, I also found this paper titled The Heliospheric Termination Shock on arxiv.org. I haven't looked at this article yet. It was later published here in a book titled Physics of Collisionless Shocks. Also, this article on collisionless shock waves seems accessible (I've only skimmed it). --Modocc (talk) 15:12, 3 September 2017 (UTC)[reply]
@Modocc: That's a great answer, but it looks like there's some work to do on Wikipedia. Bow shocks in astrophysics says the Sun has no bow shock wave, while Heliosphere says it does! Also, the latter article uses File:Heliosphere-washbasin.svg to illustrate a shock wave, but I have to ask ... isn't there still fast flow going on in the sink situation past the "boundary", just hidden by the roiling water on top? Still working on understanding why the transition is abrupt, or what determines how abrupt. Wnt (talk) 23:18, 4 September 2017 (UTC)[reply]
Recent space probe data suggests that the interstellar gases aren't supersonic relative to the solar wind at the distance where they interact (as once assumed) so no shock wave. Sagittarian Milky Way (talk) 02:31, 5 September 2017 (UTC)[reply]