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September 15

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Tungsten carbide is inorganic?

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It says at Tungsten carbide is inorganic, but I thought the definition of organic simply means that it has carbon in it. ScienceApe (talk) 00:11, 15 September 2013 (UTC)[reply]

Organic compound, first paragraph has the answer. OsmanRF34 (talk) 00:16, 15 September 2013 (UTC)[reply]
Yes, well, then it should be fixed as hydrogen carbide may just as well be the mascot for organic compounds. Plasmic Physics (talk) 10:46, 15 September 2013 (UTC)[reply]
Is there anyone in this century that still calls methane 'hydrogen carbide'? TenOfAllTrades(talk) 04:03, 16 September 2013 (UTC)[reply]
It's an extreme example to prove a point: how arbitrary, inconsistent, and useless, the designation of '(in)organic' is. Plasmic Physics (talk) 00:59, 17 September 2013 (UTC)[reply]
I don't know; if anything I would say it's the opposite. It strikes me that we call it methane and not hydrogen carbide precisely because 'organic' chemistry and behavior of the compound is so much more important and relevant and applicable than the 'inorganic'—particularly in comparison with the inorganic carbides. TenOfAllTrades(talk) 13:02, 19 September 2013 (UTC)[reply]

Plank length

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Moved from Wikipedia:Reference_desk/Humanities. OsmanRF34 (talk) 02:57, 15 September 2013 (UTC)[reply]

If one had an oak plank board that was 2 by 12 and weighted 100 pounds how long would it be (estimate)?--Christie the puppy lover (talk) 22:50, 14 September 2013 (UTC)[reply]

See Specific Gravity Weights Of Materials from READE.
Wavelength (talk) 23:48, 14 September 2013 (UTC)[reply]
See Planck length. μηδείς (talk) 00:43, 15 September 2013 (UTC)[reply]
Haah! Clarityfiend (talk) 02:46, 15 September 2013 (UTC)[reply]
I was going to say that myself. :-) StuRat (talk) 03:11, 15 September 2013 (UTC) [reply]
You mistyped the question. Plank length has worked fine for 8 years. PrimeHunter (talk) 12:08, 15 September 2013 (UTC)[reply]
Note that a board sold as "2 x 12" isn't necessarily really 2 x 12. According to Lumber#Dimensional lumber, it's really 1.5 x 11.25 inches. This will significantly affect your calculations. StuRat (talk) 03:11, 15 September 2013 (UTC)[reply]
Now that simplifies it. I'll take it from here. Thanks for the very nice answers.--Christie the puppy lover (talk) 11:48, 15 September 2013 (UTC)[reply]
You're quite welcome. StuRat (talk) 12:35, 16 September 2013 (UTC)[reply]
Heh. Depends if it's a physics question or an engineering question. --jpgordon::==( o ) 20:21, 15 September 2013 (UTC)[reply]
I tend to think most questions involving wood planks are engineering/carpentry questions. StuRat (talk) 12:35, 16 September 2013 (UTC)[reply]
Resolved

StuRat (talk) 12:35, 16 September 2013 (UTC)[reply]

My nuts are overly saturated

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Is it possible to breed or genetically engineer the saturated fats out of tree nuts and peanuts, or is there some reason why this wouldn't work ? The reason I ask is that these would seem to be an ideal source of protein, especially for vegetarians, if not for the sat fat. I would anticipate that nuts with only unsaturated fats might tend to seem greasier, since the fats would be more liquid, but that would be OK by me. StuRat (talk) 04:02, 15 September 2013 (UTC)[reply]

The purpose of the fat is to nourish the embryonic plant. If you mutate the plant, then grow it from a cutting, rather than a seed, you can breed out traits otherwise necessary for the seed's survival. Indeed, that is how they breed seedless plants, from mutated cuttings. μηδείς (talk) 04:06, 15 September 2013 (UTC)[reply]
Note that I don't want to remove all fats, just the saturated fats. StuRat (talk) 04:16, 15 September 2013 (UTC)[reply]
See selective breeding and seedless fruit and apply the notion of getting rid of the saturated fat instead of the seed as a whole. μηδείς (talk) 19:36, 15 September 2013 (UTC)[reply]
The notion of a seedless peanut is interesting. ←Baseball Bugs What's up, Doc? carrots04:44, 15 September 2013 (UTC)[reply]
I'm still waiting for the boneless chicken. And as a plus, they should be a lot easier to catch. :-) StuRat (talk) 07:38, 15 September 2013 (UTC) [reply]
It's not all that obvious the saturated fat is bad - it's not "physics". Eating the nuts seems to be beneficial overall. See [1] for example. I would put the empirical evidence that nuts are good far above the theory about any improvement that might be made. The composition of dietary fat, its uses in the body, its basis as the biochemical origin of a vast network of eicosanoids, prostaglandins and cannabinoids ... we may not know what alterations of the plants would be good for us, and it is hard to collect sufficient data. Wnt (talk) 04:13, 15 September 2013 (UTC)[reply]
Fittingly, given the title of this section, it appears that the way Rocky Mountain oysters are typically prepared, they would be right much fatty. ←Baseball Bugs What's up, Doc? carrots16:04, 15 September 2013 (UTC)[reply]
Yeah, I thought it was a request for medical advice. 86.160.220.5 (talk) 19:28, 18 September 2013 (UTC)[reply]

Solar eclipses

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These question grow out of my memory of seeing an eclipse in the 2000s, either the Solar eclipse of December 14, 2001 or the Solar eclipse of April 8, 2005: we were on the edge of the partial eclipse, and the only ways I knew that an eclipse was happening were (1) reading about it, and (2) seeing a tiny chunk missing from the bottom of the Sun when viewing it with welder's glass.

  1. In general, what percent of the Sun must be covered for the eclipse to cause a noticeable darkening of the sky, assuming a cloudless day? In other words, imagine that you're sitting on the ground, looking around at everything in front of you (but not to the point that you're obsessed and ignoring everything else); in this situation, when will you start noticing if your eyes are normal and you're not paying particular attention to the Sun?
  2. When a total eclipse occurs and you're out of the path of totality but not out of the path of partiality, part of the Sun will be obscured. Is there a relationship between (1) the portion of the Sun that's obscured, and (2) the direction from you to the location where the eclipse is total? During the eclipse in question, I was far to the north, and the "missing" chunk was at the bottom of the solar disk. If I'd been way to the south, would the chunk be missing from the top? Or would it have been missing from one side if I'd been at the same latitude but way far east or west?

Nyttend (talk) 04:35, 15 September 2013 (UTC)[reply]

1) It will vary from person to person, but note that our eyes' reactions to light levels isn't linear. That is, half as many photons barely makes a difference in the apparent light level, rather than seeming "half as bright". So, based on that, quite a large portion of the Sun might need to be obscured before the light level would change noticeably. Also note that passing clouds can have the same effect, so even if you did notice the change in light level, you might well ignore it. StuRat (talk) 07:19, 15 September 2013 (UTC)[reply]
When a large portion of the sun is eclipsed, you might notice something strange about shadows -- they seem to have less sharp edges. I noticed this in the 2001 eclipse. Dbfirs 08:14, 15 September 2013 (UTC)[reply]
I've experienced only partial solar eclipses, so I can't give you a quantitative answer. But I recall an eclipse that covered well over half the sun in my area, and the effect was not the same as simply an overcast sky. The way I like to describe is it's as though the sun were on a dimmer switch. But even at more than half the sun covered, there was no problem seeing; it was still sufficiently bright. ←Baseball Bugs What's up, Doc? carrots14:47, 15 September 2013 (UTC)[reply]
Actually, a 50% eclipsed sun is still 50 times brighter than an overcast day. SteveBaker (talk) 14:58, 15 September 2013 (UTC)[reply]
Full sunlight is a thousand times brighter than a well-lit room and it's 100 times brighter on a "sunny" day than on an overcast day. Since our eyes and brain compensate for this massive difference, we don't especially notice the brightness difference between the two unless we step from one to the other very quickly, it would be hard to tell the difference between the normal sun and (say) a 75% covered sun without looking directly at it (through welders' glass or whatever).
If you didn't happen to look at the sun, you'd never notice a 75% partial eclipse because it's still brighter than an overcast day...not much different than when a cloud passes in front of the sun.
However, in a full eclipse, the sun almost completely vanishes behind the moon (and there is no moonlight of course) - so everything looks as dark as night....often quite a bit darker because street and building lights - and most of the other sources of light pollution such as car headlights are not turned on because everyone is standing around excitedly watching the once-in-a-lifetime total eclipse. The sky is black and you can see the brighter stars. But that period of "totality" is generally quite short and as soon as the sun starts to be uncovered even a little bit, the sky turns blue again and our perception is that it's more like a cloudy day. SteveBaker (talk) 19:41, 15 September 2013 (UTC)[reply]
I observed the Solar eclipse of May 30, 1984 as an annular eclipse. As I recall, 85% of the sun was covered. Passersby kept asking me what I was doing -- they had no idea an eclipse was in progress. The only way you could tell by looking around was if you noticed that there was no glare off the chrome of parked cars. Duoduoduo (talk) 18:39, 15 September 2013 (UTC)[reply]

Is carbon tetrachloride an organic compound?

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The carbon tetrachloride says it is, however "CCl4" is in Template:Carbon compounds. Which is correct?--Jsjsjs1111 (talk) 11:20, 15 September 2013 (UTC)[reply]

To clarify, Template:Carbon compounds is specifically/only inorganic ones--naming threw me a bit. Our organic compound has some cited commentary on the uncertainty of the limits of that category, so I'm not sure there is a good answer. But we should at least be self-consistent and hopefully have WP:CITE to support whichever way we go. DMacks (talk) 11:36, 15 September 2013 (UTC)[reply]
Yes this template is confusing. Apparently it is missing cyanogen, tetrabromomethane, etc.--Jsjsjs1111 (talk) 11:47, 15 September 2013 (UTC)[reply]
Also, that template is redundant to Template:Inorganic compounds of carbon (need to merge them and kill one, once someone figures out what operational definition to use). And missing several other carbon–oxygen ones. There seem to be articles about almost thirty (!) such, but some are closer to some definitions of "organic" than others. DMacks (talk) 12:47, 15 September 2013 (UTC)[reply]
Jsjsjs1111 and DMacks, please note that I've sent the templates to TFD. Nyttend (talk) 20:11, 15 September 2013 (UTC)[reply]

Quality of light needed to charge from solar panel

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I was wondering if natural light is needed to charge devices that use solar panels. I've tried to charge a solar phone charger using indoor light from florescent bulbs, but it didn't seem to charge. Direct, unclouded sunlight was needed to charge it at all. Are there different types? I can find nothing about it in the Solar panel or solar cell articles.--Auric talk 13:16, 15 September 2013 (UTC)[reply]

While there is some relevant solar cell physics here, your problem is a universal feature that you would always encounter with any device that converts light into energy. If you reduce the intensity of the power source, you obviously won't be able to draw as much power from the device. This means that for the same current the voltage must be lower (if you can draw that current at all). Now for your battery charger, the voltage is critical, if this drops below a certain value (which changes during the charging process), it won't work.
Note that natural light is hugely more intense than indoor light, we don't notice this because the brain automatically compensates for this. Count Iblis (talk) 13:42, 15 September 2013 (UTC)[reply]
Yes, indeed! To put numbers on that, our Sunlight article says that at ground level, the noonday sun can deliver 1004 watts per square meter, which is composed of 527 watts of infrared radiation, 445 watts of visible light, and 32 watts of ultraviolet radiation. Standard solar panels only use visible light to generate power - so we should go with the 445 watt number. Compare that to a 100 watt light bulb - which (being only about 3% efficient) produces something like 3 watts total - spread over the entire room! So you'd need around 150 hundred watt bulbs, with reflectors behind them to direct all of the light onto a one square meter solar panel! Since even the best solar panels produce 160 watts per square meter - this would require 100 times as much electricity to light the bulbs as the system would produce! Even using LED lamps (which are the most efficient way to generate visible light that we can buy) - you'd need ten times as much electricity as you'd make from the solar panel.
From the point of view of recharging your phone from ordinary room lighting, the engineering standard for "normal" room lighting is 150 lux. Full sunlight is 100,000 lux. So expect your phone to charge about 650 times slower than it would in direct sunlight! The standard for office and workshop lighting is 500 lux - but you're still going to have to wait 200 times longer! Even on an overcast day, the sun delivers 1000 lux. SteveBaker (talk) 14:43, 15 September 2013 (UTC)[reply]
A good example is when driving on a sunny day. The driver and passengers in other cars are pretty much shrouded in seemingly very dim light until you get very close to the car. Vision is an amazing system. ←Baseball Bugs What's up, Doc? carrots14:51, 15 September 2013 (UTC)[reply]

Eigenstate Symmetry

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It is often said that the ground state of a system must respect the symmetries of the Hamiltonian. Is this strictly true? Even if there are multiple degenerate ground states? — Preceding unsigned comment added by 31.55.103.72 (talk) 17:00, 15 September 2013 (UTC)[reply]

Not if the ground state is degenerate, but in that case you can choose them to respect the symmetries of the Hamiltonian. Count Iblis (talk) 17:21, 15 September 2013 (UTC)[reply]
No, if the ground state is degenerate it may be impossible to find a ground state solution that respects all the symmetries of the Hamiltonian. That's how the Higgs mechanism works and leads to spontaneous symmetry breaking. Nothing surprising there. Try placing a sphere on a table and placing a second sphere on top of the first. The Hamiltonian has axial symmetry. But the ground state is achieved when the top sphere rolls to one side. This ground state is degenerate(since there are many directions for the ball to roll to) and does not have axial symmetry. Dauto (talk) 17:37, 16 September 2013 (UTC)[reply]
On the other hand, you could presumably find a linear superposition of such states which did have the axial symmetry. So why is this not physical? Jheald (talk) 18:25, 16 September 2013 (UTC)[reply]
Such a superposition would have nonlocal properties. So, if you imagine that there are observers then they will end up observing the system in a localized state. But then a Many Worlder will argue that you still have a symmetric state but this involves the observers being entangled with the system, this is then a superposition of all the possible ways the symmetry can be broken. Count Iblis (talk) 18:43, 16 September 2013 (UTC)[reply]

How is it calculated actually? It's clear that alcohol has, thermodynamically, lots of calories, and so is the case of straw and grass. But shouldn't food calories only be things that we can absorb as calories? So, if they tell you alcohol has say 1000 kc for 1 l, does it mean that you can use those calories as a form of energy? OsmanRF34 (talk) 18:04, 15 September 2013 (UTC)[reply]

Celery, made up of lots of long sugars, presumably has lots of calories, even though none of the sugars are digestable by the human body. A package of celery with Percent Daily Values (image is here, in case you're not familiar with the name) will tell you that it has no calories, so it's clear that in this context, it's only things that we can absorb as calories. Nyttend (talk) 20:16, 15 September 2013 (UTC)[reply]
As that article itself says:
direct calorimetry would give systematic overestimates of the amount of fuel that actually enters the blood through digestion. What are used instead are standardized chemical tests or an analysis of the recipe using reference tables for common ingredients[8] to estimate the product's digestible constituents (protein, carbohydrate, fat, etc.). These results are then converted into an equivalent energy value based on the following standardized table of energy densities.
DMacks (talk) 04:14, 16 September 2013 (UTC)[reply]
As I understand it, first they dehydrate the item to be tested, then they add fuel and burn it. The amount of heat produced in excess of that produced by the fuel alone is the raw number of calories. However, it then gets tricky when dealing with indigestible calories. This requires them to determine the percentage of the total calories which are digestible, which is done via a number of theoretical and experimental methods. Another wrinkle is that it also takes calories to digest food, so this should really be subtracted to calculate the net calories gained from each food, but this is further complicated by the fact that each person's digestive system has a slightly different efficiency in digestion, and certain combos of foods might be easier to digest than others. So, they don't even attempt to account for this. StuRat (talk) 04:21, 16 September 2013 (UTC)[reply]
Yes, it's a hard question. To give a concrete (if extreme) example, if you were to eat a raw potato (which I strongly do not recommend you try!!) - and swallowed it in large chunks without much chewing - then the starch that's inside the cells of the potato wouldn't be metabolized into sugars and then absorbed into the blood stream because the cell walls of the potato are made of cellulose - and humans lack the ability to digest that. The undigested potato passes right through your gut, almost unchanged. When you cook a potato, you rupture the cell walls, releasing the starch so you can digest it. The theoretical calorific value of a raw potato (calculated by the drying and burning method) is exactly the same as for a cooked potato - but the number of calories you'd absorb from eating it raw would be almost zero - compared to around 100 to 200 calories if you cooked it. SteveBaker (talk) 16:38, 17 September 2013 (UTC)[reply]

Two Astronomy Questions

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So I'm writing a science fiction story, and I'm having a lot of trouble determining where things are.

First, is there some way of telling how far a planet or moon will be from another planet or moon at a particular date? All I've been able to find is approximate distances from Earth or the sun, never exact distances. I mean it's the easiest thing in the world to find that Mars will be in Pisces, and Jupiter in Aquarius, but that tells me virtually nothing about the distance between the two.

Secondly, what directions should I be using? From google searches, I've found the directions North & South (toward & away from Andromeda), Coreward & Rimward (toward and away from the core, which is in Sagittarius from our perspective), and Spinward & Trailing (which I can't identify). Various science fiction roleplaying references give me various stars which are supposedly Spinward or Trailing, but all of these stars seem to be in the wrong places when I compare them to actual constellation maps or galactic maps. So what's the story with these directions, and are these even the directions I should be using for travel inside the solar system?

67.142.167.20 (talk) 18:44, 15 September 2013 (UTC)[reply]

Are you looking for the real locations of actual objects in our solar system at specific future dates? Or are you writing about fictional moons and planets, and trying to describe them realistically? It sounds like you're interested in travel within the solar system, so you'd want to use heliocentric ecliptic coordinates. This is a set of spherical coordinates centered on the Sun. It's similar to latitude and longitude on the Earth's surface, with a third coordinate: the distance of a location from the center of the Sun. The planets orbit in the "equator" (the ecliptic), and the poles point out of the plane of the solar system. If you want the locations of real solar-system objects at future dates, you'll want to use one of the ephemeris services you can find online. If you give a bit more info about what you're looking for, I'm sure someone here can help you get started. --Amble (talk) 22:23, 15 September 2013 (UTC)[reply]
On second thought, instead of using an ephemeris server you should probably check out Celestia. It's a free-software program that lets you see the sky from wherever you like at a future date. For example, you could ask for the view from a particular place on the surface of Ganymede on this date in 4010. --Amble (talk) 22:30, 15 September 2013 (UTC)[reply]
(ec)There is the galactic coordinate system for positions of stars in the galaxy. However, this is not a suitable system for travel within the solar system. Yes, we can compute the distance between any two solar system planets to a reasonable degree of precision (given the size and irregularities of the bodies, it makes no sense to compute the distance to meters). For coordinates in the solar system, the ecliptic coordinate system is probably easiest. --Stephan Schulz (talk) 22:35, 15 September 2013 (UTC)[reply]
You didn't tell us about your math background, so I'll just describe how I would do it and you can ask for clarification if you don't follow. Go to NASA's HORIZONS service, choose a target body (say Mars), and choose a date range. Afterwards, click "change" next to Table Settings and select Heliocentric ecliptic lon. & lat. and Heliocentric range & range-rate. This gives you the Mars' position in a spherical coordinate system centered on the Sun, so it tells you exactly where Mars will be. Finally, click Generate Ephemeris. To calculate the distance to another planet, get the coordinates for that other planet and calculate the distance between the two sets of coordinates. --Bowlhover (talk) 00:44, 16 September 2013 (UTC)[reply]

Hydrochloric acid changed the colour of plastic tube

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I had some concentrated hydrochloric acid in a 50 ml tube (polypropylene) for a few months, stored in mostly darkness. The tube seems to have changed in colour to a sort of light pink (still translucent). Why has the tube changed colour? — Preceding unsigned comment added by 78.150.23.174 (talk) 18:47, 15 September 2013 (UTC)[reply]

I don't know, but it might help if you specified the temperature at which it was stored and the pre-storage colour of the tube. Nyttend (talk) 20:14, 15 September 2013 (UTC)[reply]
The tube was clear (transparent) and the storage temperature was 18-24 C. — Preceding unsigned comment added by 78.150.23.174 (talk) 20:59, 15 September 2013 (UTC)[reply]
Well, that's pretty weird -- polypropylene is a long-chain aliphatic compound, and thus is supposed to be inert to acid-catalyzed reactions. The only thing I can think off the top of my head is free-radical chlorination, but that requires UV rays (such as from sunlight) as well as free chlorine in the solution -- so, since you kept the test tube in darkness, that can't be it either. Anyone else have an idea what's going on? 24.23.196.85 (talk) 01:03, 16 September 2013 (UTC)[reply]
My first thought is to find out what the plasticizer is. Wnt (talk) 02:25, 16 September 2013 (UTC)[reply]
Oh yeah, if there are phthalate/terephthalate esters (or any other kind of esters, for that matter) in the plastic, you get hydrolysis and possibly alfa-substitution and/or aromatic chlorination reactions as well. So yes, this could be responsible for the discoloration. 24.23.196.85 (talk) 03:16, 16 September 2013 (UTC)[reply]

Leaf fluffies

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Resolved
Leaf fluffies

Does anyone know what these fluffy things are on the leaf (two views provided at right)? DRosenbach (Talk | Contribs) 20:43, 15 September 2013 (UTC)[reply]

They look like galls. This page has an idenfification tool that might help narrow it down a bit. - Karenjc 22:15, 15 September 2013 (UTC)[reply]
I was thinking of something like that too, but the photos are not clear enough. Can you possibly get a good close up? The type of plant the leaf is from may help too. (oak?) 'Fluffy' suggested caterpillars, but they aren't are they? This page Oak Gall Gazetteer may help, "the shaggy-looking HAIRY OAK GALLS produced on the upper leaf surface by A. villosa;". (Hairy Gauls, miniature Tribbles? ;-) )-ө- 220 of Borg 03:30, 16 September 2013 (UTC)[reply]
This flickr picture looks very similar Oak Leaf Hairy Gall. ӧ 220 of Borg 03:46, 16 September 2013 (UTC)[reply]
Yes -- thank you all! DRosenbach (Talk | Contribs) 23:58, 16 September 2013 (UTC)[reply]

Relativistic speed

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Is there anything in this universe that travels at relativistic speeds, aside from subatomic particles? Is there anything macroscopic, aside from possible alien technology? For convenience, let's arbitrarily define relativistic to mean v > 0.2c. --140.180.246.129 (talk) 21:08, 15 September 2013 (UTC)[reply]

Galaxies located far away (due to the expansion of the universe). Also, closer to home, in case of supernova and hypernova explosions, matter gets ejected at extreme relativistic velocities. Count Iblis (talk) 21:15, 15 September 2013 (UTC)[reply]
Here I've got to say I'm only guessing, but I thought that you could NOT include the Red Shift of galaxies, as this comes about from creation / dilution of space between the galaxies from when they were first formed. This kind of recessional velocity could be faster than the speed of light, I had thought. Indeed, if our universe is not confined to that which is "observable", then we must admit that there are galaxies beyond the observable ones, and these must be travelling faster than c, relative to us, (as we do with respect to them). Myles325a (talk) 03:21, 16 September 2013 (UTC)[reply]
There are stars orbiting the supermassive black hole in the center of the galaxy (Sagittarius A*) that travel pretty quickly, but the fastest is S2 (star) at 1.67% of the speed of light. That doesn't meet your 0.2c criterion. The stars of the double pulsar PSR J0737-3039 only move at about 0.1% of c. As the Count points out, distant galaxies have recessional speeds that can be very large, but I assume you want something that's traveling with v > 0.2c relative to its surroundings. The polar jets of supermassive black holes and even stellar-mass black holes can be very relativistic, well over your threshold. They are jets of plasma, not just particles, and they're pretty well macroscopic. --Amble (talk) 22:15, 15 September 2013 (UTC)[reply]
Velocity is relative! A velocity cannot be ascribed to an object! Velocity is a description of a relationship between two objects!! — Preceding unsigned comment added by 86.189.1.198 (talk) 23:11, 15 September 2013 (UTC)[reply]
That's true, but most of the time we assume a velocity is measured "relative to its surroundings", so the question makes good sense if we assume "travels at relativistic speeds" means relative to a reference frame in which the other objects of significant size have velocities very much less than c. Dbfirs 08:21, 16 September 2013 (UTC)[reply]
The numbers in the infobox for PSR J1748-2446ad, the fastest known millisecond pulsar give a surface velocity of about .128c assuming the radius is 16km (the upper bound in the infobox). Katie R (talk) 14:40, 16 September 2013 (UTC)[reply]
Actually, it looks like I read 16km as the diameter when it is actually the radius - the article itself mentions that the surface velocity is about .24c. It really shows how incredibly dense a neutron star is when you consider that gravity manages to keep it from flying apart even though it is rotating that quickly. Katie R (talk) 15:01, 16 September 2013 (UTC)[reply]
Relativistic Jets. Dauto (talk) 15:34, 16 September 2013 (UTC)[reply]

Mycoremediation of Sarin for Gas Masks

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I've read that most of the information regarding the mycoremediation of Sarin; including the species involved is classified by the United States defense department; however in theory could the mushrooms that bio remediate sarin be used as part of a cheaply distributed gas mask? — Preceding unsigned comment added by CensoredScribe (talkcontribs) 22:07, 15 September 2013 (UTC)[reply]

That would depend not on how efficiently they process (detoxify) sarin, but how efficiently they absorb it from the air. And I don't think anything currently available (including mushrooms) can beat activated charcoal in terms of absorption efficiency (well, technically adsorption efficiency in this case, but you got the point, right?) In other words, it's one thing to detoxify sarin in the ground, and another thing entirely to remove it from the air we breathe. 24.23.196.85 (talk) 00:50, 16 September 2013 (UTC)[reply]
Mycoremediation relevant link. ж 220 of Borg 03:40, 16 September 2013 (UTC)[reply]
The problem is all a matter of speed. A human doing moderate exercise needs around 40 liters of air per minute - extracting the toxin from that volume of air in that amount of time will be the issue here. The Mycoremediation studies probably do things like planting a bunch of mushrooms in a container with maybe a liter of contaminated soil, waiting a few weeks and seeing if the amount of toxin was reduced by 10%. We need 40 liters to be reduced by close to 100% within one minute - using only enough mushrooms that could fit in a 10cm cube - and for them to be able to continue to do that for hours. That's asking a hell of a lot! SteveBaker (talk) 14:46, 16 September 2013 (UTC)[reply]

Powerful singing voice before infections?

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This is really weird...right before I get a bad ear infection or a case of strep, my usual 2.5 octave (my bottom sounds like Rick Astley, so it's not as high as you would think) singing voice with excessive vibrato turns into a well managed 4 octave powerhouse. This is 2-3 days before I get a throat or ear infection. It's even weirder because i'm a guy. Any ideas why?
I am not requesting medical advice as (a.) this is not a condition that requires treatment and (b.) I am asking why, not how or how to fix it. --Sean R. Kurth (talk) 22:56, 15 September 2013 (UTC)[reply]

(EDIT: Both my school and church chorus teachers have confirmed this odd occurence) — Preceding unsigned comment added by Sean R. Kurth (talkcontribs) 23:29, 15 September 2013 (UTC)[reply]
The voice lowering as a result of a throat infection is a common effect. Not quite sure what causes it, though, perhaps a thicker than normal layer of mucous on the vocal folds ? StuRat (talk) 04:27, 16 September 2013 (UTC)[reply]
Treating the vocal cords as vibrating strings, we get:
where is the tension, is the linear density (that is, the mass per unit length), and is the length of the vibrating part of the string.
Which means that the heavier the vocal cord gets, the larger becomes and the lower the frequency it vibrates - and as StuRat points out, swelling and increased mucous both contribute to that. SteveBaker (talk) 14:28, 16 September 2013 (UTC)[reply]