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January 19[edit]

About AHL[edit]

Is AHL special for each kind of bacterium? — Preceding unsigned comment added by Viacoha (talk • contribs) 02:06, 19 January 2013 (UTC)[reply]

What does the American Hockey League have to do with bacteria? ←Baseball Bugs ^{What's up, Doc?} carrots→ 02:12, 19 January 2013 (UTC)[reply]

[1] Gzuckier (talk) 08:50, 19 January 2013 (UTC)[reply]

AHL → acyl-homoserine lactones (enables many gram-negative bacteria to engage in quorum sensing) — Preceding unsigned comment added by Viacoha (talk • contribs) 02:16, 19 January 2013 (UTC)[reply]

Yes, different bacterial species generally have different AHLs. 24.23.196.85 (talk) 02:38, 19 January 2013 (UTC)[reply]

Astronauts talking in space[edit]

You know that sound just needs only medium to travel. So, if two astronauts touched their helmet together and are talking, will they be able to listen each others voices?27.62.9.222 (talk) 12:31, 19 January 2013 (UTC)[reply]

Yes, that should work if the helmets touch at points that are made of some hard material (glass, plastic, metal, etc - rubber would likely work poorly). 88.112.41.6 (talk) 14:02, 19 January 2013 (UTC)[reply]

Also, if they only touch at a single point, that might not be enough. That is, the volume level might be too low to hear. StuRat (talk) 19:02, 19 January 2013 (UTC)[reply]

They did that in First Men in the Moon (1964 film). Bubba73 ^{You talkin' to me?} 02:59, 20 January 2013 (UTC)[reply]

And in Homeward Bound. Whoop whoop pull up ^{Bitching Betty | Averted crashes} 05:07, 20 January 2013 (UTC)[reply]

When you consider how well a Tin can telephone works, I'd think it would work extremely well. An old garage mechanic's trick is to use a length of wood to listen to sounds from deep within a running car engine by placing one end of the wood on the bone behind your ear...I've done that lots of times - and it works well. So a large cross-section isn't needed...and the astronauts can always shout. SteveBaker (talk) 21:22, 22 January 2013 (UTC)[reply]

Aurora[edit]

Are auroras harmful? What will happen if any flying object entered in it, e.g. a plane?27.62.9.222 (talk) 12:42, 19 January 2013 (UTC)[reply]

No. Auroras are caused by energetic charged particles (solar wind) colliding with atoms in the Thermosphere, far above the heights that most planes fly at. Even if the space shuttle were to fly through it, it's very unlikely that any harm or damage will occur due to the interaction being at the atomic level. douts (talk) 14:05, 19 January 2013 (UTC)[reply]

Two bar magnets tied together[edit]

If two bar magnets are tied with a string such that one is placed over other, and North pole of first lies above South pole of second, and S pole of first lies above N pole of second. Both magnets are tied for a long time, e.g., for one month. Their poles will interchange or the poles will remain same as initially they are. Britannica User (talk) 13:13, 19 January 2013 (UTC)[reply]

If not heated they are likely to remain the same. Ruslik_Zero 18:31, 19 January 2013 (UTC)[reply]

For the record, they don't even need to be tied together -- opposite poles attract, so they'll stick together naturally. BTW, in this scenario they'll actually keep their magnetism BETTER than if they are stored separately -- they will form a magnetic circuit, which will reduce the leakage of magnetic energy. (The same thing is often done with horseshoe magnets, by either letting two of them cling together end-to-end, or else by placing a keeper bar across the poles.) 24.23.196.85 (talk) 05:12, 20 January 2013 (UTC)[reply]

If they were tied the other way round, then they might lose a bit of magnetism over a month, especially if dropped, hit with a hammer or heated, but I can't think of any conditions under which they would exchange poles. Dbfirs 17:35, 20 January 2013 (UTC)[reply]

Gas from posterior side of body[edit]

Please, don't take this question as funny one. What is the common term for gas released from posterior side of our body ? It has foul smell. Why ? It also produces sound sometimes. Why ? Sunny Singh (DAV) (talk) 13:32, 19 January 2013 (UTC)[reply]

See Flatulence. Mikenorton (talk) 13:34, 19 January 2013 (UTC)[reply]

Please don't take this answer as a funny one, but the common term is a fart. This, however, is somewhat rude, so don't go round saying it to people you don't know. A common, more polite term is just "gas", as in "I can't eat chillies, they give me gas". Again, it's not something that people discuss much, unless needed. IBE (talk) 15:14, 19 January 2013 (UTC)[reply]

I think "gas" can either mean farting or burping. That is, how the gas escapes the body isn't specified. StuRat (talk) 18:59, 19 January 2013 (UTC)[reply]

The medical term for gas generated in the digestive tract is flatus. Gandalf61 (talk) 17:32, 19 January 2013 (UTC)[reply]

One of the stranger medical devices is the flatus bag, designed to collect farts: [2]. Apparently the flatus gases are sometimes analyzed [3], while at other times they are just collected to prevent the patient and medical staff from being exposed to unpleasant and potentially toxic levels of flatus gas. StuRat (talk) 19:15, 19 January 2013 (UTC)[reply]

One can find almost anything on the Internet. http://www.fartnames.com/ has a list of euphemisms, including Trouser trumpet, Message from the Interior, Under-thunder, and my favorite, Step on a Duck. --Guy Macon (talk) 18:09, 19 January 2013 (UTC)[reply]

Semiautomatic cook off[edit]

As the article on Cooking off mentions, it can potentially happen with semiautomatics that fire from a closed bolt position. If a magazine was loaded with the first cartridge being full of a pyrotechnic such that it heated the chamber intensely, could it then just unload the rest of the rounds as if by fully automatic fire? 210.210.129.92 (talk) 16:29, 19 January 2013 (UTC)[reply]

If the following conditions are met, it could happen:

The chamber would have to be hot enough but not the magazine. If, for example, the entire firearm is in a fire, it is likely that the rounds in the magazine will cook off first (less thermal mass, so they get hot first).

As you mentioned, the firearm would have to fire from a closed bolt.

The action would have to be able to cycle at that temperature despite parts expanding and lubrication burning off.

There would be a delay as each new round heats up. This delay would have to be shorter than the time it takes for the chamber to cool down enough so that it doesn't set off a round.

The trigger mechanism would have to be such that an unpulled trigger or the safety (if on) only stops the firing pin from engaging. If it is designed so that an unpulled trigger also locks the cycling, the second round would not make it into the chamber. I don't know if any actual firearms are designed this way. See Trigger (firearms). --Guy Macon (talk) 17:29, 19 January 2013 (UTC)[reply]

If the chamber is hot enough it doesn't matter why it's hot enough. But that said, Guy's last point is kind of interesting... I don't know enough to answer that offhand, but I'll pay attention to that point in the future. Shadowjams (talk) 20:58, 19 January 2013 (UTC)[reply]

Is there relation between yank and work ?[edit]

Is there relation between yank and work ? Is there any formula involving both yank and work ?Sunny Singh (DAV) (talk) 18:35, 19 January 2013 (UTC)[reply]

Well, "yank" is the derivative of force with respect to time, "work" is force moved through a distance. How "related" is that? Not much - a force can vary over time without producing any motion - so the work done is zero. (Imagine a battery powered electromagnet stuck to your refrigerator...as the battery runs down, the force changes (a "yank") - but no actual mechanical work is produced. I'm sure there are plenty of formulae that incorporate both terms - nothing immediately comes to mind though. SteveBaker (talk) 19:12, 19 January 2013 (UTC)[reply]

I'm not doubting Steve's derivative answer above, except to comment that "yank" is not a precisely defined term. In its original (Scottish or mid-nineteenth-century American) meaning, I think it included a connotation of some displacement, and so would involve some work being done. I can't think of any formulae either. You would need to specify some parameters of a "yank" before you could deduce anything at all about the work done by it. A derivative with respect to time has no simple connection with an integral (of force) with respect to displacement, except that it's the same force. The effects of the force depend on how it is applied. A large yank on an immovable object will do no work, but a small yank over a large distance might do much work. Dbfirs 17:30, 20 January 2013 (UTC)[reply]

The definition of "yank" that I use is mentioned in our article "jerk (physics)"...sorry, no reference though. SteveBaker (talk) 15:20, 21 January 2013 (UTC)[reply]

Yes, the "mass times jerk" sense is also mentioned in Wiktionary, but it is a modern sense not mentioned in older dictionaries. Dbfirs 23:04, 21 January 2013 (UTC)[reply]

One definition from the jerk (physics) article: third derivative of position suggests that work is performed, since there is force × distance component (∫F•dx). ~:74.60.29.141 (talk) 00:03, 22 January 2013 (UTC)[reply]

Yeah - that's the definition of "Jerk" - it's not the same as "Yank". SteveBaker (talk) 21:09, 22 January 2013 (UTC)[reply]

The two modern definitions of yank: ("mass times rate of change of acceleration", and "rate of change of force") are equivalent by Newton's second law (assuming the force is in the direction of the jerk which seems to be implicit in "yank"). In "theory", neither necessitates an actual displacement in the rate of change (e.g. if applied to an immovable object), but in the real world at least a small displacement (and therefore at least a small amount of work) must be involved. Dbfirs 07:47, 23 January 2013 (UTC)[reply]

How images of galaxies are taken?[edit]

I want to know that how astronomers take images of galaxies ? For example, how do images of Milky Way are taken although we are inside it ? It is just like taking the image of a house by sitting in a room inside that house.Parimal Kumar Singh (talk) 19:09, 19 January 2013 (UTC)[reply]

Photographs of other galaxies are taken with cameras (or comparable sensors) attached to telescopes. If you see a "photo" of the Milky Way galaxy that looks complete, like this, it's either a photo of some other galaxy (that hopefully is similar to the Milky Way galaxy) or it's a simulation. -- Finlay McWalterჷTalk 19:16, 19 January 2013 (UTC)[reply]

Or more accurately, it is like making a world map without a satellite: [4] --140.180.253.61 (talk) 20:40, 19 January 2013 (UTC)[reply]

That's a great example. It wasn't until 1972 that we had an actual photograph of the earth as opposed to smaller photos stitched together. --Guy Macon (talk) 20:56, 19 January 2013 (UTC)[reply]

That is not what our article claims ("The Blue Marble was not the first clear image taken") Rmhermen (talk) 23:40, 19 January 2013 (UTC)[reply]

This photo from 1967 is described as the first one. File:ATSIII_10NOV67_153107.jpg RudolfRed (talk) 00:50, 20 January 2013 (UTC)[reply]

The answer for Parimal Kumar Singh question is simple. You're right that we are inside the Milky Way but we can see things from all directions from Earth so basically we can see all the parts of the Milky Way not the Milky Way as the whole. With our technology right now, we can easily match them up (just like putting the puzzle together) to create an accurate image of what Milky Way would look like as a whole. By the way, your analogy of taking a picture inside the house can't comparable to taking a picture of the Milk Way because the house has walls to block your view unlike the Milky Way.184.97.244.130 (talk) 05:43, 20 January 2013 (UTC)[reply]

Actually, we cannot see all parts of the Milky Way - a large part is obscured by interstellar dust clouds, and the galactic center. We only fairly recently found out that we most likely live in barred spiral galaxy, not a plain spiral. --Stephan Schulz (talk) 11:23, 20 January 2013 (UTC)[reply]

if light is just electromagnetic spectrum, why don't we have electric web cams with equal precision?[edit]

What is the exact reason that I can't replace a web cam with an electric version (that works in the electric part of the spectrum) and replace a light source with a source of electrons. Do electrons not bounce off people etc the same as light does?

Basically, other than color (due to the research into it), what's so special about light versus electricity when it comes to cheap 'imaging'? 178.48.114.143 (talk) 21:51, 19 January 2013 (UTC)[reply]

Electrons will interact with the air, not travel through it, which is why tubes like TV tubes are evacuated. If you up the energy enough to get through the air, you make a particle beam weapon. -- Finlay McWalterჷTalk 21:56, 19 January 2013 (UTC)[reply]

Fair enough. And if we imagine a room without air, then can a web cam work over the electric part of the spectrum just as well as light, as long as we 'illuminate' the room with a source of electrons? Or, in addition to not traveling through air, do they also not bounce off of objects nicely? 178.48.114.143 (talk) 22:23, 19 January 2013 (UTC)[reply]

You mean one of these, only larger? No reason why that wouldn't work. A bit like killing a fly with a bulldozer, but it should work. --Guy Macon (talk) 22:29, 19 January 2013 (UTC)[reply]

For the ordinary things you'll find in a room, the electrons will be absorbed by the surfaces of many of them, yielding no picture. When you see a picture of an object like a fly imaged with an electron microscope, it has probably been plated with a thin layer of gold so that it reflects the incoming electrons - see Electron microscope#Sample preparation. If you did that, you'd still only get a monochrome image. -- Finlay McWalterჷTalk 22:37, 19 January 2013 (UTC)[reply]

Do you mean I would get a monochrome image because I've just plated everything with gold? (after sucking all the air out). But, yeah, question answered. 178.48.114.143 (talk) 23:10, 19 January 2013 (UTC)[reply]

I think one thing should be clarified first. Light is an electromagnetic wave, or, alternatively, a stream of photons - see Wave-particle duality. Photons are uncharged particles with zero rest mass, and hence travel at the speed of light. They typically only interact with matter by being absorbed or emitted whole. Electrons, on the other hand, are charged particles with a rest mass of roughly 1/1800s u. As a result, electrons can travel at any speed, are influenced by electric and magnetic fields, and can interact with other charged particles in ways photons cannot. I'm not aware of any material that is transparent to electrons in the same way glass is to visible light. While electron beams are also called beta rays, they are not part of the electromagnetic spectrum, and have quite different properties from electromagnetic waves. In particular, beta rays are ionizing radiation, and hence quite unhealthy. --Stephan Schulz (talk) 23:27, 19 January 2013 (UTC)[reply]

That's interesting. Could you explain where radio itself (like am/fm, wifi, bluetooth, etc) falls here? Is it like light, but goes through stuff? Is it (as the name implies) part of the electromagnetic spectrum? If so, why doesn't light go through stuff the same way? I probably had more this in mind than electrons, I didn't realize that electrons are so different from radio waves - I thought they were similar, thanks for explaining. 178.48.114.143 (talk) 00:26, 20 January 2013 (UTC)[reply]

You need to read electromagnetic spectrum which puts all of the various types of light into relation with each other. Radio is the term for a range of wavelengths of light (colors, if you will) which are significantly lower in energy than light which our eyes are tuned to see. All light interacts with some matter, but different light interacts with different matter. For example, glass is generally transparent to most light in the visible range, but it is opaque to most wavelengths in the ultraviolet range. Also, radio doesn't just "go through stuff". Radio can be blocked if it lacks a straight path to reach the receiver (known as a line-of-sight path). Radio is reflected by parts of the Earth's atmosphere, so it can be reflected around smaller objects and hills and things, but many places exist in a "radio shadow", especially in mountainous areas where large mountains can block effective line-of-sight to radio sources. The article Radio propagation covers some of this. --Jayron32 00:43, 20 January 2013 (UTC)[reply]

I'm sure this is oversimplified (the solid state physics lecture I took for my minor is too many years in the past), but I'll try. As I wrote above, photons can only be absorbed or emitted whole - light is travelling in quanta. So in order to be absorbed, a photon needs to interact with something that can absorb exactly the amount of energy provided by the photon. In a single, isolated atom, electrons occupy shells (or orbitals), which correspond to very precise energy levels. Each shell can only be occupied by a given numer of electrons. A photon can only be absorbed by the atom, if either it knocks an electron completely off the atom, or if there is an electron in some shell, and an empty spot in a higher shell, and the difference in energy between the two shells corresponds to the energy of the photon. For most materials and visible light, the first case happens rarely, if at all, because it requires a lot of energy. For X-rays and gamma rays, this does happen, hence these are also ionizing radiation - knocking an electron off a neutral atom makes it into an ion. This is, BTW, the mechanism behind the Fraunhofer lines in spectroscopy - the black lines correspond to a valid state transition in some atom, so the energy can be absorbed. In a solid body, neighbouring atoms interact, and as a result, the simple energy levels get spread out into energy bands. But electrons still can only be in one of the bands, and the number of electrons in a band is limited. Bands are normally filled from lower energy levels to higher energy levels. In a conductor, the highest band with electrons is only partially filled, so electrons can absorb very small quanta of energy, and simply move a bit up in the band. That is why metallic conductors are usually completely opaque. In an isolator, the lowest filled level is (nearly) full, and the next higher level is (nearly) empty. Thus, for an electron to absorb an photon, the photon must provide enough energy to pass the band gap to the next higher band. If the band gap is large enough, visible light photons cannot provide this energy. Thus they cannot be absorbed, and hence the material is transparent. UV light has lower frequency and hence higher energy photons, so it is absorbed by material that is transparent for visible light, like e.g. window glass. There is a slight complication: Except at absolute zero, not all electrons are in their ground state - some are excited, and hence in a higher band than usually . Thus, they can absorb some low energy photons - no real material is completely transparent. But the number of excited electrons usually is low enough that this happens rarely. If on the other hand, the material is completely ionized, we have a plasma, in which free electrons and ions can interact directly with light. As a result, plasmas are opaque. At the time of the big bang, as the universe expanded, it cooled, and at a critical temperature, the Hydrogen and Helium in the universe turned from plasma to a neutral gas. Only at that time did the universe became (mostly) transparent. We can still see that moment in time - the phenomenon is usually called the surface of last scattering if we talk about how it came about, or the cosmic microwave background if we talk about what we observe today. --Stephan Schulz (talk) 10:10, 20 January 2013 (UTC)[reply]

There is no such thing as "the electric part of the spectrum". The whole spectrum, light included, is electromagnetic, but it does not involve electrons. It sounds as if you think the ordinary process we call electricity (i.e. the flow of electric currents) lives somewhere on the electromagnetic spectrum: it doesn't, it's an entirely different process. --ColinFine (talk) 21:34, 20 January 2013 (UTC)[reply]

Consider also the cost and difficulty of focussing, reflecting and such. I do a lot of work with infrared light. To focus a beam of visible light requires nothing more than a 50 cent chunk of glass. To focus a beam of infrared light (which is right next to the visible range), you need a highly exotic Nickel-Selenium lens costing $300! SteveBaker (talk) 21:07, 22 January 2013 (UTC)[reply]

Depends on what part of the infrared spectrum you're working with. I also work with infrared light at times, and my 50-cent chunk of glass works just fine. The difference is that I'm working in the near infrared, while I suspect you're working with mid- or long-wave infrared. --Carnildo (talk) 03:17, 23 January 2013 (UTC)[reply]