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Does an adult need Dihydrotestosterone? What will happen if he won't have it? — Preceding unsigned comment added by 87.68.222.18 (talk) 00:13, 12 February 2012 (UTC)[reply]

Does our article Dihydrotestosterone answer your questions? Comet Tuttle (talk) 00:21, 12 February 2012 (UTC)[reply]

Hi,

I would like to know, what is the distance of the hair matrix to the scalp, in an averge adult person. — Preceding unsigned comment added by 87.68.222.18 (talk) 00:15, 12 February 2012 (UTC)[reply]

Not an answer, but a note that our article Matrix (hair) is a tiny stub despite the enormous category template it's got. Comet Tuttle (talk) 00:36, 12 February 2012 (UTC)[reply]

Who first suggested the name "alkali metal" for the group 1 elements? What was the etymology of the name? Double sharp (talk) 03:24, 12 February 2012 (UTC)[reply]

The word alkali comes from the Arabic word meaning "ash", see [1]. Some of the earliest common compounds featuring alkali metals were things like soda ash and potash, which are sources of sodium and potassium respectively, and which provide the names for both of those items. Today, alkali means the same as "basic", in the "acids and bases" sense, and alkali metals will, after reacting with water, produce solutions of very high pH (bases) like sodium hydroxide and potassium hydroxide. I'm uncertain who specifically came up with the name; but a good guess would be Humphrey Davy, who is credited with being the first chemist to isolate many metals, including the alkali metals. --Jayron32 03:57, 12 February 2012 (UTC)[reply]

Why do apple seeds smell like almonds? Whoop whoop pull up ^{Bitching Betty | Averted crashes} 04:37, 12 February 2012 (UTC)[reply]

Because they contain hydrogen cyanide. --Jayron32 04:39, 12 February 2012 (UTC)[reply]

Then would not the smell of appleseeds contain a sufficient level of hydrogen cyanide to be lethal? Whoop whoop pull up ^{Bitching Betty | Averted crashes} 04:42, 12 February 2012 (UTC)[reply]

No, the ability of one's nose to detect trace amounts of HCN is significantly lower than toxic amounts. HCN is certainly very poisonous, but the amount one gets from appleseeds is trivial. It doesn't occur in the seeds directly, but rather it "outgasses" from them due to the slow decomposition of organic cyanide compounds. Lots of fruits seeds and pits, including almonds themselves, will release very small, trace amounts of HCN. You'll notice that people aren't dropping dead from eating cherries or apples, so empirically we know it mustn't be a problem. --Jayron32 04:51, 12 February 2012 (UTC)[reply]

People do not typically eat appleseeds. Whoop whoop pull up ^{Bitching Betty | Averted crashes} 04:56, 12 February 2012 (UTC)[reply]

No, but even if they did they are harmless. --Jayron32 04:58, 12 February 2012 (UTC)[reply]

You guys must have different apples. When I was at primary school, we were taught not to eat apple seeds, and I'm sure I've read that eating one or a few on rare occaisons won't hurt you, but routinely or often eating them will. The cyanide present in the seeds does not get into the apple flesh. Keit121.221.97.9 (talk) 05:38, 12 February 2012 (UTC)[reply]

Snopes says "the body can detoxify cyanide in small doses, and the number of apple seeds it takes to pack a lethal punch is therefore huge". The BBC gives the lethal dose as 30,000 seeds. So pip, pip, cheerio. Clarityfiend (talk) 09:36, 12 February 2012 (UTC)[reply]

By the way, the Snopes article gets a couple of things wrong. It says that cyanide kills by preventing the blood from carrying oxygen. Close but not quite — it does bind to hemoglobin, but the actual deadly effect comes from its binding to cytochrome oxidase a, inside the cells using the oxygen. And there is a recognized antidote, sodium thiosulfate if I remember correctly, although it doesn't always work. --Trovatore (talk) 10:37, 12 February 2012 (UTC)[reply]

Our article on cyanide poisoning has a pretty thorough overview of therapeutic options. TenOfAllTrades(talk) 16:20, 12 February 2012 (UTC)[reply]

I vaguely remember reading somewhere that "one man died from eating as little as a cup" of apple seeds. (Or was it an eighth of a cup? I think it must have been a cup, because an eighth of a cup is an ounce, and they'd have said that instead).

Anyway, I thought it was amazing that anyone would have the patience to gather together a cup of the things to eat them, but that's probably a lot fewer than 30,000.

Personally, when I eat an apple (which is rarely — not my favorite fruit at all), I usually peel and chew one or two of the seeds for the pleasant almond flavor. Hasn't killed me yet. --Trovatore (talk) 09:56, 12 February 2012 (UTC)[reply]

Cyanide poisoning#Chronic exposure states that "Exposure to lower levels of cyanide over a long period (e.g., after use of cassava roots as a primary food source in tropical Africa) results in increased blood cyanide levels, which can result in weakness and a variety of symptoms, including permanent paralysis, nervous lesions, hypothyroidism, and miscarriages. Other effects include mild liver and kidney damage." Presumably the same thing could happen if you regularly eat appleseeds. Whoop whoop pull up ^{Bitching Betty | Averted crashes} 20:18, 12 February 2012 (UTC)[reply]

A little pedantic, but an eighth of a cup is a fluid ounce, not an ounce. The mass will depend on what it is an eighth of a cup of. --Tango (talk) 18:00, 12 February 2012 (UTC)[reply]

"A pint's a pound / The world around." That sort-of works, if the liquid in question is primarily water. As regards cyanide, the article suggests the answer to a question I was about to ask - as to whether it's possible to "build up an immunity" to it. (As with the fictional "iocaine powder" in The Princess Bride.) Near as I can tell, the answer is no. ←Baseball Bugs ^{What's up, Doc?} carrots→ 21:59, 12 February 2012 (UTC)[reply]

Actually, the last time I was over there, a pint was closer to 90p. (Yeah, it's been a while.) --Trovatore (talk) 22:05, 12 February 2012 (UTC)[reply]

Travatore - yes a long while: Average cost of a pint of bitter hits £3.

Baseball Bugs - in the Imperial system, "A pint of pure water / Weighs a pound and a quarter". So if your "world around" only includes the USA, (a bit like the World Series, I suppose) then you're quite right. Alansplodge (talk) 00:04, 13 February 2012 (UTC)[reply]

I wonder if the phrase originated at a time when it was true the world around. Apparently the Imperial gallon is fairly recent. Anyone know when the phrase started? Wnt (talk) 09:35, 13 February 2012 (UTC)[reply]

See Comparison of the imperial and US customary measurement systems; "The American colonists adopted the English wine gallon of 231 cubic inches (3.78541178 litres), and used it for all fluid purposes. The English of that period used this wine gallon, but they also had the ale gallon of 282 cubic inches (4.62115205 litres). In 1824, the British adopted the British imperial gallon, defined as the volume of 10 pounds of water at a temperature of 62 °F, weighed in air with brass weights, by calculation equivalent to about 277.42 cubic inches (4.5461 L) - much closer to the ale gallon than the wine gallon." Measure in the UK before 1824 was a crazy hotch-potch of units, many specific to individual trades. There were also Scottish measurements, Irish and even Cornish measurements like the "Cornish apple gallon" of 7lb weight. Alansplodge (talk) 19:11, 14 February 2012 (UTC)[reply]

Hmmm, the wine gallon article says it was meant to be 8 troy pounds at some point, so indeed, a (wine) pint was a (troy) pound the world around. But somehow the wine pint was super-sized to an avoirdu... non-Troy pound, and the ale pint/Imperial pint became an oddity of 1.25 pounds. Wnt (talk) 14:58, 15 February 2012 (UTC)[reply]

I wonder - cyanide smells like almonds, but do almonds smell like cyanide? Is that the strongest smell most people detect from them, and if so, is it pleasant or unpleasant to them? I see that the essential oil of bitter almonds is "almost pure benzaldehyde" which has "a pleasant almond-like odor". For me almonds were a very peculiar food - until I was 16 or so, they had a truly awful smell, one of the most repulsive foods I knew, I think it was like some kind of petroleum product or something; but over just a few years I lost the ability to detect that entirely and started eating them as a pleasant food. (It's remarkably hard to remember a smell you can't smell; I remember that it was second only to cigarette butts in sticking to my fingers and not washing off if I so much as touched one) But I never sniffed any cyanide for comparison. ;) Wnt (talk) 09:46, 13 February 2012 (UTC)[reply]

And in Luminosity article, in the astronomy section. They says "In measuring star brightnesses, visible luminosity (not total luminosity at all wave lengths), apparent magnitude (visible brightness), and distance are interrelated parameters. If you know two, you can determine the third." What exactly the formula they are talking about? Is visible luminosity in this case they actually meant absolute magnitude? Thanks!Pendragon5 (talk) 06:16, 12 February 2012 (UTC)[reply]

Two versions of the formula are given at the end of the first subsection of the "In astronomy" section. The simplest is m_star = - 2.72 - 2.5 · log₁₀(L_star/dist_star²), where L_star is in terms of solar luminosities, and dist_star is in light-years. Since absolute magnitude is the apparent magnitude a star would have if it were 32.6 light-years (10 parsecs) away, it can be obtained from visible luminosity by plugging 32.6 light-years for dist_star in the formula. Doing this should simplify to M_star = 4.83 - 2.5 · log₁₀(L_star) where 4.83 is the absolute magnitude of the Sun. 98.248.42.252 (talk) 16:22, 12 February 2012 (UTC)[reply]

What does Novim do? Why do they say they are for science without advocacy? How is that different from awareness without action? Meow Will Always Love You (talk) 09:20, 12 February 2012 (UTC)[reply]

Why in the observation data of the this article, there are so many question mark? Many info about this star is not in the observation data. Is it because we don't know them yet or we haven't update it? So if possible can someone update the date please. It is maybe outdated.Pendragon5 (talk) 09:21, 12 February 2012 (UTC)[reply]

I'm a little confused as to how you think the question marks can be changed to values as the article says it's a triple star system, and the question marks refer to single stars only. I'm not familiar with the way Wikipedia handles triple star systems though. --TammyMoet (talk) 14:31, 12 February 2012 (UTC)[reply]

Actually it's binary star system. And i don't know if it's possible to have values of the ? marks.Pendragon5 (talk) 19:10, 12 February 2012 (UTC)[reply]

The article says "The T Tauri system consists of at least three stars" so if the article is wrong, correct it if you have sources to back your statement up. Maybe those sources will also give values for you. --TammyMoet (talk) 19:50, 12 February 2012 (UTC)[reply]

Ops, it's three lol. Anyway i don't have any sources that state the value so yea that's why i asked in the first place.Pendragon5 (talk) 00:29, 13 February 2012 (UTC)[reply]

Are there any major physical or behavioral differences between lab mice and wild mice? Would a lab mouse put into a litter of wild mice babies grow up as a wild mouse? Could a wild mouse be kept as a pet and learn to be handled the way lab mice can? Would a wild mouse run in a wheel? Boredtolife (talk) 15:31, 12 February 2012 (UTC)[reply]

As far as I know lab mice are not domesticated to the same extent that say cows and sheep are. So yes, a lab mouse let into the wild would become Feral. For a wild mouse (if raised from a pup) in captivity with lots of human handling, it would be susceptible to Imprinting – thereby appearing tame. A captive wild mouse in a small cage (raised from a pup -so it was not afraid to be observed), would still want the stimulation of running in a wheel if one was to be made available. In very large Vivariums, the attraction of wheels fades, as there is plenty of space to run around in. The Homo sapien show something similar to 'wheel behaviour' when forced to live in cities. They spend an awful lot of time hammering away at a keyboards, to get stimulation from something they refer to as the world wide net. This behaviour makes lab mice seem almost human.--Aspro (talk) 19:31, 12 February 2012 (UTC)[reply]

A closer comparison might be people who ride on a stationary bicycle or use a tread mill. StuRat (talk) 19:40, 12 February 2012 (UTC)[reply]

As for physical differences, lab mice are often bred to be white, presumably so they are easier to see, and to differentiate an escaped lab mouse from a wild mouse that happens to get into the lab. Different breeds of lab mice also are bred with specific genetic problems, to aid in studying humans with the same problem. For example, diabetic mice might be bred. StuRat (talk) 19:39, 12 February 2012 (UTC)[reply]

I'm not actually sure why albino rats became standardized, other than the fact that the famous Wistar rat, which was the original model rat organism produced for labs (e.g. a rat breed with reasonably standardized characteristics that was sold to experimenters) was albino, and that the scientist who produced them found them more appealing for reasons he doesn't seem to have elucidated. I have read in various places that the white rat was perceived as more appealing than darker rats, because the latter were associated with the "pests" one really is not very fond of. (This is probably why albino rats are common as pets, as well.) They were fairly and purposefully domesticated, so I wouldn't necessarily assume they are just wild rats that happen to be albinos. Some strains of lab rat are very docile compared to wild rats and are preferred for that reason. --Mr.98 (talk) 22:21, 12 February 2012 (UTC)[reply]

Generally, laboratory mice are purchased from disease-free stocks and are reared under disease-free conditions, except for any experimental reasons. I suppose a mouse let out into the wild would therefore be very susceptible to disease. Thincat (talk) 23:14, 12 February 2012 (UTC)[reply]

Some strains of lab mice have a reputation for being nastier than others, but in general wild mice are much more aggressive. [2] Twenty generations is usually enough to get a selective effect based on the underlying genetic variation of a population, and with mice that can pass in under two years. Additionally, the mice may have mutations of large effect which, though not viable in a wild population, might appeal to human owners. Bear in mind that the lab strains of mice started off from mouse fanciers back in the 1800s; they were pets before they were subjects. Wnt (talk) 09:23, 13 February 2012 (UTC)[reply]

These are different from regular fatty acids. I'm wondering if the same mechanism is used to metabolise trans fats. Do bacteria eventually metabolise it? Are alkenes and alkanes taken into the bloodstream? If so, how are they excreted? 216.197.66.61 (talk) 17:58, 12 February 2012 (UTC)[reply]

My guess is that alkanes probably pass through the body without anything interesting going on - though n-hexane has known toxic effects so something must be going on there. I'd also suspect that alkenes get converted to alcohols in the strong acid of the stomach via Hydration reactions. LukeSurl ^{t c} 19:06, 12 February 2012 (UTC)[reply]

If the stomach were substantially effective at simple aqueous-acid hydration of alkenes, I don't think unsaturated fats would have as noticeable a metabolic distinction from saturated fats (or more importantly, that cis vs trans unsaturated fatty acids would be different from each other at all). DMacks (talk) 20:23, 12 February 2012 (UTC)[reply]

An important metabolite appears to be 2,5-hexanedione and cyclic derivatives 2,5-dimethylfuran and gamma-valerolactone. Just how the body metabolise an alkane without any functional groups? [3][4] — Preceding unsigned comment added by Nothing gold can stay (talk • contribs) 00:40, 13 February 2012 (UTC)[reply]

Hmmm, it looks like alkanes and the like can be taken up quite nonspecifically via chylomicrons. (PMID 6330487) Once they get into the body, they are prone (like so many other compounds) to run into the cytochrome P450 system, (PMID 2009082) which reacts them with molecular oxygen. Oxygen, when wielded by the body's enzymes to detoxify chemicals or kill invading organisms, is so incredibly reactive that there are only a few things, like the carbon particles in India ink or the mineral particles in asbestosis and silicosis, which are able to resist it. But the P450 enzymes don't interact with everything, and they aren't always fully active, so sometimes a second compound (methyl iso-butyl ketone in the second reference) can greatly increase metabolite formation. Weird thing about P450s is, as in this case, they can cause more trouble than they cure. But I suppose that on average stuff has to be water soluble in order for the body to wash it out effectively. Wnt (talk) 09:10, 13 February 2012 (UTC)[reply]

${\displaystyle {\frac {L}{L_{\odot }}}\sim {\left({\frac {M}{M_{\odot }}}\right)}^{3.9}}$ What is the ~ mean in this case? This is a formula taken in luminosity article.Pendragon5 (talk) 19:09, 12 February 2012 (UTC)[reply]

Based on the linked Mass–luminosity relation article about that equation as well as the general idea of it being a formula, I'd say Tilde#Mathematics: proportional or approximately equal. DMacks (talk) 20:19, 12 February 2012 (UTC)[reply]

In this case, it is probably intended to mean "approximately equal to". The idea of dividing by the luminosity and mass of the sun is to cancel out any constants of proportionality. For this formula, that doesn't quite work because the relationship isn't exactly proportional so if the star is sufficiently more or less massive than the sun, the "constants of proportionality" (which aren't really constant) won't exactly cancel out. That is why the relationships for stars less than 0.43 solar masses or more than 2 solar masses in the Mass-luminosity relation do have constants in them. An exponent of 3.9 is about right for stars near the mass of the sun, so there is no need for a constant out the front. --Tango (talk) 21:00, 12 February 2012 (UTC)[reply]

In fact, I'm sufficiently confident that it should be "approximately equal to" that I have changed the symbol to ≈. The article is unreferenced, so it's difficult to know why ~ was used in the first place. --Tango (talk) 21:05, 12 February 2012 (UTC)[reply]

Mathematical_symbol#Symbols says ~ is used for an approximation that is poor and I suspect in this case it may well be poor! Thincat (talk) 23:20, 12 February 2012 (UTC)[reply]

In astronomy, we use typically use "~" to indicate that the functional dependence is approximate, so it means "is approximately proportional to" (I'd read it as "goes as" or "varies as"). The normalizations are superfluous, the formula could be written as ${\displaystyle L\sim M^{3.9}}$. "≈" is used for approximate numeric equality. All this is from my personal experience, I'm not aware of these symbols being formally defined anywhere. --Wrongfilter (talk) 08:13, 13 February 2012 (UTC)[reply]

If that was the intended meaning of ~ then yes, the normalisations are superfluous. Once it's normalised, it ceases to be proportional and becomes equal (or, in this case, approximately equal. That's why I changed the symbol. --Tango (talk) 00:59, 15 February 2012 (UTC)[reply]

Let use Proxima Centauri as an example. In the data, section "observation data" and "characteristics". Each of them has an apparent magnitude. What is the difference between them? What is the actually apparent magnitude of that star?Pendragon5 (talk) 19:45, 12 February 2012 (UTC)[reply]

If I understand correctly (and it's not linked, which is surprising) - one is apparent magnitude(V), the other apparent magnitude(J). They're the magnitudes when a range of the light is considered - photometric system gives the bands that correspond with the different letters. -- Finlay McWalterჷTalk 20:10, 12 February 2012 (UTC)[reply]

Which apparent is more universal in astronomy? I meant like people use that apparent magnitude more in formulas that calculate distance to star, absolute magnitude...Pendragon5 (talk) 00:26, 13 February 2012 (UTC)[reply]

It always has to be stated over which wavelength range (which "filter") an apparent magnitude is measured, otherwise the value of the apparent magnitude is useless. Popular astronomy books may skip those details, then they typically mean visual magnitudes. An absolute magnitude derived from an apparent magnitude taking only distance into account will be in the same filter. There is also the concept of bolometric magnitude, which is the magnitude taken over the entire wavelength range, but that is not directly measurable. --Wrongfilter (talk) 08:20, 13 February 2012 (UTC)[reply]

So visual magnitude is magnitude (v) right?Pendragon5 (talk) 19:16, 13 February 2012 (UTC)[reply]

I have a wound cylindrical inductor of 500 turns with negligible resistance. The coil is 200mm long and 100mm dia. One end is connected to ground. The other end is fed from an ideal current source. How long will it take a change of current at the input to travel to the grounded end. This is for a project- not for homework.--92.28.91.245 (talk) 22:00, 12 February 2012 (UTC)[reply]

It's going to be almost instantaneous. Unless you are doing something incredibly precise, you can just assume it is instantaneous. --Tango (talk) 22:22, 12 February 2012 (UTC)[reply]

See Speed of electricity. If the coil were to be unwound then the flow would take about 0.66 nanoseconds to travel its length (i.e. be measurable at the "ground" end assuming that the current source is at the opposite end). Since however the conductor is coiled then the steady state current would take longer than that. As shown in the inductor article there are several variables that need to be considered. Are you asking about when would the first electron flow show up at the grounded (output) end or when steady flow happens? hydnjo (talk) 00:30, 13 February 2012 (UTC)[reply]

This is a more complex question than you may think. You have not given the wire diameter, but any real coil must have wire of a certain diameter, and that wire must have resistance. Also, in practice the coil may have nearby metalwork. All these things add distributed capacitance to the coil turns. This capacitance and the resistance mean that the inductor is a transmission line (like coax cable), with a velocity of propagation determined by transmission line formula. This velocity will be well below the Sped-Of-Light / speed of electricity value. However you will need to carefully select any instrumentation as the rate of change in current at the grounded end will be swamped out by the rate of change imposed by the total self inductance. Keit124.178.131.79 (talk) 01:26, 13 February 2012 (UTC)[reply]

I see, an L c circuit. hydnjo (talk) 02:04, 13 February 2012 (UTC)[reply]

Not exactly. With the diminsions the OP specified, it would behave as an LC circuit as far as the continuous (sine or whatever) voltage and current measured at the "hot" end of the coil is concerned, but the OP asked about the time to propagate a change from hot end to the grounded end - this is quite a different thing. To evalaute this correctly, you need to model it as a transmission line - an inductance with capacitance distributed along it, not just one lumped capacitance. And, in practice, distributed resistance as well. A single lumped LC (or LCR) circuit just shifts the phase, with phase shift varying with frequency. A distributed LC (or distributed LCR) circuit introduces a delay substantually independent of frequency, as well as the phase shift. Keit121.215.135.56 (talk) 03:05, 13 February 2012 (UTC)[reply]

The assumption of "negligible" resistance implies "negligible" propagation time. The equation for an R-L circuit's step response will yield an instantaneous response for R=0. For a small but finite R, we can calculate the rise-time. Nimur (talk) 01:32, 13 February 2012 (UTC)[reply]

Negligible resistance does not imply negligible propagation time (think superconductivity). hydnjo (talk) 01:38, 13 February 2012 (UTC)[reply]

No, it certainly does not. Putting R=0 into the time constant formula gives a zero time value, but this is misleading. A circuit containg L without R will start with zero current, with the current increasing constantly with time. Keit124.178.131.79 (talk) 01:43, 13 February 2012 (UTC)[reply]

Please note I said fed by a current source.--92.29.198.2 (talk) 02:51, 13 February 2012 (UTC)[reply]

Yes - you did. That makes correctly taking into account the distributed capacitance even more important. Keit121.215.135.56 (talk) 03:10, 13 February 2012 (UTC)[reply]

I find it ironic that we are disecting Lc circuits ;-) hydnjo (talk) 04:08, 13 February 2012 (UTC)[reply]

Can't have one without the other. Keit121.215.142.221 (talk) 06:03, 13 February 2012 (UTC)[reply]

The answer seems to have emerged. It seems obvious to me now that the equivalent circuit is a distributed inductance with distributed capacitance to ground.(ie a transmission line) Keit was right in that there must be a ground return to which the inductor will have some capacitance. Indeed my project necessitates the inductor being in close proximity to a ground plane. So the velocity will be 1/sqrt(LC) regardless of whether a voltage or current step is applied at the input end. thank you all esp Keit. Also the equ cct can now have resistance and shunt conductance (the other 2 properies of a TL) BTW whos this joker hydnjo and what does he know beyond what he read in the inductor article?? Its not a resonant circuit--92.29.194.107 (talk) 08:52, 13 February 2012 (UTC)[reply]

The current entering the first turn will also induce current in the last turn. So a change at the bottom end will take about the time for an EM wave to travel that 200 mm which is about 0.2/300000000 seconds or 0.6666 nanosecond. If you want to make a delay line a zig zag may be a better shape. If it was really unwound the distance would be 500xpi×0.1 or time of 523 nanoseconds. Graeme Bartlett (talk) 10:08, 13 February 2012 (UTC)[reply]

Not true. The mutual inductance between the first and the last turn is very small,as is the self inductance of one turn. Immediate voltage rise across the last turn is at first pretty much prevented while the capacitance to ground is charged. It has to be that way as otherwise the transmision line formula would be invalid. The transmission line formula appilies to ANY system of distrubuted capacitance and distributed inductance, regardless of configuration, though admittedly you could create a situation where the mutual inducatnce between the fisrt and last trurns is significant - That's not a straight coil that the OP is asking about. And a zig zag will give less delay because the inductance will be less. Keit121.221.99.129 (talk) 12:16, 13 February 2012 (UTC)[reply]