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April 15

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Congenital absence of appendix

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The article on the appendix says that congenital absence of the appendix does not affect anything, but this case study published online shows differently. I'm looking for a second opinion. http://74.125.113.132/search?q=cache:yvGgkxsdJgEJ:www.bhj.org/journal/2008_5002_april/download/page-293-294.pdf+congenital+absence+of+appendix&cd=1&hl=en&ct=clnk&gl=ca —Preceding unsigned comment added by 99.227.94.24 (talk) 02:03, 15 April 2009 (UTC)[reply]

It's hard to make much of a single case study, described in somewhat broken English in the Bombay Hospital Journal. Looie496 (talk) 03:04, 15 April 2009 (UTC)[reply]
The authors of the paper have not explicitly stated the reason for the patient's symptoms (i.e. the diagnosis). One could debate whether this patient has "Congenital absence of appendix" or "Rudimentary appendix". If the latter, perhaps the diagnosis is indeed appendicitis of the rudimentary appendix? The authors do not describe their further management of the patient. Did they remove the rudimentary appendix? Did the patient improve?
The main problem here is with the definitions of "congenital absence of the appendix" and "rudimentary appendix". If they were defined more clearly, we could give a more helpful answer. Axl ¤ [Talk] 08:28, 15 April 2009 (UTC)[reply]

What is a naturally occurring isotope?

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How are Antimony-121 and Antimony-123 two naturally occuring isotopes? Also, why is Antimony-121 more abundant than Antimony-123? —Preceding unsigned comment added by 174.6.144.211 (talk) 02:41, 15 April 2009 (UTC)[reply]

You might have a look at Isotope#Occurrence_in_nature and then let us know if something specific is unclear. --Scray (talk) 03:14, 15 April 2009 (UTC)[reply]
(ec)You can start by looking at Isotope#Occurrence in nature and Nucleosynthesis. In general, the various processes that transform the elements yield nuclei with various combinations of protons and neutrons. Only certain combinations are stable. Others will almost immediately decay into other elements. Still others will be relatively stable and may take years millennia, or longer to decay. An element is defined by the number of protons contained in the nucleus, and a given element may only be stable or have a measurable life with a certain number of neutrons. For example, Carbon (see also Isotopes of carbon) have three isotopes with a half life of more than a few seconds, but those three isotopes – with six protons, and six, seven, or eight neutrons – have very long lives (in human terms). Those are the isotopes general identified as naturally occurring. -- Tcncv (talk) 03:26, 15 April 2009 (UTC)[reply]
(Sorry, I guess I got off track) To generalize, there is no reason why an element cannot have more than one stable isotope. hydrogen has two, oxygen has three, and as you found antimony has two. As to why they have different abundances, I expect that is due to the different isotopes being formed through different mechanisms, or if formed through the same method, that method (such a supernovas) just happens to produce more of one isotope than another. -- Tcncv (talk) 03:56, 15 April 2009 (UTC)[reply]
Does the varying longevity of isotopes of an element have to do with the curve of binding energy? Edison (talk) 04:41, 15 April 2009 (UTC)[reply]
It does, to some extent. Loosely, radioactive decay can be thought of as a Quantum tunelling phenomena and the tunneling probability is dependent on the released energy. However, the transition rate is more heavily influenced by various selection rules governing the decay mode. For example, in beta decay there are selection rules which arise out of conservation of angular momentum and parity. The influence of these selection rules can be huge - it is energetically possible for Calcium-48 to undergo single beta decay but it has never been observed to decay via. single beta decay (it has been observed to undergo double beta decay though).
Going back to the original question, it is possible to have two stable (or effectively stable) isotopes of an element as long as there is no significant decay mode to another element or isotope. A nuclear decay chain can be thought of as a waterfall, where nuclei keep "falling" into lower states until there's nothing to fall into. It just so happens that both Antimony-121 and Antimony-123 are at the bottom. Someone42 (talk) 09:09, 15 April 2009 (UTC)[reply]

Precambrian craton formation

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Hi, I've been studying a little about the geological history of the Earth preceding the ordovician period, and I've come across some information on the cratons that are known to have existed, as well as an animation of how they were believed to have been positioned at the point when the crust first formed into plates. There is an animation on Wikipedia here (warning, may enduce seizures), but I find the morphing process is much easier to see in this video clip (youtube, starts at about 2:15), though I can only assume they are working from similar data.

I'm specifically concerned with the peculiar string of "islands" linking the Siberia craton and the Baltica craton. By the Carboniferous period they seem to have all been gobbled up by the larger cratons as pangea is formed. I assume that the reason they have been separated from Siberia and Baltica in the first place is because of some difference in the mineral/fossil record that indicates separation before Carboniferous, but I'm curious about the positioning of these mini-cratons, and the size. Are they being placed in a string simply because it is impossible to tell exactly where they were located, or is there some reason to believe that they were arranged in such a manner? And if the land fossil record for that period is as sparse as I think it is, how can they rule out that these mini-cratons were not in fact much larger, or merged together into a larger mass?

I don't know if anybody can help with these questions, but opinions are welcome! Thanks ahead! 124.154.253.25 (talk) 03:35, 15 April 2009 (UTC)[reply]

It sounds like you are talking about a volcanic arc. I don't know why geologist believe that there was an arc link the cratons you are talking about. Volcanic arcs, like the one at the caribean or aleutian islands for instance, are actually quite common and they do indeed align along a string of islands as you described. Dauto (talk) 04:36, 15 April 2009 (UTC)[reply]

Yes, it does look very like a volcanic arc, and if it is, the real question is what kind of evidence led scientists to their conclusion? I thought perhaps that they may have evidence for a certain number of small cratons separate from Siberia and Baltica, likely created by a similar event (such as a volcanic hot spot under a floating plate) assuming they are geologically similar. Fossil evidence could then further show that they were separated by water, or alternately the cratons could have been discovered very far from each other. I'm not educated enough in any field to make wild guesses like that though! 124.154.253.25 (talk) 06:35, 15 April 2009 (UTC)[reply]
The evidence nowadays would be a line of acid volcanics such as dacite or andesite from that period, possibly embedded in marine sediments. There is likely to be folding in marine (or back arc) sediments, as the volcanic arc was fused onto a craton. In this particular case I don't know the specific evidence. Graeme Bartlett (talk) 06:46, 15 April 2009 (UTC)[reply]
This paper by Robin Cocks and Trond Torsvik [1] includes a fairly detailed discussion of all the 'bit and pieces' or terranes caught up between Baltica and Siberia. I'm assuming that you're mainly talking about 'Kazakhstania'. I think there is some faunal evidence, i.e. an analysis of high v. low latitude types of fossil and possibly palaeomagnetic data allowing the construction of an apparent polar wander path for at least some of the terranes. Remember that geoscientists have difficulty reconstructing continents where there are no 'magnetic stripes' preserved to help close up intervening oceans (that's pretty much anything before the middle Jurassic or about 200 million years ago); several distinct reconstructions may be possible from the same data. Mikenorton (talk) 10:22, 16 April 2009 (UTC)[reply]
Thanks for the link! Really interesting and there's loads there to delve into. 124.154.253.25 (talk) 02:00, 21 April 2009 (UTC)[reply]

Pharmaceutical naming

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How do drugs get their common or generic names? I'm thinking here of lorazepam, minoxidil, sildenafil, etc. They're somewhat random syllables, although I'm pretty sure that anything ending in -mab is a monoclonal antibody, so there's some correlation sometimes. Is there a convening body to dispense these names, is there some centralized random generator everyone can go to, or what?

Related question, when the actual corporations do their branding, are they wide open, or do they obey some rules also? For the examples above, how do we arrive at Ativan/Temesta; Rogaine/Regaine; Viagra/Revatio?

Do we already have an article on this? I tried some possibly lame searches, but I'm stumped. Thanks! Franamax (talk) 07:24, 15 April 2009 (UTC)[reply]

Medicines will often have more than one name:
  • a generic name, which is the active ingredient of the medicine
  • a brand name, which is the trade name the manufacturer gives to the medicine.
The generic name is the official medical name for the active ingredient of the medicine.
The brand name is chosen by the manufacturer, usually on the basis that it can be recognised, pronounced and remembered by health professionals and members of the public. An example would be Viagra - this is the well-known brand name given by Pfizer to the generic medicine sildenafil. (Brand names are capitalised; generic names are not.) More here. Cuddlyable3 (talk) 09:56, 15 April 2009 (UTC)[reply]
We've discussed these issues before on the ref desk, e.g. generic names and trade names. These are just 2 of the first few hits I got after I searched the RefDesk archive (above) for "drug names". BTW, I was told 5 or 6 years ago that focus group research had indicated that "v", "x", and "z" were "strong" letters to include, and since then we've seen "Vioxx", "Zyvox", and the list goes on... --Scray (talk) 11:28, 15 April 2009 (UTC)[reply]
Franamax, your own screen nama also kind of sound a little bit like a brand name. Dauto (talk) 12:24, 15 April 2009 (UTC)[reply]
Hey, you noticed! :) The marketing committee was up all night on that one. It's a combo of my given name and "max", connoting large or major. If the Pope can be pontifex maximus I figured I could follow the same strategy. Sadly, I am not yet considered infallible. ;) Franamax (talk) 18:31, 15 April 2009 (UTC)[reply]
Heh, I've occasionally thought of using the handle Antonissimo. —Tamfang (talk) 19:06, 15 April 2009 (UTC)[reply]
Ooohh, an absolute superlative in singular masculine form - that's good! The other strategy is to go for self-deprecation. When I first saw User:Mild Bill Hiccup, I laughed for quite a while. (For those not familar with American culture, the contrast is here). Franamax (talk) 20:03, 15 April 2009 (UTC)[reply]

Thanks for the pointers! The World Health Organization Expert Panel on the International Pharmacopoeia and Pharmaceutical Preparations handles the generic names - why didn't I think of that. Boy, the United Nations has an article on everything! Franamax (talk) 18:31, 15 April 2009 (UTC)[reply]

resistivity of vacuum

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what is the resistivity of vacuum? is it zero? --harish (talk) 11:20, 15 April 2009 (UTC)[reply]

No, it's infinite. Since ρ is defined as electric field divided by current density, and there is no current in a vacuum (since there are no charge carriers), ρ is infinite. Note that this is only hypothetical, since nobody has ever created or discovered a perfect vacuum. --Heron (talk) 13:07, 15 April 2009 (UTC)[reply]
Are you sure? If the electric field was strong enough I would expect electrons to be ripped off the negatively charged object and be attracted to the positively charged object, so a current would flow. (Although, I guess technically it wouldn't be a vacuum then.) --Tango (talk) 14:46, 15 April 2009 (UTC)[reply]
How about electron flow through a CRT? Glowworm. —Preceding unsigned comment added by 98.17.38.45 (talk) 14:19, 15 April 2009 (UTC)[reply]
A CRT has electrons flowing through it, and is therefore not a perfect vacuum. — DanielLC 16:08, 15 April 2009 (UTC)[reply]

A "perfect vacuum" is merely" a philosophical construct that is never observed in practice" per the article. Outer space is 1×10-6 to <3×10-17 torr, compared to zero torr for a perfect vacuum. Intergalactic space could have as little as 1 molecule per cubic meter. But a heated filament in that region would emit electron toward a positively charged plate and the current would be as great as in a common vacuum tube(valve) or cathode ray tube, since these devices do not depend on residual gases to carry the current. That current, known as the "Edison effect" or thermionic emission, can be characterized by a curve relating current and voltage. It does not have a constant resistance, and the space does not have a constant restivity. It does not follow Ohm's Law any more than a semiconductor does. The tubes are evacuated to the greatest extent possible. Do electrons travelling through a region of otherwise pure vacuum count as "particles" and how much pressure would each electron in, say, one cubic centimeter create? Edison (talk) 16:08, 15 April 2009 (UTC)[reply]

Electrons have mass, so when an electron strikes a surface it exerts a force – a pressure. The amount of force depends on the speed of the electron. In the case of a radio vacuum tube the situation is special because the electrons are not moving at random. They are all moving in one direction – toward the anode. – GlowWorm. —Preceding unsigned comment added by 98.17.38.45 (talk) 22:01, 15 April 2009 (UTC)[reply]
A perfect vacuum is not merely philosophical, since I can define some arbitrarily small volume of space which contains no particles at the time I measure it. But at the next instant, those damn physicists insist it will get filled up again - nature abhors a vacuum and Casimir keeps putting in new particles.
Nevertheless, if the definition of a perfect vacuum is zero particles and you send a particle (or wave that may also be a particle, whatever) into that space - it is definitely not a perfect vacuum anymore!
And "pressure" is a statistical concept, so I'm not sure how valid it is to talk about a single particle exerting a "pressure" - wouldn't we instead term that an "impact force"? Franamax (talk) 22:25, 15 April 2009 (UTC)[reply]
To mend the gaping flaw in my reasoning, for "particle" substitute "particles possessing mass". EM waves pervade the universe and are indeterminate in location. Erk! Franamax (talk) 22:41, 15 April 2009 (UTC)[reply]
I spoke of a single electron, but I had in mind the statistical effect of many electrons. – GlowWorm.
While it's true that it is not possible to create a perfect vacuum, that does not make a merely philosophical construct. It is a physical idealisation. There is nothing wrong with talking about the properties of physical idealisations. They are very usefull for thought experiments, for instance.
The OP asks about the vacuum resistivity. Paradoxically, the answer to this question can be both zero and 'infinity' depending on how we chose to understand the question. If you shoot electrons creating a streem, these electrons would continue to move unoposed even in the absence of an external field, zero resistivity! I seriously suspect that this was the setup that the op had in mind. But that's not what people usually understand by measument of resistivity. The normal understanding is to apply an external field and see what happens. If you do that, you get no current because there are no carriers, infinite resistivity! —Preceding unsigned comment added by Dauto (talkcontribs) 03:45, 16 April 2009 (UTC)[reply]
It's zero. The more fundamental understanding of resistance is not the ratio of potential difference to current (for one thing, that works only for DC), but rather a measure of the energy dissipated when a given current flows through the medium (power dissipation is I2R, where I is the current). Since vacuum dissipates no energy, it has no resistance. --Trovatore (talk) 04:01, 16 April 2009 (UTC)[reply]
I disagree with the "zero resistance" claim. A superconductor will carry a certain amount of current with no voltage across it and no energy dissipate, so zero resistance applies. A perfect vacuum (no electrons or ions) can have substantial voltage across it with no current. Don't tell me a vacuum tube diode with the filament cold has zero resistance when I apply 12 volts across the cathode to anode gap and no current flows. Edison (talk) 19:05, 16 April 2009 (UTC)[reply]
I stand by what I said. The vacuum has no resistance at all. That does not, of course, make it a superconductor, or even a conductor; it means only that no energy is lost to scattering when charge carriers flow through it. --Trovatore (talk) 18:36, 17 April 2009 (UTC)[reply]


Regarding your difficulty in making the current actually flow from the cold filament, well, the fact that it depends on the temperature of the filament should give you a clue that this is not a property of the vacuum. If the issue were related to the vacuum, then heating the filament wouldn't change it.
Rather, the discrepancy between potential difference and current is properly addebited to the non-ohmic contact between the metal and the vacuum. --Trovatore (talk) 19:48, 17 April 2009 (UTC)[reply]


I don't think the resistivity will be infinite even in the absence of carriers, because beyond a certain field strength the vacuum itself will become a carrier: electron-positron pairs will be created out of the void and split apart, going in opposite directions and producing a net current. This phenomenon is a consequence of quantum electrodynamics. Looie496 (talk) 05:11, 16 April 2009 (UTC)[reply]
According to the indentation, you're responding to me, but I didn't say the resistivity was infinite. I said it was zero. --Trovatore (talk) 05:47, 16 April 2009 (UTC)[reply]
Looie, that which you talk about is called the dielectric strength. It's not the samething as resistivity. You are right. Vacuum's dielectric stregth is not infinite. A simple back of envelop calculation (Or better yet, a dimentional analysis) gives the vacuum dielectric strength which is 9 to 10 orders of magnitude higher than the dielectric strength of any material. Dauto (talk) 17:57, 16 April 2009 (UTC)[reply]
There is a voltage drop across a resistor if another resistor is in series with it to form a voltage divider. (Without the second resistor, the full voltage of the power supply will appear across the resistor, and this would not be called a voltage drop.) A voltmeter will show a voltage drop between the filament and plate of a radio vacuum tube. This indicates that the vacuum in a vacuum tube is a resistor, even though it dissipates no power.(There is a resistor in the plate circuit to form a voltage divider.) – GlowWorm. —Preceding unsigned comment added by 98.17.41.25 (talk) 10:35, 16 April 2009 (UTC)[reply]

If you have two electrodes in a perfect vacuum with some constant potential difference then there will be a nonzero tunnel current. Count Iblis (talk) 13:25, 16 April 2009 (UTC)[reply]

formation of pure alkyl nitrile

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Hi, I need to find a way to synthesise the following product: it's a n-hexane with a CN group attached to the 2-carbon. My "base" reactant has to start with a hydrocarbon, though presumably I can employ a whole host of special reagents after that. I can't find a suitable process? I must obtain a single pure product, no isonitrile stuff (unless you can tell me an easy way to purify it without jumping through hoops). I don't think Kolbe nitrile synthesis will give me that, and all the other processes I've looked up either seem to apply to aromatic compounds or they leave oxygen groups on the carbon chain. John Riemann Soong (talk) 11:28, 15 April 2009 (UTC)[reply]

Working forwards, what reactions do you know that convert an alkane (is it correct to assume "hydrocarbon" means alkane?) into something-else (i.e., what kinds of functional groups can you make in Step 1)? Working backwards, what are ways you can make a nitrile (what would be a precursor) and what are ways you can add a "prebuilt" nitrile or nitrile-containing structure onto some other structure (i.e., what could be the precursor for the final step)? Do you have to start with a 6-carbon hydrocarbon? DMacks (talk) 04:23, 16 April 2009 (UTC)[reply]

Invasive species

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(Not homework.) Dog (something)bine. Southern Ontario. Grows in underbrush of dead pine forest. A short human-height plant, wood-like or stalk-like (don't remember which). Nearby red pine dying from fungual infection. Any ideas? Thanks. ~AH1(TCU) 11:33, 15 April 2009 (UTC)[reply]

Are Woodbine#Climbing_plants or "dogwood vine" or "creeping dogwood" helpful? --Sean 16:43, 15 April 2009 (UTC)[reply]

Do scientists have to pay to have their work published in a peer-reviewed journal?

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I was wondering if scientists have to pay to have a paper published in a peer-reviewed journal? The Open Chemical Physics Journal says that they "aim at providing the most complete and reliable source of information on current developments in the field" but charge anywhere from $450 to $900 to have a paper published. This sounds like a vanity press. The fees are listed towards the bottom of the page. [2] A Quest For Knowledge (talk) 12:18, 15 April 2009 (UTC)[reply]

No they don't and that's a vanity journal - seems to popular with 9/11 cranks. --Cameron Scott (talk) 12:25, 15 April 2009 (UTC)[reply]
Yes, and that's partially why I'm asking. In particular, Bentham's Open Chemical Physics Journal published an article [3] authored by 9/11 conspiracy theorists. The most prominent is Steven E. Jones who was relieved of his teaching duties and placed on paid leave at Brigham Young University after he published an article on BYU's web site promoting 9/11 conspiracy theories.[4] They claim that the destruction of the World Trade Center was not the result of terrorist attacks. Instead, they claim that the US government had planted explosives inside the WTC prior to 9/11 and that the towers were brought down by controlled demolition. The article Bentham published claims that active thermitic material was discovered in the dust from the 9/11 World Trade Center remains which 9/11 conspiracy theorists argue are the result of a controlled demolition. If Bentham is a vanity press, that would explain how this paper was published. If, however, this is a legitimate peer-reviewed scientific journal, how did it get past a peer review? A Quest For Knowledge (talk) 13:07, 15 April 2009 (UTC)[reply]
While I know nothing in this case it is rather an open secret that peer review depends very heavily on the opinions of the editor, who is in charge of assigning reviewers, shepherding the article through to publication, etc. Peer review is not without its problems. (And not all important articles are peer reviewed.) --98.217.14.211 (talk) 01:15, 16 April 2009 (UTC)[reply]
I don't know whether this journal is vanity or not, but with open access journals often the costs are put upon the authors of the articles (more usually their institutions), and the scholars in question usually have money for this written into their grant proposals for this purpose if they are planning to go in this direction. For example, the Public Library of Science, which is very reputable, works this way. --98.217.14.211 (talk) 12:31, 15 April 2009 (UTC)[reply]
Journals – open access and otherwise – often defray some of their costs through page and color charges (extra fees for color figures). Page charges are higher for open access journals which enjoy little to no other income from subscription fees. As the anon above notes, the very reputable PLoS family of journals charges substantial publication fees: [5].
That said, it's worth looking closely at any journal which charges page or publication fees — some of them are vanity presses. Reputable journals will usually have an indication that they waive fees for authors who can't afford them; the PLoS statement is in the second paragraph of that link, above the prices:
We offer a complete or partial fee waiver for authors who do not have funds to cover publication fees. Editors and reviewers have no access to author payment information, and hence inability to pay will not influence the decision to publish a paper.
Be very wary of any journal that isn't prepared to make a similar guarantee. TenOfAllTrades(talk) 12:43, 15 April 2009 (UTC)[reply]
The journal(s) in question (Bentham) began in 2007. To quickly fill up about 300 "open" journals, Bentham spammed universities looking for papers. Anything that was thrown at them that appeared remotely interesting was "peer reviewed" and published online as quickly as possible. The main goal was to get a lot of hits. The second goal was to fill the journals with articles as quickly as possible. The third (perhaps fourth, fifth, or sixth) goal was to ensure the articles had a respectable peer review. -- kainaw 13:29, 15 April 2009 (UTC)[reply]

I think there are plenty of legitimate journals with page charges. Not the most high profile journals, but nonetheless respectable places for researchers in specialty fields to publish their work. I believe my colleagues who work in leather chemistry pay page charges. Some fields are very small, and their journals just don't sell a ton of subscriptions. I don't think this necessarily says anything about the quality of the work or the journal. ike9898 (talk) 13:39, 15 April 2009 (UTC)[reply]

There seems to be an inherent conflict of interest there. That is, if the journal is short on funds and can get them by publishing articles, they may be tempted to publish total crap just so they can meet their payroll. StuRat (talk) 15:01, 15 April 2009 (UTC)[reply]
If its a peer reviewed journal, then that shouldn't happen as total crap should not pass review. Besides, journals don't chase money that way, because its ultimately self defeating. What they really chase is impact factor. The higher your IF, the more people want to publish in your journal and the more people that want to read it, which means more money for you. If you publish crap, your IF will decrease which means less people will want to publish in your journal which means less money for you. There is possible COI in a peer reviewed journal, though, as publishers can manipulate IF's in devious ways. Rockpocket 19:15, 15 April 2009 (UTC)[reply]
What you say makes sense for the long-term future of the journal, but short-term concerns like avoiding immediate liquidation tend to outweigh lofty long-term goals. StuRat (talk) 16:14, 19 April 2009 (UTC)[reply]

FWIW, I'm more concerned about the fact that they published an article promoting 9/11 conspiracy theories. A Quest For Knowledge (talk) 15:06, 15 April 2009 (UTC)[reply]

I found a second Bentham article promoting 9/11 conspiracy theories in a discussion from the archives of the Reliable Sources Noticeboard regarding another Bentham journal, The Bentham Open Civil Engineering Journal. [6] Unfortunately, the discussion of the journal's reliability doesn't go very far. A Quest For Knowledge (talk) 15:52, 15 April 2009 (UTC)[reply]


The existence of page charges does not provide any evidence of a journal being a less than reliable source. Googling "journal page charge" showed "article processing charges" of $1400-$1550 (US) for BiomedCentral, which in turn tabulates the page charges for a number of journals, including Cambridge University Press ($2700), Blackwell ($3000), American Physiological Society ($3000), Elsevier ($3000), Wiley ($3000), and Springer ($3000). Some journals have lower page fees. If you do not like Jones and do not believe his analysis of the building collapses on 9/11 (notably the only modern highrises which ever collapsed from a fire or impact), do not try to discredit him or his analysis by claiming that page charges or "processing charges" make a journal "vanity press." Edison (talk) 15:53, 15 April 2009 (UTC)[reply]

It should be noted that it costs more to publish if the figures are in colour. Agree with Edison that a journal should not be considered a vanity press due to page charges. David D. (Talk) 15:59, 15 April 2009 (UTC)[reply]
Note that Nature (journal) (which is about as respectable as you can get) charges fees for colour pages (though not non-colour pages), but its rather difficult to publish a paper in Nature without colour. Scientific publishing has a strange business model. Most journals have a very limited circulation, so to cover costs (and make a profit) the publishes charge at both sides - they charge the scientists to publish and they charge the scientists to read. Amazingly, the scientists also provide most of the content (the manuscripts), yet the publishers often demand the scientists hand over the copyright of that work also!
For such smart people, the scientists appear complicit in getting screwed over every which way. Why? Well, its because the scientific publishing system is so important to progressing in a scientific career, that scientists are willing to do whatever it takes to get their papers published. One could take a principled stand and refuse to comply, but ultimately your paper will not get published in a top journal and your career will not progress. There are efforts to change the system, such as the PLoS and BioMed Central journals, where you typically pay for publishing only. These are similar to Wikipedia is spirit, freely disseminating knowledge is the goal. What distinguishes them from vanity publishing is, as other point out above, they all have clauses saying you only have to pay if you can. If you genuinely don't have the grants funds to cover the costs of publishing, then you don't have to pay. Rockpocket 16:27, 15 April 2009 (UTC)[reply]

If I were trying to get a job in academia or to get tenure or to get a research grant renewed, $3000 for a publication would seem like chump change, besides the fact that the grant is likely to pay it. How many hours and dollars went into the research and writing, do you suppose? Anything to pad the resumé. Edison (talk) 18:52, 15 April 2009 (UTC)[reply]

Many would argue is that publication is just one of the many costs of doing research. Just like presenting your work at conferences - some people get invited and have their way paid, but others are on their own to find funds they can use for this. ike9898 (talk) 19:50, 15 April 2009 (UTC)[reply]

Just to clarify, I've never heard an employed scientist spending their own personal money to publish. Sometimes you can use grant money or department funds or something else. It's just that you'd rather save that money if you could for supplies, equipment and salaries. ike9898 (talk) 20:03, 15 April 2009 (UTC)[reply]

When I was an unemployed scientist, I would certainly have paid page charges to get a publication in a good journal. Maybe you cash in some U.S. Savings bonds, or hit up the parents for some cash. It could lead to becoming an employed scientist. The alternative might be writing off many years of college and graduate school and accepting a less desirable career path. Do not underestimate the value of a publication if the resumè is a bit anemic. Edison (talk) 18:59, 16 April 2009 (UTC)[reply]
I agree. It also demonstrates true dedication to your work. ike9898 (talk) 13:56, 17 April 2009 (UTC)[reply]
Many journals that have page charges are willing to waive them in cases of need. Looie496 (talk) 16:34, 19 April 2009 (UTC)[reply]

Rate Constant, k

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Hi all I after having read the article on rate constant I am still a little confused over what k is. Is it the rate of the reaction when the reactants are at a concentration of 1 mol dm-3 I've also heard that k isn't even a constant anyway as it if affected by the likes of temperature. How bizarre to call it a constant then! Another thing which particularly confuses me is writing the rate equation for a zero order reaction. I believe it would be written: Rate = k. Which means rate = rate constant. What does this mean? I hope what I've written makes sense - I find chemistry very confusing. Thanks in advance to all who help. —Preceding unsigned comment added by 92.10.162.104 (talk) 12:29, 15 April 2009 (UTC)[reply]

The arrhenius equation states what k is. Yes k is not really a constant, rather a coefficient. If the order of reaction is 0 with respect to all reactants, then yes, rate = k (mol dm-3 s-1).
Clearly, if the total order of reaction (m + n) is 0, then changing the concentration of reactants will have no effect on how fast/slow the reaction is (because any number to the power of 0 equals 1). --82.21.28.65 (talk) 12:58, 15 April 2009 (UTC)[reply]
They may call it a "constant" since, in this particular equation, it is constant with respect to reaction progress. When you use the rate equations, you pick a value of k and stick with it. But yes, it's not really constant - I think maybe "rate coefficient" would probably have been a better name, but I think we're stuck now :) --Bennybp (talk) 13:34, 15 April 2009 (UTC)[reply]

Armour made from tantalum-niobium alloy - how effective?

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Any metallurgists here? In Terminator: The Sarah Connor Chronicles, it has been stated that the armoured parts of the more recent Terminator T-8xx models are constructed from an alloy of tantalum and niobium, as it was discovered that this was more effective on the battlefield of the future than the previously-used titanium, due to its greater heat-resistance (in response to the development of laser cannons and particle beam weapons, I suppose, as the war has been going on for decades in the story), comparible resistance to bullets and lighter weight (I think).

So, in the world of Real Life, how would armour plating made from an alloy of tantalum and niobium really measure up in a war situation? Not nescessarily when attached to a robotic soldier, just in general, I mean. Thanks. --81.77.155.81 (talk) 15:13, 15 April 2009 (UTC)[reply]

My first thought is that the armour would be vastly more expensive than titanium, since there appears to be three orders of magnitude difference in crustal abundance.
Looking at the various properties: Ta/Nb is much more dense than Ti, so a given thickness would be much heavier; the specific heat capacities are comparable, but Ta/Nb has greater conductance, so it would be better at absorbing laser input; also Ti becomes brittle on heating in presence of oxygen, if you can get a good laser shot in, that bit will turn blue and fracture on mechanical impact; Ta/Nb would probably be better with particle beams, due to its higher density; the hardnesses are comparable but I have a suspicion that Ti might be more susceptible to brittle fracture than Ta/Nb, not sure there - recall that the SG-1 team beat the replicators by using bullets, which shattered them.
If you're giving me unlimited funds, I'd probably pick the Ta/Nb solution. In fact, I'm not sure why you made us use Ti in the first place. In the end, it depends on what you want to use the armour against, and how fast you want the thing you're protecting to move. The best way to withstand an enemy shot is to move out of the way, heavy armour restricts that ability. If you're going to stand and take fire, there's a lot to be said for the sandwich technique using depleted uranium.
And when it comes right down to it, I'll take whatever Cameron is made of - those are very impressive construction materials! :) Franamax (talk) 20:33, 15 April 2009 (UTC)[reply]
The high density of tantalum and uranium as well makes it a problematic alloy to use in large scale, for a humanoid cyborg, but if you do not care if one terminator is only capable to run ultra hard surface and simply would get stuck in a normal madow it is good. Normally a superalloy similar to the stuff used in turbines, a alloy based on nikel or cobalt with aditions of other refractory metals like tantalum and rhenium gives the best performance when heated. But a woven fabric of carbon nanotubes as interior of a composit material would be my favorite. For a particle beam weapon the exact sort of particle is important, protons, neutrons, electrons, alpha particles are positrons would all need a material capable to with stand the negative effects of nuclear conversions of the material. For Protons the answer can be found in the normal satelite business, because solar wind is made mostly from protons.--Stone (talk) 21:01, 15 April 2009 (UTC)[reply]
The OP is asking about the present-day performance of the armour, so perhaps I've led us astray discussing particle-beam weapons. Laser and microwave weapons are reality or close to it, so that's where we should focus, along with projectile and heat munitions (and vacuum bombs, which render the issue moot if you don't have a secondary air source). I think we agree on the tradeoff between weight and mobility.
As to a nanotube inner lining, would this be part of the breathable atmosphere inside the armour? There is growing evidence that carbon nanotubes are as toxic as asbestos, so I think maybe I'd rather have a chrysotile blanket instead (not amphibolic!) At least then I could quantify the risk. Franamax (talk) 21:50, 15 April 2009 (UTC)[reply]

Hair & Facial Hair Question

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Hello, if this has been asked before, I apologize, I couldn't find where it had been asked. Anyway on to the question: How is it possible for some body hair to be a completely different color than all the other hair on someone? For example I am blonde most of my arm and leg hair is blonde (dirty blonde to be exact). But my facial hair is RED. I'm not really upset about it and I'm not looking for answers on how to dye or change the color of my hair, I'm just curious about the origins and reasons behind this.

Thanks! 12.204.178.35 (talk) 15:33, 15 April 2009 (UTC)[reply]

Us blondes have hair of many colours, on our heads and elsewhere on our bodies. You will probably have red hair on your head as well as blue (honestly), black and red. It's just that the blonde is more prominent on your head, whereas the red is probably more prominent on your face. Nothing to worry about - it's a quirk of nature. For the record, the only blonde hair on my body is on my head!--TammyMoet (talk) 15:48, 15 April 2009 (UTC)[reply]
Yes I've noticed the array of colors I seem to have! I wasn't too sure if it was kosher to talk about but my pubic hair isn't blonde..that's for sure. It's good to hear this isn't uncommon, but I'm still wondering why this is? Aside from artificial coloring you don't seem many dark haired people with such a wide variety of colors. 12.204.178.35 (talk) 16:19, 15 April 2009 (UTC)[reply]
Blue?! —Tamfang (talk) 19:17, 15 April 2009 (UTC)[reply]

I'm afraid TammyMoet's response isn't quite true. The gene responsible for red hair in most people is MC1R. When this was first being studied, the scientists found a strange correlation. They found that if you have two null alleles (two copies of a gene variant that doesn't function fully) then you have an increased likelihood of having red hair. However they also found that if you have just one null allele (what we call heterozygous for that gene), then you may have a different hair colour on your head, but that you have an an increased likelihood of having red body hair. I was one of the scientists working on that project, and having a reddish beard myself, I decided to sequence my own MC1R gene to see if I was a heterozygote myself. Turns out I am.

Quite why being a heterozygote results in this effect isn't quite clear. However it may be something to do with different effects of haploinsufficiency on different hair types. More simply, that the hairs on your head manage to make enough eumelanin with just a single copy of MC1R, but the hairs on the rest of your body need more MC1R. If you want to read the scientific data behind this, let me know and I'll find it for you. Rockpocket 17:25, 15 April 2009 (UTC)[reply]

I don't see how what you said contradicts what I said: indeed, it seems to be the science behind the phenomenon. I was sure there was a genetic component to it, however I was also sure there'd be a geneticist along in a while - and I was right! Thanks.--TammyMoet (talk) 20:45, 15 April 2009 (UTC)[reply]
A "Blue Heeler"
A "Blue Picardy Spaniel"
The thing I took issue with was the suggestion people have blue hair. There are no blue pigments naturally occurring in mammals, so (healthy) humans don't make blue hairs. Rockpocket 21:35, 15 April 2009 (UTC)[reply]
However, healthy humans might look at a hair and perceive it as being blue. I run into that terminology problem all the time. Franamax (talk) 21:58, 15 April 2009 (UTC)[reply]
There are plenty of dog breeds called "Blue somethingorother" - they aren't literally blue either - but the problem (again) is that blue is just a word - and it means whatever people want it to mean in some specific context. SteveBaker (talk) 22:25, 15 April 2009 (UTC)[reply]
Even blue jays lack blue pigment. -GTBacchus(talk) 22:31, 15 April 2009 (UTC)[reply]
In fact, for some people there's no such separate concept as blue. The world is painted in shades of grue! Franamax (talk) 22:45, 15 April 2009 (UTC)[reply]
Yikes, I better light a torch then! ArakunemTalk 22:50, 15 April 2009 (UTC)[reply]
Just to be clear, the preceding reflects two very different things. Blue jays have no blue pigment, but are coloured blue due to a process referred to as structural colour. Grue refers to the different ways languages classify colours. See here for the actual article on the topic. Matt Deres (talk) 13:33, 16 April 2009 (UTC)[reply]

Calculating Kc equilibrium constant

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Hi I am really stuck with the following problem. The reaction needed for the question is: Cu2+ + H2A (a fictional substance)(equilibrium arrows - don't know how to insert them) CuA + 2H+. The question states that at the beginning there are 0.005 moles of Copper ions and 0.005 moles of H2A. At equilibrium there is 0.0049 moles of CuA. Therefore: Kc = [CuA] [H+]2/ [Cu2+] [H2A] The question asks me to calculate the number of moles of all substances at equilibrium. So I already know that equilbrium the concentration of CuA is 0.0049 moles. Therefore at equilbrium there is (0.005-0.0049) 0.0001 moles of Copper ions left. Now comes the part I'm not too sure on. Looking at the answer the concentration of H+ = 0.00492 = 2.401x10-5 mol. My question is how can it be known for sure that the concentration of H+ ions is the square of the concentration of CuA? Is it not possible that the concentration of H+ ions would not be the square of 0.0049 moles at equilibrium? How do we know this is the case? Sorry for asking silly questions and thanks in advance to anyone who can help. —Preceding unsigned comment added by 92.8.198.137 (talk) 18:51, 15 April 2009 (UTC)[reply]

I don't think squaring the concentration of CuA will give you the concentration of protons. Check the stoichiometry of the reaction (and note the link I provided), specifically the ratio of CuA to H+ in the equation, and ask yourself: for every mole of CuA, how many moles of H+ will be generated? Looks like AFTER you have answered this question and determined the concentration of protons, you may want to square it in calculating Kc. I hope this helps. --Scray (talk) 22:10, 15 April 2009 (UTC)[reply]
As Scray said. Rethink the meaning of the formula: [H+]2 does not mean you have to square [H+] to figure out the concentration, but the other way round: [H+] is the concentration and it needs to be squared to obtain the equilibrium constant. See here. Steipe (talk) 20:23, 23 April 2009 (UTC)[reply]

Hearling loss from high frequencies

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Is it possible to have hearing damage if you listen to imperceptibely high frequencies (20 khz~ish) at high volumes? 206.176.119.180 (talk) 18:58, 15 April 2009 (UTC)[reply]

I can't answer your question, but I just wanted to help you understand the relationship between volume and intensity. Volume is a human phenomemon and the correct word is called loudness and is measured in the decibel system while you are asking a question that will probably get answered using the terminology of sound intensity which is power per square meter and is related to the strength of a sound wave. A very intense sound wave is, for example, has a very high value of watts per square meter and vice versa. Even if you can't hear the sound wave you speak of in your question, it will still have intensity, although it will not have loudness because if you can't hear a sound, that means it has zero decibels, but not necessarily zero watts per square meter and if I can further request to whoever answers the OP's question: is my understanding of the relationship between intensity and loudness correct? is loudness, or intensity--specifically, what causes hearing damage? The article goes into generalities such as "caused by a wide range of biological and environmental factors... etc...". I suggest read the Hearing impairment article because it gives some good info, but doesn't specifically answer what you are asking--which is a good question, because I imagine those who read the article would like to know as well. 71.55.186.77 (talk) 20:28, 15 April 2009 (UTC)[reply]

It is incorrectly said above that if you can't hear a sound, that means it has zero decibels. Without a specified reference sound pressure, a value expressed in decibels cannot represent a sound pressure level. With any specified reference sound pressure, a sound pressure that is equal to the reference has zero decibels relation to it. The "loudness of inaudible sound", regardless of reference sound pressure, is actually minus infinity decibels. Cuddlyable3 (talk) 21:29, 15 April 2009 (UTC) [reply]

There is evidence from studies of the effects of ultrasound scans on unborn children that this does not harm their hearing - even though the intensities (watt/sq.meter) are quite high and the amniotic fluid (which is mostly water) transmits the sound very efficiently. I assume that this is because the eardrum has inertia and elasticity and at sufficiently high frequencies is simply not moving measurably in the tiny amount of time before the sound wave reverses direction. This (admittedly sketchy) information suggests that the transmission of sound energy from outside of your ears to inside (where high energies could certainly wreck your hearing) must be a function of frequency. That doesn't mean that a super-loud burst of 20kHz wouldn't harm you though...it just says that we can't simply measure the raw intensity and nothing else. SteveBaker (talk) 20:47, 15 April 2009 (UTC)[reply]
Agree. The possibility of damage to hearing depends on the efficiency of the amplification mechanism in the human ear. Quite simply, if the eardrum and the little bones can't transmit the vibrations to the cochlea because they don't respond fast enough, those little hairs (cilia) won't get damaged. Where the exact cutoff line is, I dunno. It's very likely that the human ear has evolved to efficiently filter only the frequencies that matter most. I'd speculate that ~20KHz might still be in the danger zone, but I don't know of any detailed mechanical studies on the amplification and coupling factors in that region. I do know that I cover my ears as soon as I hear feedback rising in frequency, since it quickly becomes very painful. Actually, as a former soundman, I would leap for the volume knobs first. Intense high-frequency sound is double-plus-ungood. Franamax (talk) 21:10, 15 April 2009 (UTC)[reply]
Yes on 20KHz not being "imperceptibely high frequencies"--many people can hear in that range quite well. Of course not all loudspeakers are capable of producing sound in that range. Pfly (talk) 05:25, 17 April 2009 (UTC)[reply]

honey

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What is the chemical or chemicals in honey that help reduce the pain of a sore throat? 65.121.141.34 (talk) 20:03, 15 April 2009 (UTC)[reply]

I don't think that is known. [7] Rockpocket 20:10, 15 April 2009 (UTC)[reply]
I don't know either - but I suspect it may not be a chemical effect so much as a lubricant. SteveBaker (talk) 20:41, 15 April 2009 (UTC)[reply]
I don't know what has the analgesic effect, but the antiseptic effect is due to the way the high concentration of sugars absorbs water, killing the bacteria. --Tango (talk) 20:43, 15 April 2009 (UTC)[reply]
Anti-inflammatory compounds in Honey. Live long and prosperous \V/ --Grey Geezer 22:15, 15 April 2009 (UTC) —Preceding unsigned comment added by Grey Geezer (talkcontribs)
But avoid non-pasteurized honey or you might encounter these: Botulism. Particularly dangerous for small children. 76.97.245.5 (talk) 08:39, 20 April 2009 (UTC)[reply]

Microbial growth media

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What is the correct term for the class of growth media ingredients that includes yeast extract, peptone, beef extract, casamino acids, tryptone, etc?

Microbiological media ingredients? Amino acids? I'm not convinced there is a specific classification. Rockpocket 20:13, 15 April 2009 (UTC)[reply]
Please look here or do you mean Lysogeny broth (Luria-Bertani Broth)? Cheers --Grey Geezer 21:43, 15 April 2009 (UTC) —Preceding unsigned comment added by Grey Geezer (talkcontribs)
Sorry, that doesn't really answer my question at all. ike9898 (talk) 22:20, 15 April 2009 (UTC)[reply]
D'Oh! I overlooked the word ingredients [don't misguide old men by a "poorly formulated title" of your request ;-)]. Coming from the media you would call them something like non-defined (or poorly defined) protein complements. If you search for them in the Sigma-Aldrich Catalogue (they should have an interest to label this "Class" correctly, right?), they call them "Microbiology, Basic Ingredients, Protein Sources", or "Peptones" (which has another more specific definition). In my view (as suggested above): No generally accepted "Class" designation. Better? ;-) --Grey Geezer 07:30, 16 April 2009 (UTC) —Preceding unsigned comment added by Grey Geezer (talkcontribs)
Yeah. The reason I ask is because not too long ago a microbiologist got all bothered when I referred to this sort of thing as a "nitrogen source". I never got the chance to ask her exactly why she thought that was wrong or what term she would prefer. It is possibly because these ingredients do more than just supply nitrogen, they also contain growth factors and minerals, and provide some energy as well. However, I haven't been able find to good single word to describe these things. Maybe I should come up with one (remember Sniglets?). ike9898 (talk) 13:28, 16 April 2009 (UTC)[reply]

Very small creatures in my fish tank?

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I have something very small living (uninvited) in my freshwater tropical aquarium. It is circular/oval, white/grey in colour and about 1mm-2mm across in size. It moves very slowly across the surface of the glass or the leaves of the plants, as if it were a tiny mollusc or similar. There are several of them in the tank and the numbers seem to be steady. They don't appear to be damaging the plants although it's possible they are attacking the fish. Every so often a fish appears to have a twitchy fit for about a day and then dies, but this is rare. Is it likely they are harmless or could they be some kind of parasite?Popcorn II (talk) 20:51, 15 April 2009 (UTC)[reply]

Planaria? --Dr Dima (talk) 21:25, 15 April 2009 (UTC)[reply]

Hmmm. Possibly but I don't think so. These things are about 2mm at their largest. Many of them are less than 1mm. They also seem to move as a whole, solid form without stretching, as a Limpet might.Popcorn II (talk) 22:55, 15 April 2009 (UTC)[reply]

Might be freshwater ostracods.--Eriastrum (talk) 23:07, 15 April 2009 (UTC)[reply]

Possibly but they really don't look like Crustacea. There's no detail on them, just a smooth shape.Popcorn II (talk) 12:49, 16 April 2009 (UTC)[reply]

Ostracods are enclosed in a little shell, and that is all you usually see: they look like little beans, usually whitish. The crustacean-looking parts (jointed legs and mouth parts) are enclosed inside.--Eriastrum (talk) 19:38, 16 April 2009 (UTC)[reply]
A photo would be helpful. ;-) Axl ¤ [Talk] 08:36, 17 April 2009 (UTC)[reply]

Biological Function of Grief in Humans

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Feelings (affection, humor, fear, even belief etc.) can be associated with "biological functionality". However, what is the "functionality" of grief (e.g. after the loss of a child, partner, parent)? A functional reaction should be "feeling strong" or even "euphoria" to start up again (to find a new partner, have another child etc.). Griefing for a loss (which will not be returned) does not really make sense, right? If the loss is by accident, predator, or disease, the learning effect (how to avoid) is low: the griefing person is depressed, inactive, sometimes suicidal, has no appetite, which should increase the chance for an accident, falling prey to a predator or even getting weak and sick: So exactly the opposite = grief makes things worse. The question is not: How to overcome grief but "what is its function in the big game"? Any suggestions? --Grey Geezer 22:08, 15 April 2009 (UTC) —Preceding unsigned comment added by Grey Geezer (talkcontribs)

It probably has something to do with the fact that we've evolved in tribal settings with small groups of a dozen or two people. We have evolved a sense of community. Just as we've evolved responses such as pain so that we know that when we've been bitten in the leg by a rampaging marmosette we should avoid letting that happen again - so taking a 'wound' to the community is something we need to try to avoid - so we have a different 'pain' response to that. If there were no emotional consequences to losing tribe members ("friends" and "relations") perhaps we'd be less cohesive as a group? SteveBaker (talk) 22:16, 15 April 2009 (UTC)[reply]
Its very beneficial to be part of a group, from a group of two people (couple relationship) to a family/tribe group, to a polical/national group. The emotions try to make you stay part of that group no matter what. These emotions fire even if the group doesn't physically exist anymore. Remember that the whole idea of being in a group is a mental idea - so its sort of an idea organism that isn't going to know if the group physically exists - for most of your life the group will exist and these emotions will keep you in-line with group which is beneficial biologically. Also the reason can be non-biological. There is no biological reason for a person to kill themselves for the group (that doesnt include their offspring), but the idea organism that permeates all minds in the group has alot to gain from someone doing this - its strenghtens the idea of group, and thus the group is likly to expand and beat other group ideas. So basically you need to understand that once we had evolved a mind with certain capabilities, we also had 'ideas' that could evolve, and these virtual ideas can very often beat out the biological requirements of their hosts (us). It is beneficial for the rest of the group (which is really just an idea of what the group is) if you are seen to endure great hardship from leaving or someone else leaving the group. Likewise people are typicalyl very moved if they see someone else who has lost their partner, often it motivates them to form even strong group connections with there still existing partner, this is how an idea (like marraige) can be a duplicate of groups, an idea that copies itself as small groups everywhere. So while its pointless for us to greive, it helps everyone else with the idea of marraiage, and it likely results in the idea of marraige being further conveyed.--Dacium (talk) 03:28, 16 April 2009 (UTC)[reply]
Like other aversive things, its main function is probably preventative: the knowledge (obtained by observation of other people) of how terrible grief is makes you work very hard to avoid it. Looie496 (talk) 05:21, 16 April 2009 (UTC)[reply]
Thanks for the comments! So grief, which is often used as a "personal problem" could be a "group signal" ("Look how valuable members of this group are to me!". And like "Look, I suffer to have all those piercings / brandings done to show that I am really part of your group!") and this signal would be "more worth" than the "temporary inefficiency" of the grieving group member. Makes kind of sense. However, grieving for pets then seems to be a "side effect". It seems that Inuit or other native groups are very pragmatic (more realistic?) with the dogs who guarantee part of their survival (also very important!), e.g. killing off "weaker" female pups or disposing of the corpses of long-time sleigh dogs, but skinning them and using their fur. --Grey Geezer 07:51, 16 April 2009 (UTC) —Preceding unsigned comment added by Grey Geezer (talkcontribs)
Yes, I think that's true. We've "adopted" pets - they are a part of the "tribe" - they are treated like children and grieved for in the same way. When animals are "working animals" they are more like tools than members of the family - we can be more hard-hearted about the whole thing. When your old car is beat up and rusty - you get a new one - you throw the old one out...unless you can salvage something from it. SteveBaker (talk) 12:56, 16 April 2009 (UTC)[reply]

Grieving is a retreat into a passive state when one finds that previously held assumption(s) about one's safety, posessions or relationships have been broken. Its survival value lies in 1) the retreat from action to allow time for, hopefully, the sufferer to evaluate their changed situation and 2) the person's grief signals can bring aid from other(s). Babies use 2) as feeding strategy. Cuddlyable3 (talk) 23:20, 16 April 2009 (UTC)[reply]

Achilles tendon and stairs

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I find it easier to climb flights of stairs if i do not fully relax my Achilles tendons after each step. Why would that be? —Preceding unsigned comment added by 79.67.15.85 (talk) 22:36, 15 April 2009 (UTC)[reply]

Tendons (and all soft tissue in general) has elasticity. If you keep the tendon tight, it will stretch a bit when you apply weight to it and release some of that energy to propel you up to the next step. Of course, this only works if you go up the stairs on the balls of your feet, not the heels. -- kainaw 12:14, 16 April 2009 (UTC)[reply]

Ear canal bandpass

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What is the bandpass of the human ear canal in Hz? (not homework) —Preceding unsigned comment added by 79.67.15.85 (talk) 22:41, 15 April 2009 (UTC)[reply]

Is this what you're looking for? ArakunemTalk 22:48, 15 April 2009 (UTC)[reply]
Well no. Im looking for the actual transfer function of the human ear canal (including the pinna), but excluding the ear drum, choclea etc —Preceding unsigned comment added by 79.67.15.85 (talk) 22:53, 15 April 2009 (UTC)[reply]
Does this help? Cuddlyable3 (talk) 22:59, 16 April 2009 (UTC)[reply]
This book has what you want. You will also find a lot of papers on the subject if you search for "head related impulse response" or "HRIR" on google scholar. Here are some from the Acoustical Society of America [8][9][10]. There is also a lot published by the Audio Engineering Society and the IEEE, but sorry, I don't know if they are exactly what you want as I can mostly only read the abstracts. SpinningSpark 13:48, 18 April 2009 (UTC)[reply]