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May 29

A superacid under extreme conditions

In theory, is it possible to create the superacid cyclotrihydrogen(1+) tetrahydridoborate by pressurising a mixture of dihydrogen and borane in stoichiometrically amounts? If so, how much pressure could be adequate? (MPa/GPa?) Plasmic Physics (talk) 00:35, 29 May 2013 (UTC)[reply]

Perhaps you can find out the heat of formation of these ions, and then see what pressure would be needed to impart that much energy to your gas mixture. It sounds very unlikely to be stable to me! -- much more likely to be diborane and hydrogen molecules. Other terms to use are borohydride or tetrahydroborate. The trihydrogen cation is always a triangle under normal conditions (could be linear under extreme magnetic field), so the cyclo bit seems redundant. An alternative for this bit is hydrogenonium. You may be interested in boron pentahydride BH₅ which has an energy to lose a proton of 332.4 kcal mol-1.[1] Graeme Bartlett (talk) 10:32, 29 May 2013 (UTC)[reply]

I was not expecting it to be stable relative to STP conditions. Certain compounds behave in this way under extreme pressure. I think iodine behaves like this, turning into iodylium iodide under pressureYet another term is boranuide. Plasmic Physics (talk) 16:16, 29 May 2013 (UTC)[reply]

Berylium inhalation when needle-scaling.

I am trying to find references to recent reports that the cause of a number of cases of lung cancer has been traced back to operators of needle-scalers inhaling beryllium dust from the break-down of the beryllium-hardened needles. Can someone give me some advice on how to find that information please? Thanks in advance. 122.108.189.192 (talk) 04:56, 29 May 2013 (UTC)[reply]

Try a search like http://www.ncbi.nlm.nih.gov/pubmed/?term=beryllium%20cancer to get general information on beryllium and cancer. I don't know if you can find a source that directly evaluates needle scaling (honestly I have no idea what that is, so I may have missed the papers) - you'll probably need to divide the problem into a question of how much exposure the workers get and how much that exposure contributes to cancer. From the references it looks like even the idea that it causes cancer is still not well settled. You might want to follow up with more specific questions here. Wnt (talk) 06:10, 29 May 2013 (UTC)[reply]

Perfect answer. Thanks a lot Wnt. 122.108.189.192 (talk) 06:32, 29 May 2013 (UTC)[reply]

How can I measure the weight of a plant without unrooting it, or cutting it into pieces?

Any ideas?--朝鲜的轮子 (talk) 05:40, 29 May 2013 (UTC)[reply]

Is there just one plant involved, or do you need to determine the weight of a large number of plants of the same species, as in a lab setting, in which case you may be able to sacrifice one plant for the purposes of making density measurements? And are the plants in the ground, or within containers whose volume can be measured? Also, what type of plant is it, which might make a difference because for example if they are trees, the densities of various types of wood are known, which might be helpful. Red Act (talk) 09:04, 29 May 2013 (UTC)[reply]

To make it more clear, I just found a picture of one cactus growing from the edge of rooftop, and I wonder how much weight it can support before it falls.--朝鲜的轮子 (talk) 00:34, 30 May 2013 (UTC)[reply]

Very much has been written on the topic, see tree allometry for starters, and search for "[your species] allometry" on google scholar. For certain species and locations, you can find published equations that let you estimate weight from diameter at breast height. Much of the work is done in the interest of carbon sequestration, but they still give biomass. SemanticMantis (talk) 12:30, 29 May 2013 (UTC)[reply]

mathematics helps us to estimate ,when we have resultant of any piece of a matter we can generalize it to total.A. mohammadzade--78.38.28.3 (talk) 15:17, 29 May 2013 (UTC)[reply]

Easy. Look up the mass of the earth from a source before the tree was planted (about 5.972x10-24 kg), then look up the mass of the earth today. Subtract the two and that will give you the mass of the tree.63.153.198.142 (talk) 00:13, 30 May 2013 (UTC)[reply]

No. because the mass of the plant is persumably taken from the mass of earth. Also there are many factors that changes the earth's mass e.g. space shuttles, meteors.--朝鲜的轮子 (talk) 00:33, 30 May 2013 (UTC)[reply]

solar wind

any research report or reference is needed about mathematics formula of solar wind direction and movement path curve : 1) inside planetary disk 2)at outer part --78.38.28.3 A. mohammadzade (talk) 06:15, 29 May 2013 (UTC)[reply]

Are you asking us to do your homework again, Mr Mohammadzade? Tell us what you have done to resolve this yourself, then we may assist if you are stuck. Have you looked at our Solar Winds article? Chased up the references listed and seen where they lead? Tried googling "solar wind model" and such like? The Good Lord helps those who help themselves. Wickwack 124.182.169.60 (talk) 07:21, 29 May 2013 (UTC)[reply]

excuse me not home work.

   The kappa of  solar wind  path  curvature is :

κ=|dT/ds|

R(t)=(ri(t), rj(t), rk(t))

the velocity is tangent to curve :

V(t)=(rʹi(t), rʹj(t), rʹk(t))

Then T will be:

T=V(t)/|V(t)|

The normal vector is :

N=Tʹ/|T|

B=TxN

Taw formula :

Ʈ=-(dB/ds).N

And we will use velocity vector for finding ds :

dT/ds=(dT/dt). 1/(ds/dt)=Tʹ.(1/|V(t)|)

and dB/ds=dB/dt .dt/ds=dB/dt .(1/|V(t)|)

A. mohammadzade--78.38.28.3 (talk) 08:03, 29 May 2013 (UTC)[reply]

for finding movement path curve ,for example inside supposed plate we need its equation,

Ax2+Bxy +Cy2+Dx+Ey+F=0

Θ=1/2arc tan (C/B-A)

rotated coordinates is

│■(x@y) │=│■(cos Θ&-sin Θ@sin Θ& cos Θ)││■(xʹ@yʹ)│

A. mohammadzade--78.38.28.3 (talk) 08:06, 29 May 2013 (UTC)[reply]

Physical attractiveness of men

I need to know what features women look in an ideal physically attractive man:

Eye size: 24 mm or more or less?
eyebrow size?
chest size?
skin color and skin texture? --Yoglti (talk) 08:14, 29 May 2013 (UTC)[reply]

It will differ tremendously from woman to woman. Have you read Physical attractiveness? Rojomoke (talk) 08:20, 29 May 2013 (UTC)[reply]

Depends a lot also on what for. Dmcq (talk) 14:26, 29 May 2013 (UTC)[reply]

And on what culture the woman is from. 24.23.196.85 (talk) 00:21, 30 May 2013 (UTC)[reply]

Also depends on where in their monthly cycle the woman is. When fertile, women tend to be attracted to men who look more like Neanderthals, while they like less testosterone-filled specimens the rest of the time. StuRat (talk) 08:53, 30 May 2013 (UTC)[reply]

Makes me wonder why the Neanderthals died out ;-) Dmcq (talk) 14:46, 31 May 2013 (UTC)[reply]

4) Skin texture should be smooth (not counting hair/stubble). Pimples, warts, moles, wrinkles, etc., tend not to be attractive, although some women might like men with scars. StuRat (talk) 08:56, 30 May 2013 (UTC)[reply]

Size of the nucleus

If we expand an atom to the size of the earth, then what would be the size of the nucleus (assuming that the atom and its nucleus are expanded in the same ratio)? 27.63.224.93 (talk) 08:47, 29 May 2013 (UTC)[reply]

Your question is rather vague, for a couple reasons. First, it doesn't specify which element the atom is, which makes a considerable difference. Second, quantum mechanical "fuzziness" makes atoms and nuclei not have well-defined boundaries, so to be precise you'd need to specify how the sizes involved are to be measured. For a discussion of some of that complexity, see Atomic radius. However, as a very rough order-of-magnitude approximation, for some elements atoms have a radius that's roughly 10⁵ times bigger than the radius of their nucleus, and 10⁵ times smaller than the Earth's roughly ~~6km~~ 6Mm radius is ~~6cm~~ 60m. Red Act (talk) 09:32, 29 May 2013 (UTC)[reply]

The Earth has a radius of only 6 kilometres? -- Jack of Oz ^[Talk] 09:49, 29 May 2013 (UTC)[reply]

(penny drops) Oh, you must mean 6K (= 6,000) miles, I guess. -- Jack of Oz ^[Talk] 09:51, 29 May 2013 (UTC)[reply]

Sorry, I'm half asleep. It's 6 AM here, and I haven't gone to bed yet. I've corrected my above post. Red Act (talk) 10:01, 29 May 2013 (UTC)[reply]

If it is a Rydberg atom, the nucleus can be regular-sized (however you choose to interpret that); and the atom can, in principal, be planet-sized (however you choose to interpret that). The probability of finding such an atom is low because that is not an equilibrium enery configuration; but out in the vastness of the universe, there are probably a few such specimens, very weakly bound, but drifting among the otherwise empty and non-interacting vacuum. This is not meant to be facetious, just to remind the OP that atom-size is not trivial to define; nor is there a magical "inflation operator" that can make everything twice as big - because it's nontrivial to define "inflation" - which physical parameter(s) do you actually scale? Length? Energy? Physical constants? If you choose to scale some of these, you might produce an unphysical system that cannot exist; if you choose to scale all of them, you might have accomplished nothing but an arbitrary change of units. Nimur (talk) 14:14, 29 May 2013 (UTC)[reply]

Guys, you are completely confusing the OP with answers to questions they didn't ask. Pick a fairly medium-sized atom, like say iron. The Van der Waals radius is as good of a measure of size as anything, and for Iron that is 126 picometers. That's 1.26 x 10^-10 meters. The nuclear radius of iron is about 3.7 femtometers (3.7 x 10^-15) meters) per this The ratio of the size of of the atom to the size of its nucleus is thus 1.26 x 10^-10/3.7 x 10^-15, or 3.4 x 10⁴ times as big (34,000 times). Take the radius of the earth: 6,371 kilometes (6.371 x 10⁶ meters). Divide the radius of the earth by 34,000 and you get 190 meters. Thus, if you scaled an iron atom up to the size of the Earth, it's nucleus would be a ball about 190 meters in radius. A medium sized skyscraper is about that tall, while the largest skyscrapers are about twice that tall, so if you imagine a ball that contains the Empire State Building, that's a good ballpark-estimate size. --Jayron32 15:28, 29 May 2013 (UTC)[reply]

Actually 380 meters across. Ruslik_Zero 16:46, 29 May 2013 (UTC)[reply]

So corrected. --Jayron32 17:23, 29 May 2013 (UTC)[reply]

So a ballpark would be a good ballpark estimate then? SteveBaker (talk) 14:03, 31 May 2013 (UTC)[reply]

Do the same problem for a star and compare the answer to the size of a neutron star. Count Iblis (talk) 18:13, 29 May 2013 (UTC)[reply]

Which Scientific name is correct

There seems to be confusion in the scientific name of the species. In the page Eld's deer, the species is referred to as Panolia eldii whereas in Sangai page it is shown as Rucervus eldii. Which one is correct?--Somesh Tripathi (talk) 17:34, 29 May 2013 (UTC)[reply]

There does seem to be some confusion. Based on http://arts.anu.edu.au/grovco/Pitra%20deer.pdf, rucervus was the original classification but is clearly out of date -- the correct genus should either be cervus or panolia. Lots of classification details are in flux nowadays, unfortunately. Looie496 (talk) 18:51, 29 May 2013 (UTC)[reply]

Question regarding the statistical approach to define entropy

Hi,

First, I'd like to apologize if I am doing this wrong - I have never used a talk page before. Please note that before bothering you I've contacted a few other folks (a NIST guy being my best hope), but they all claim ignorance on the subject. Anyway. in my old age I’ve started to think about stuff and the depths of my ignorance astound me. My current topic of confusion is entropy, which I thought I had at least some handle on, but I’ve come up with an observation that baffles me.

Let’s start with the classic mixing of two ideal gasses (say Argon and O2 for illustration). They are in a box and are separated by a divider. They are at equal pressure & temperature, etc. etc. You remove the partition and the two gases mix and entropy increases by the well known formula. OK, I think I understand that fairly well from a statistical point of view – there is only one arrangement where the gases are unmixed and a gazillion where they are intermixed to some degree so the intermixed configuration is much more likely and thus has higher entropy from a statistical point of view

Now, let’s redo the same problem where the two gases are very, very similar (say Argon-39 and Argon-40) – again you get a significant entropy change when you remove the partition and the two gases mix

Finally, let’s redo the same problem where the gases in both sides are identical. Suddenly, there is no entropy change. Why is that? If I had a very powerful microscope and there were relatively few atoms it seems like I could (at least theoretically) track the individual atoms and it is very unlikely that they would spontaneously separate back to their original sides, so hasn’t the entropy of the system increased when the two sides mix? Note that if tracking atoms seems impossible, stick in the large gaseous molecule of your choice

So, it appears the definition of entropy of the system hinges on our ability to distinguish one atom from another. If that is true, let’s look at the second case again – do we really believe we could tell where the Argon-39 atoms are in the box relative to the Argon-40 atoms? If we can’t is the entropy of that system unchanged when we remove the partition?

If the answer to the above question is well known, I’d appreciate hearing it. If it is not, I hope somebody enjoys thinking about it. I get the feeling this relates back to some quantum mechanics weirdness that won’t be at all satisfying.

Thanks for your help, KentLyons (talk) 17:45, 29 May 2013 (UTC)[reply]

This is a very astute question indeed. I don't have time for a proper answer but will point you to the articles that do (or should). Read Gibbs paradox, and Identical particles. Dauto (talk) 18:00, 29 May 2013 (UTC)[reply]

Thank you very much for the links - those were very helpful articles and concepts

KentLyons (talk) 18:39, 29 May 2013 (UTC)[reply]

Let me add that when you consider observing the atoms this has an impact on the entropy. If we choose units such that kB = 1/log(2), then the entropy is equal to the amount of information measured in bits needed in addition to what is already given, to specify the exact physical state of the system. This means that in thought experiments where you consider following some atoms as they mix after you remove the partition, you need to take into account the amount of information about those atoms that you have. Count Iblis (talk) 18:47, 29 May 2013 (UTC)[reply]

Proto-languages and proto-genomes

If they can reconstruct proto-languages by comparing the features of the descendant languages, why can't they reconstruct "proto-genomes" of extinct species that spawned large clades by comparing the genomes of the species in that clade?

I came up with this question while researching on wikipedia to find out when the common ancestor of both me and my cat lived. This would have been the common ancestor of the clade boreoeutheria. That ancient animal spawned many other descendant lineages in addition to the one that led to carnivores and primates. Similarly, the Proto-Indo-European language has many modern daughter languages, and linguists have (mostly) reconstructed PIE by comparing them.

Could the same techniques be applied to the genomes of all the animals in this clade, comparing all their shared sequences and extrapolating back to deduce which ones were present in the ancestor, to come up with a complete (or near complete) proto-genome? If so, would it also be possible to insert that genome into an ovum and create a living animal? — Preceding unsigned comment added by 209.182.120.111 (talk) 18:24, 29 May 2013 (UTC)[reply]

People have been doing that kind of thing for a long time for specific genes. Doing it for whole genomes is not quite on the cards yet, because whole-genome sequencing is still relatively new and expensive, so the number of individual animals whose whole genomes have been sequenced is still pretty limited. But you can expect to see more of it as time goes on. (Of course some parts of an ancestral genome may simply have been lost in all descendents and therefore can't be inferred.) Looie496 (talk) 18:34, 29 May 2013 (UTC)[reply]

There was a study within the last decade (I think I came acrost it Scientific American) that showed a comparison of the whole genomes of various mammalian orders, with various retained areas on the chromosomes lined up for comparison in a color-coded chart. This was at the genome scale, and it implied a proto-mammalian genome. Unfortunately I have not been able to locate the article, and have looked many times before your question today. μηδείς (talk) 18:45, 29 May 2013 (UTC)[reply]

Here's a link to a similar study with a chart that shows a comparison of the superorders' chromosomes. You may have to register to see the information. http://www.sciencemag.org/content/309/5734/613/F2.expansion. μηδείς (talk) 18:55, 29 May 2013 (UTC)[reply]

Are there similar things to ruin marble?

Just discovered ruin marble, and my mind is rather blown. Curious to know if there are other natural materials that tend to be artistic like this. I'm not talking abstract or geometric art (or flowers that look like floral still lifes). And colour-shifting animals don't count. I've seen octopuses "paint" some amazing things, but so temporary. Anything else (rock, fungus, tree, dog) that naturally and commonly resembles a portrait, landscape or notably "human" design (an ampersand, for example) would be nice to know of. Wouldn't mind starting a collection. InedibleHulk (talk) 19:39, 29 May 2013 (UTC)[reply]

There is stone that has inclusions of Dendrite (crystal) that gives the impression of landscapes. Type dendrite stone into Google image search for examples. Dominus Vobisdu (talk) 19:58, 29 May 2013 (UTC)[reply]

The term for objects resembling other objects is simulacra. --TammyMoet (talk) 20:45, 29 May 2013 (UTC) Here is a list of photos of such things. --TammyMoet (talk) 20:48, 29 May 2013 (UTC)[reply]

Thanks for the vocabulary tips. Those should help. That's a great list! InedibleHulk (talk) 00:00, 30 May 2013 (UTC)[reply]

This is sort of open ended. Desert rose, Butterfly Alphabet, even the legendary Indian face on Mount Rushmore might qualify. I don't know where the reality ends and the hallucination begins, and I'm not sure there is a criterion to determine that. Wnt (talk) 00:08, 30 May 2013 (UTC)[reply]

Sort of, yeah. Sorry. I wouldn't count the desert rose crystal, since flower shapes are far more common in and associated with nature than human art. The butterfly alphabet is more what I'm looking for. The big Indian face is more of a freak occurence. I'm looking for minerals and organisms that fairly routinely look a certain way. Not every piece of ruin marble is as "good" as the example in our article, but they all seem to follow the basic cityscape template. You know? InedibleHulk (talk) 18:58, 30 May 2013 (UTC)[reply]

Um, none of the faces on Mount Rushmore were Indians, and they are definitely all man made. Perhaps you're thinking of the Old Man of the Mountain (c. 8000 BC - AD 2003) (R.I.P.). Maybe the article Pareidolia is a worthwhile read here as well. --Jayron32 00:15, 30 May 2013 (UTC)[reply]

Or perhaps he's thinking of Crazy Horse Memorial. ←Baseball Bugs ^{What's up, Doc?} carrots→ 12:13, 30 May 2013 (UTC)[reply]

Per [2], etc. No doubt in my mind the mountain would look prettier undefaced (or unfaced), and have better character whether seen as stone or face. Wnt (talk) 00:36, 30 May 2013 (UTC)[reply]

You (and the writer of that article) definitely needs to read Pareidolia. --Jayron32 00:38, 30 May 2013 (UTC)[reply]

The BBC did a piece on that just today. Weird. InedibleHulk (talk) 18:02, 31 May 2013 (UTC)[reply]

Did I not say that I didn't know where reality ends and hallucination begins? Nor does a desert rose look like a rose, for that matter, in any meaningful objective sense. Wnt (talk) 00:40, 30 May 2013 (UTC)[reply]

Like SemanticMantis (talk) 01:35, 30 May 2013 (UTC)[reply]

Just about everything in the pictures that get sent back from Mars. Gzuckier (talk) 04:17, 30 May 2013 (UTC)[reply]

See asterism (gemology) for stones that look like they have stars painted on them. Here's a better pic: [3]. StuRat (talk) 08:46, 30 May 2013 (UTC)[reply]

Do a websearch for Snowflake obsidian and see what you find! --TammyMoet (talk) 09:59, 30 May 2013 (UTC)[reply]

Another thing to search for is moss agate (a common name for some stones containing the dendrites mentioned by Dominus Vobisdu above). Our article doesn't show any examples, but these can sometimes exhibit quasi-landscape or quasi-organic forms. [4] [5] —Deor (talk) 11:08, 30 May 2013 (UTC)[reply]

I've come across a lot of cool minerals at GemSelect.com. Opal in particular seems to fairly frequently "design" stormy seascape-like pictures. A tad abstract (and pricey), but pretty cool. Ammolite is pretty sweet, but in a surreal way. But I'll look into those other keywords, too. Thanks!

To be clear, I'm not really looking for "freak" things (like regular granite that happened to erode into an unusual shape, or one of those oak trees with faces). Those are interesting, too, but I'm mainly looking for things that tend to look a certain "artistic" way. The routine smiley face on the Death's head hawk moth, for instance, or the moss agate above. The star gems don't really count, since they need to be helped by a human. They look nice, though. InedibleHulk (talk) 18:49, 30 May 2013 (UTC)[reply]

There's also cotham marble (known as landscape marble), which looks like an English landscape - a result of stromatolite growth - images here. Mikenorton (talk) 07:13, 31 May 2013 (UTC)[reply]

Very cool. Thanks. InedibleHulk (talk) 18:02, 31 May 2013 (UTC)[reply]

Radiation and motion sickness

A friend who's undergoing radiation for throat cancer is stuck with bad motion sickness. How does radiation cause nausea? Nyttend (talk) 22:45, 29 May 2013 (UTC)[reply]

I'm not allowed here to give any specific advice, but researching the general question I found a review here (not a famous journal though) which discusses overall alterations in motility, but sounds pretty confused (2007). One mechanism (I don't know how often though!) is that the altered motility allows bacterial overgrowth [6]. It turns out that there are actually two mechanisms - "early" and "delayed" - the former being due to direct effects on the epithelial cells and barrier function, and the latter due to scarring (fibrosis and sclerosis) which is actually affected by the enteric nervous system, though it doesn't look like this review [7] discusses whether the scarring causes the aberrations in motility down the line. This is only the briefest look I've made, and I am still likely missing very major aspects of the biology ... Wnt (talk) 23:50, 29 May 2013 (UTC)[reply]

As far as I can tell, the answer is not solidly understood. There is clear evidence that the ability of radiation to induce nausea depends on the site that is irradiated, with the intestines and brain being the most sensitive; there is reasonably solid evidence that the mechanism is release by radiation-damaged cells of some substance or substances into the bloodstream, which is detected by the chemoreceptor trigger zone linked to the area postrema. The most popular hypothesis is that the substance in question is serotonin, but that isn't proven. Looie496 (talk) 00:10, 30 May 2013 (UTC)[reply]

I didn't mention before, but 5-HT inhibitors do have therapeutic use ([8] (I didn't figure out if it inhibits cause or symptom, though) Wnt (talk) 00:30, 30 May 2013 (UTC)[reply]

The OP is not asking for diagnosis or treatment. The inner ear is nearby and serves as a balance organ. I'd still ask the doctor though, since he's paid for such things and a lot more knowledgeable than we. μηδείς (talk) 00:42, 30 May 2013 (UTC)[reply]

Thanks for the links. I made sure to ask something that would pass User:Kainaw/Kainaw's criterion (my question is modelled on the "beta strep infection" question); my friend is definitely getting the treatment he needs from professionals, and anyway I'm only hearing news from a distance without anything close to explanations of why it happens. Nyttend (talk) 01:19, 30 May 2013 (UTC)[reply]

These links might be helpful.

Ask the Experts (radiation safety)
Radiation Information - Questions and Answers | Radiation Information and Answers

—Wavelength (talk) 04:28, 30 May 2013 (UTC)[reply]