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February 3[edit]

gravity was not the only force that formed the universe[edit]

Notice to the shape of galaxies they don’t show the spiral formation because of the rotation and Thermo dynamical interactions with gravity . if there was only gravity cased to shape the galaxies they could condensed in all around the central mass . we know some of galaxies are so, with gathering comas in outer zones .also for the solar system we have planetary surface with sun with lens shaped placement .after finding out the properties of sun and planets , that appears to us the rule and mechanism of creation of this system . the gravity cases to create general force and movements in space and also it case general weightless in all over the world for galaxies and stars . --78.38.28.3 (talk) 09:11, 3 February 2011 (UTC) this means that. however the space objects have inertial and gravitational mass . a. mohammadzade[reply]

Hi. This page is for questions about science. Do you have a question?--Shantavira|^{feed me} 09:22, 3 February 2011 (UTC)[reply]

He's been here before, he was pushed away because his "questions" seemed to be theories that he thought of himself. aka original research —Preceding unsigned comment added by 216.120.192.143 (talk) 16:18, 4 February 2011 (UTC)[reply]

Help identifying mineral[edit]

Any suggestions about what the blue (or blue-green) mineral shown in this photo might be? There is an outcropping of this stuff in the East Bay hills behind Berkeley. I have no reason to suspect that it's anything remarkable, just would like to figure out what it actually is. Looie496 (talk) 00:20, 3 February 2011 (UTC)[reply]

A close-up photo, of a freshly chipped of piece would help. The sparseness of veg looks like the soil suffers from metal poisoning.--Aspro (talk) 01:04, 3 February 2011 (UTC)[reply]

I have some smaller pieces, but it's difficult to take a useful photo -- basically the blue stuff looks powdery and nonreflective -- however it also sometimes seems to associate with what looks like very dark blue or dark brown quartz. The lack of veg is a consequence of the fact that this rock lies in the middle of a fire road. Much of the road looks the same. Looie496 (talk) 01:41, 3 February 2011 (UTC)[reply]

Sorry if I'm stating the obvious here but I thought 'copper' as soon as I saw it. Is this some kind of cuprous salt? Is the rock blue inside or just the surface? Richard Avery (talk) 13:43, 3 February 2011 (UTC)[reply]

There are outcrops of the Franciscan Complex in the area, which includes blueschists (at least in some places) that have visible glaucophane and lawsonite, minerals that give it the characteristic blue colour (mainly the glaucophane), but I can't tell if this is what is in the picture. Mikenorton (talk) 22:45, 3 February 2011 (UTC)[reply]

Nothing about geology is obvious to me :-). The underlying rock is sort of opaque-quartz-like, so I think it is more likely to be igneous than metamorphic. The blue is mainly on the surface, but in a few places the quartzish stuff takes on a dark blue tinge. I would readily believe it is a cuprous salt if they can be that color. Looie496 (talk) 01:13, 4 February 2011 (UTC)[reply]

Like Richard Avery, I immediately thought "copper". Given the color variation shown in the photo, my guess is a mixture of secondary copper minerals formed by surface weathering of a copper-containing rock, perhaps including chrysocolla and azurite. As has been intimated, it's hard to tell from just the photo, though copper ores are known to occur in Alameda County. Deor (talk) 03:47, 4 February 2011 (UTC)[reply]

I think most green rocks in NorCal are serpentine. Just a random tidbit from my head; I don't claim this is particularly reliable. --Trovatore (talk) 09:42, 4 February 2011 (UTC)[reply]

Insoluble sodium compounds[edit]

Are there any insoluble sodium ionic compounds? --75.15.161.185 (talk) 00:33, 3 February 2011 (UTC)[reply]

Uranyl zinc acetate is usually highlighted in qualitative inorganic analysis texts as perhaps the only way to precipitate Na⁺ from aqueous solution. DMacks (talk) 00:44, 3 February 2011 (UTC)[reply]

Sodium#Compounds lists a few, but all the compounds I can find include other metals in addition to sodium. (As our article notes, the insoluble precipitate formed with uranyl zinc acetate is uranyl zinc sodium acetate – UO₂)₂ZnNa(CH₃COO)-6H₂O – which contains uranium and zinc in addition to sodium.) If you're restricting your search to compounds which contain sodium as the only metal, you're probably out of luck; if you're willing to accept more than one metal, then you have a lot of choices. TenOfAllTrades(talk) 02:34, 3 February 2011 (UTC)[reply]

Tesla Roadster[edit]

About how much does it cost to fully recharge the Tesla Roadster? ScienceApe (talk) 03:00, 3 February 2011 (UTC)[reply]

The cost of electricity varies considerably across the world, so we'll need a bit more information about where you are. Perhaps first one should ask "how much electricity" is required to recharge? And even that may vary depending on the circumstances. Mitch Ames (talk) 03:48, 3 February 2011 (UTC)[reply]

$10.04 US dollars[1], using the average US price of 11.62 cents/KwH[2]. Ariel. (talk) 03:50, 3 February 2011 (UTC)[reply]

(EC) The 135 Wh/km usage times 393 km range would imply a full charge of 53.055 kWh. I don't know where you live, but I pay my electric company $0.1064 USD per kWh during off-peak hours, so it would cost me $5.65 USD to fully recharge a Roadster from empty during off-peak hours. Red Act (talk) 03:53, 3 February 2011 (UTC)[reply]

Oops, my calculation was based on the 135 Wh/km battery-to-wheel efficiency, but I should have used the plug-to-wheel efficiency, which is around 174 Wh/km. That would change my answer to $7.28 USD. Red Act (talk) 04:00, 3 February 2011 (UTC)[reply]

(ec) From Tesla Roadster#Battery system: "A full recharge of the battery system requires 3½ hours using the High Power Connector which supplies 70 amp, 240 volt electricity..."; that's 58.8 kWh to charge. (The actual amount of available juice is given in the article as roughly 53 kWh; losing about 10% in charging/discharging doesn't sound too unreasonable.) If you're paying the $0.1162/kWh provided in Ariel's response, it will set you back just shy of $7 per 'tank'. The actual answer will depend on your local electricity price; in some jurisdictions you may receive a significant discount late at night and very early in the morning—just when your car is most likely to be in the garage. TenOfAllTrades(talk) 04:06, 3 February 2011 (UTC)[reply]

The section "Recharging" in article says 4 hours and 90A not 3½ hours and 70A, so the article contradicts itself. (The 48hours for 120V and 15A works out to the same amount of energy.) Ariel. (talk) 04:29, 3 February 2011 (UTC)[reply]

The section you're looking at indicates that the high-power charger is connected to a circuit with a 90-amp breaker, but the actual current drawn is the lower 70 amps: [3]. TenOfAllTrades(talk) 13:21, 3 February 2011 (UTC)[reply]

Interesting numbers. Here in the UK we're on a nominal 240V supply, with standard wall sockets on a 30A-fused ring main via a (non-consumer-replaceable) 100A "consumer fuse" which serves the entire premises. Wall plugs are individually fused at a maximum of 13A . My current electricity tariff is about £0.22 ($0.354) per kW/h. So for a start I'd need a custom power supply for the 70A I'd need. 53-ish kW is around £11.60 ($18.97). And none of this is "pollution-free". I can look out of my window and see the power stations that are generating the electricity for "pollution-free" electric cars. Tonywalton ^Talk 03:12, 4 February 2011 (UTC)[reply]

The point is they can be pollution free as I can look out my car window and see spinning wind turbines. Ok, so technically, pollution was created in the manufacturing of the wind turbines and the car and less then 10% of the electricity in the US comes from wind, but you know what I mean. Googlemeister (talk) 16:07, 4 February 2011 (UTC)[reply]

What is the Emergency oxygen system bag for?[edit]

Hi - Concerning Emergency oxygen system

What is the plastic bag for? The standard demonstration says 'bag will not inflate during normal operation'

Thanks — Preceding unsigned comment added by InverseSubstance (talk • contribs) 04:01, 3 February 2011 (UTC)[reply]

It serves as an oxygen reservoir -- see the fourth paragraph in this section: Bag_valve_mask#Components. DRosenbach ^{(Talk | Contribs)} 05:09, 3 February 2011 (UTC)[reply]

The Straight Dope : When you wear an airline oxygen mask, why doesn't the plastic bag inflate? Hope this helps. APL (talk) 06:30, 3 February 2011 (UTC)[reply]

The trick is, they want to blow oxygen at you, as fast as possible, but not so fast that it could be hazardous. So, we need a pressure regulator of some sort. SCBA and SCUBA, (two other cases of breathing air out of a "tube") solve this problem by including a pressure regulator valve. A quick look at the diagrams in types of diving regulator readily demonstrate how sophisticated those systems are: the valves are multiple stages, they have many moving parts, and the breathing apparatus is actually a bit tricky to use (well, you just breathe in and out - but if you've ever used one, you'll know that it feels a little weird breathing over a pressure-differential between inhalation and exhalation - and you must make a tight pressure seal with the air supply, or else the intake valve won't actuate). So, the "bag" on an aircraft is much simpler - it allows for reasonably high-pressure oxygen to be jetted at your lungs without popping them (by acting as a pressure "reservoir"), and also allows gas to flow without requiring a pressure seal. It's the safest, fewest-moving-parts, fewest-things-that-could-break, easiest-to-operate, way to get air out of a hose and into your lungs. Nimur (talk) 17:32, 3 February 2011 (UTC)[reply]

I2Cl6[edit]

What is the shape and structure of I2Cl6? —Preceding unsigned comment added by 122.169.220.58 (talk) 06:30, 3 February 2011 (UTC)[reply]

See iodine trichloride. Someguy1221 (talk) 06:33, 3 February 2011 (UTC)[reply]

Strongest Allergy Medicine[edit]

I am having a hard time finding out what the strongest allergy pill is, both over the counter and prescription. Does anyone know? —Preceding unsigned comment added by 76.169.33.234 (talk) 07:42, 3 February 2011 (UTC)[reply]

That sounds suspiciously like a request for medical advice. We don't do that here. You probably have 10 to 20 minutes to claim that you're researching medicobabble for your next screenplay, before someone summarily deletes your question... –Henning Makholm (talk) 10:03, 3 February 2011 (UTC)[reply]

It's also a question that does not have a good answer. Different pharmacological substances have different effects for different people and different allergies. There very likely is not "the" strongest pill. --Stephan Schulz (talk) 11:00, 3 February 2011 (UTC)[reply]

Any particular reason you can't go to a pharmacy and ask? Seems like the best way. Especially since they'll also be able to limit their answers to what they have in stock. 90.193.232.171 (talk) 11:06, 3 February 2011 (UTC)[reply]

Agree on this. I've done this very thing with pharmacies — shown up and said, "hey, this other thing isn't working for me, what else you got?" and they found me something that did what I wanted it to. This is exactly the sort of thing a pharmacist is good at, potentially better even than a general practitioner. --Mr.98 (talk) 13:30, 3 February 2011 (UTC)[reply]

What happened to the assumption of good faith? I'm not the OP, but I find the question to be an interesting scientific one. Apparently the OP had heard that Claritin is "only 11% more effective than a placebo", whatever that means. So some measure of effectiveness has been applied. If the OP's question is not well defined enough, it would be more helpful if you could explain the nuances and suggest a better formulated version of it. --98.114.146.27 (talk) 10:51, 4 February 2011 (UTC)[reply]

I'm not sure it is an assumption of bad faith to suggest that if they are looking for medication (which it does sound like they are), that they consult someone who can be more useful than we can be. --Mr.98 (talk) 01:37, 5 February 2011 (UTC)[reply]

I thought I could get an easier and faster answer this way. The only reason I asked was because I keep hearing that claritin is great, but then I read that its only 11% more effective than a placebo. I know benadryl is strong, I was just wondering if any anti-histamine was stronger. I don't even use allergy medicine, I was just wondering. —Preceding unsigned comment added by 76.169.33.234 (talk) 07:06, 4 February 2011 (UTC)[reply]

The trick about Claritin is that it is meant to keep you from falling asleep. Benadryl is not — and will make people who take it extremely drowsy in some cases (it makes me feel awful). All of this just illustrates that there's something of an apples and oranges comparison here: "strongest" is not what Claritin is trying to be, at all. It's trying to be "just enough to keep you from sneezing, but not enough to make you feel any other effects." I'm not sure if there is a metric for measuring the "strength" of an antihistamine, but I wouldn't know much about that. --Mr.98 (talk) 01:37, 5 February 2011 (UTC)[reply]

Hamilton, Ichthyologist, active in 1822[edit]

Hi all,
The usual. One Hamilton (possibly Robert) is the binomial authority for Gagata cenia. According to the ITIS, went on a bit of a taxonomic spree in 1822. Who is this guy?
--Shirt58 (talk) 09:45, 3 February 2011 (UTC)[reply]

Francis Buchanan-Hamilton, maybe? Apparently the Hamilton, 1822 notation is used for him. Although there are a surprising number of Hamiltons associated with fish, so... --Kateshortforbob _talk 11:34, 3 February 2011 (UTC)[reply]

<JHS cigar> "I love it when a plan comes together" </JHS cigar>--Shirt58 (talk) 14:01, 3 February 2011 (UTC)[reply]

single hydrogen atom emitting light[edit]

constructive interference (and destructive interference) make much more sense to the eye than to the brain!

when the e- falls it emits one photon. another jump up, then down, another photon. so, we get successive distinct portions of emitted energy. how can this become a wave with a wave front and continuous (harmonic) variation in intensity? (don't tell me about duality right now) should we have many sources (atoms) that have to emit in perfect phase to get a spatial wave front? should we consider many sources nearly in phase that by superposition give continuity? — Preceding unsigned comment added by M121121121 (talk • contribs) 11:32, 3 February 2011 (UTC)[reply]

A single hydrogen atom only emits single photons, but a real-world light source emits A LOT of photons. A photon of red light has energy of 2.84 x 10^-19 joules[4]. A one-watt light source will emit one joule per second, or around 3.5x10¹⁸ photons per second. At this sort of level, light appears to be smooth and continuous. A 40W incandescent bulb emits about 4W light (source: Incandescent light bulb). --Colapeninsula (talk) 11:52, 3 February 2011 (UTC)[reply]

When you drop a stone into a pond, you get more than one wave. Same, for different reasons, with the electron dropping down a shell of an atom. (Don't read that rigth now: Wave–particle duality). 95.112.210.107 (talk) 13:20, 3 February 2011 (UTC)[reply]

Also relevent, quantum mechanical properties are best understood only in bulk, not in isolation. Understanding the quantum behavior of electrons and photons in atoms makes much more sense than trying to work out the quantum behavior of an electron and a photon in a atom. Indeed, you can't even definitively say where one electron is located around an atom. You can say how electrons will behave when the atoms of a material are subjected to a light source. But talking about what happens to one electron as a result of one photon strike is only useful as an illustrative model, not as an accurate one. --Jayron32 13:46, 3 February 2011 (UTC)[reply]

Jayron, bite your tongue before you start talking about things you don't fully understand. There is nothing wrong with talking about the behavior of a single isolate atom. Dauto (talk) 15:53, 3 February 2011 (UTC)[reply]

In good faith, Jayron's comments are consistent with the ensemble interpretation. Perhaps not the currently most popular interpretation, but still reasonable. SemanticMantis (talk) 17:15, 3 February 2011 (UTC)[reply]

What makes you think there is always going to be a perfectly smooth in phase wave front? Dauto (talk) 15:56, 3 February 2011 (UTC)[reply]

You might find Coherence (physics) useful. Dauto (talk) 15:58, 3 February 2011 (UTC)[reply]

If I understand correctly, I think that the original poster is confused as to why a given emission adds up (constructive interference) with previous emissions. After all, you'd think that there would be a 50/50 chance it would be out of phase with the existing light and cancel it out. But light is typically incoherent, coming from many sources and going in many directions. It is like tossing two pebbles into a pool - even though you'd think that the waves could as easily interfere destructively as constructively, the net result is still a bigger commotion. Now the reasons for this are complicated and involve actual math, but the video I've added above should show you the basics: wherever the two sources interfere destructively, they are still adding up constructively somewhere else. There's simply no way for destructive interference of waves to squirrel away energy under the carpet and forget about it - it always comes out somewhere. Wnt (talk) 18:38, 3 February 2011 (UTC)[reply]

Photons do not destructively interfere in a way analogous to waves in a pond (rather their wave functions do). If your source emits n photons, regardless of their interference you will receive n photons in your detectors (energy is conserved). The interference is manifest in their distributions, not their numeracy. The interference patterns will effect the probability of a photon being found in a given place, but in an incoherent source their is no mean deviation from the probability distribution of a single photon.

there is a law about the quantified levels of energy, but no one for quantified moments of emission of photons. so, no correlation, no coherence (unless stimulated emission - not our case). we have a flood of un-correlated photons. no wave front - not smooth, not rough. what is frequency in this case? just looking for the one measured as the number of complete cycles per second, not that value computed from difference of energy of e-. M121121121 (talk) 23:03, 3 February 2011 (UTC)[reply]

I'm not sure I'm interpreting the original question right, but it sounds like perhaps you're expecting the wave nature of light to arise from multiple instances of an electron changing state back and forth, one pair of transitions for each wavelength of light. That isn't what happens. One single transition of one electron between two atomic orbitals produces one photon that all by itself consists of a wave packet that's many wavelengths long. The electron takes as long to make the one transition as the duration of the photon's entire wave packet. The wave nature of the photon arises directly from the wave nature of the electron. Red Act (talk) 00:38, 4 February 2011 (UTC)[reply]

Yes, I think the OP is confusing quantized photon emission at the atomic level with classical, electrodynamic effects that can cause coherent photon emission as a result of macroscopic movement of electric charge. We should be very clear: photons are emitted from an atom via a different physical process than the way that large amounts of electric current can create a macroscopic time-varying electromagnetic field in a conventional antenna. Nimur (talk) 02:01, 4 February 2011 (UTC)[reply]

The slow decay of deuterium atoms via the process of neutrino absorption releases one proton from its nucleus in addition to a cone of blue light'. This is analogous to a hydrogen-1 atom's light emission. ~A H 1^(TCU) 19:41, 6 February 2011 (UTC)[reply]

Magnets- where's the potential energy?[edit]

This probably sounds like a homework problem, but it's not. It's just an example I concocted with which to form my question about a concept. Imagine the following setup: a strong magnet with mass m is connected to a horizontally rolling trolley hanging from the ceiling at a height of h. Just for the sake of simplicity, imagine the wheels of the trolley are frictionless. A block of iron of mass M sits on the ground but far enough from the magnet that it stays on the ground. The total energy apparent to me is just the mechanical potential energy because of the magnet at height h, or mgh. Now you move the magnet to where it's above the block (which didn't take energy to do because of the frictionless wheels and no change in height). Suddenly the block jumps up to the height h' so that the total energy is mgh + Mgh'. So where was the potential energy in the amount of Mgh' in the before case? Did the energy to jump up come from slight slowing down of all the jiggling of the electrons in the iron as it aligned with the magnetic field or something? As much layman's terms and simplification without losing the essential information to explain this would be greatly appreciated. Thanks. 20.137.18.50 (talk) 15:35, 3 February 2011 (UTC)[reply]

The final configuration has some (negative!) magnetic potential energy that must be included in the energy budget equation. Potential energy can often be negative. Dauto (talk) 15:46, 3 February 2011 (UTC)[reply]

As in 'in the before case, the iron block has the potential to align with any given magnetic field that comes its way, but in the after case, it doesn't have as much potential to do so because it's already used much of that potential to wrap up with m'? 20.137.18.50 (talk) 15:55, 3 February 2011 (UTC)[reply]

Just to correct one thing: It does take energy to move the stationary magnet from one location to another even if the wheels are frictionless. As long as the magnet has mass, it will take a force to move it, and Work = Force * Distance. Not sure how it affects the final answer, but thats something that needs to be fixed. Even in my self-admitedly retarded understanding of physics, I am pretty sure that you cannot move a real object without some energy, even in a frictionless environment. --Jayron32 16:09, 3 February 2011 (UTC)[reply]

You only need a force to accelerate it, but in the end you decelerate it, hence you can regain the kinetic energy transferred to the magnet. You can also move it arbitrarily slowly. How you do that in practice is another matter (in practice there is no frictionless environment), but in theory there's no problem. --Wrongfilter (talk) 16:37, 3 February 2011 (UTC)[reply]

(ec) Under the usual physics approximations (frictionless, vacuum, etc.) you can have a displacement of a mass without any net energy consumption. Force and displacement are actually properly treated as vector quantities (see Work (physics)), so you counter the work done to start the object moving with an equal amount of 'negative' work in the opposite direction to bring it to a halt in its new location. The object starts and finishes with the same amount of energy (kinetic and gravitational), so no net work is done.

(I will note parenthetically that the particular case the original poster describes is a bit more complicated. One has to account for the change in potential energy due to the motion of a ferromagnetic block relative to a magnetic field and so there is work done there, but this is a separate consideration from the 'horizontal displacement on a frictionless track' issue.) TenOfAllTrades(talk) 16:41, 3 February 2011 (UTC)[reply]

The fact that the magnet is "above" the track is irrelevant. The potential-energy due to gravity, plus the potential energy due to the magnetic field, result in a local minimum of energy at that location. This question is essentially equivalent to asking "if a ball rolls down a hill, where did it get the energy from?" Well, it took energy to put the ball at the top of the hill in the first place. Similarly, the fact that the iron block was far away from the magnet means that it started with potential energy, and used that energy to counteract gravity, and is now stuck at the bottom of the potential energy well (the "bottom" of the well happens to be physically located above ground-level). To get the magnet back down, you'll have to do work! It may seem a little strange, but it will require a force to pull the iron box back downward, even though we typically like to think that gravity causes things to fall down by themselves. In this case, gravity alone isn't strong enough, so you'd have to do work to counteract the magnetic potential energy. Nimur (talk) 17:44, 3 February 2011 (UTC)[reply]

I think I see, thanks for that explanation. How would one represent that potential energy which is equal in magnitude to Mgh' in the before case? In some kind of inverse squared thing in terms of the linear distance from the block to the magnet, the flux density of the magnetic field, and some properties of the iron block? 20.137.18.50 (talk) 18:11, 3 February 2011 (UTC)[reply]

Concocting a model of magnetic potential energy that is both theoretically valid and also experimentally useful is difficult; (it is for this reason that most introductory physics courses gloss over the notion of magnetic potential energy). Magnetic potential is a vector quantity; and to make matters worse, ferromagnetism is pretty non-ideal in that ferromagnetic materials have a very nonlinear magnetic permeability (in other words, they interact with the magnetic field - which we expect - but it's hard to write a simple equation describing how they react, because so many statistical processes are happening with all the crystal lattices and electrons in atomic-orbitals). But here's a rough paintbrush stroke of how we would do it: we could assume the magnetic field is caused by a perfect magnetic dipole; and we could model the field B at all points (x,y,z). And we could assume we have a block of iron with a straightforward, effective (constant-value) μ. Then we could estimate an effective magnetic vector potential A by inverting the equation-set:

\mathbf {B} =\nabla \times \mathbf {A} ,

\mathbf {E} =-\nabla \phi -{\frac {\partial \mathbf {A} }{\partial t}},

\mathbf {B} =\mu \mathbf {H}

And we could estimate a potential energy with value equal to magnitude of |A|. All of this is possible; it is reasonable to expect that a moderately intelligent physics student could figure out how to do this; but the math is very hard, and the resulting answer depends on how well our field is modeled with a dipole, and how close the ferromagnetism is to a "constant" permeability (... neither of which are good assumptions - in fact, μ is a nonlinear function of B). Attempting a better model would require more complicated (probably numerical) inversion of those equations. Or, we could perform a simple experiment and measure the force required; but because magnetism is described by a vector field, the orientation and velocity of approach can change the effective force felt. So it's not easy to give a "straightforward" answer to "how much" magnetic potential energy is present. Nimur (talk) 18:28, 3 February 2011 (UTC)[reply]

I don't understand why you accept the gravitational potential energy (which you call mechanical potential energy) without question, but want an explanation of the magnetic potential energy. There's not much difference between them. In either case, the potential energy is stored in the field itself. You can assign an energy density to an electromagnetic or gravitational field (I'm ignoring general relativity here). The energy is a function of the lengths of the field vectors, but when you combine two fields (by the superposition principle) it's the field vectors that add, not their lengths. Therefore, the energy of the sum of two fields may be different from the sum of the energies. That difference is called the potential energy. The difference may be positive or negative, though the total field energy is always positive. -- BenRG (talk) 08:54, 4 February 2011 (UTC)[reply]

Land[edit]

copied from humanities reference desk. Itsmejudith (talk) 16:35, 3 February 2011 (UTC)[reply]

Hi, you know every land mass on earth like Eurasia, Africa, The Americas etc are they all like embedded in the seabed or do they just bob like a cork so they can move gradually?— Preceding unsigned comment added by Hadseys (talk • contribs)

Plate tectonics has the explanation. Moving this to science reference desk. Itsmejudith (talk) 16:32, 3 February 2011 (UTC) Copying it there, look for answers there not here. Itsmejudith (talk) 16:35, 3 February 2011 (UTC)[reply]

Plate tectonics go deeper than the seabed. From the point of view of plates, land and seabed are the same thing; the only difference is that one is above the waterline. —Preceding unsigned comment added by 205.193.96.10 (talk) 17:02, 3 February 2011 (UTC)[reply]

Actually, I've been wondering something about this myself in the context of the Appalachian mountains. According to what I've read, there is a sole thrust fault at the level of weak Cambrian strata which has displaced the land by kilometers relative to the underlying strata, folding up the upper layers. (See e.g. [5] - I could redraw a figure following an offline source I have lying around somewhere if need be) It is like "continental drift" on a continent! What I don't get is, where did the "excess" of pre-Cambrian strata go? Was it subducted, did it replace ocean crust - it couldn't have emerged out the far side with nothing on top of it, right? - it just confuses me. (Note that whatever goes on with that, it is only a very minor effect relative to the thousands of miles that continents and oceans move as single plates on top of the mantle) Wnt (talk) 18:51, 3 February 2011 (UTC)[reply]

At least one scientist seems to believe the "excess" crust is now buried underneath the overburden: Thin-skinned tectonics in the crystalline southern Appalachians... (with seismic profiling pictures, if you have access to the full article). Some of the crust has steeply dipped and is thus subducted; much of the crust is simply lateral with new material on top. The effects of erosion in the Appalachians should not be understated - they are very old, geologically, and a lot of crust material has been removed by wind, water, and other erosive forces. Nimur (talk) 19:31, 3 February 2011 (UTC)[reply]

That's not entirely true. The crust associated with the land is necessarily less dense than oceanic crust, which explains why the land "floats" so much higher on the mantle than sea floor does. The formation of low density crust is a consequence of multiple episodes of chemical fractionation that occurs via melting (e.g. magma formation) that is associated with the cycles of crustal formation and subduction. The bulk of this process actually occurred in Earth's distant past (see: cratonization), but the volcanic activity at subduction belts today is a small-scale analog of the processes that built continents in the past. Dragons flight (talk) 22:33, 3 February 2011 (UTC)[reply]

Continental crust is what you're looking for. It sits above continental shelf and is much thicker and more "porous" than the oceanic crust, if you will. ~A H 1^(TCU) 19:38, 6 February 2011 (UTC)[reply]

Slimming tip[edit]

I keep reading "Cutting down a bit of the belly each day is easy by simply using this 1 weird old tip. Click here to see what it is...". What is it all about? Kittybrewster ☎ 17:10, 3 February 2011 (UTC)[reply]

"There's a sucker born every minute." The tip in question is just "buy lots of acai berry, it's a miracle food! You'll lose weight, instantly!" It's patent nonsense and the FDA has issued warnings to many of these websites for claiming miraculous health and diet effects that are entirely unproven. --Mr.98 (talk) 17:27, 3 February 2011 (UTC)[reply]

Thank you. I knew you would know. I won't buy any. Kittybrewster ☎ 18:11, 3 February 2011 (UTC)[reply]

(Computing Desk visitor says:) I recommend the use of Firefox and AdBlock. I no longer actually see those annoying belly-pinching ads. Comet Tuttle (talk) 21:50, 3 February 2011 (UTC)[reply]

If the "weird old tip" has the effect of "eat fewer calories per day" then there are lots of "tips" that actually could reduce your weight. For example, "eat 5 raw carrots with every meal" would probably reduce your calorie intact because you'd be full of carrots, and hence you'd lose weight. Some weird old tips are like this. However, if the tip is "buy this ancient miracle product" (which is the typical "tip" associated with web adverts) then it is more than likely a scam. Dragons flight (talk) 22:23, 3 February 2011 (UTC)[reply]

The "weird old tip" is just a strange way to make it sound like homey, authentic knowledge, as opposed to "new", "scientific" knowledge that scares the type of people they want to sell to (unfortunately quite common these days — something not helped by having a completely FUBAR health care system). Acai berry isn't even an ancient remedy or whatever; it's just another berry. It's just a fad. By the way, there's nothing wrong with acai berry food products. But (as I understand it), they are no different nutritionally than eating blueberries. They're good for you, sure. But they're not medicine, and they're not miraculous. And probably not worth the additional cost premium added by all of this advertising and nonsense. --Mr.98 (talk) 00:32, 4 February 2011 (UTC)[reply]

Wikipedia has an article on Enforcement actions against açaí berry supplement manufacturers.--Shantavira|^{feed me} 10:19, 4 February 2011 (UTC)[reply]

I never clicked on such an ad. I just assumed that the weird tip was either to eat fewer calories, or exercise more. Googlemeister (talk) 16:02, 4 February 2011 (UTC)[reply]

I used to see adverts promising "one mum's discovery" or whatever for a brighter smile. I'd always assumed it would be something like 'brush with bleach, which we happen to sell' or 'brush with bleach, and while you're reading this we'll upload something to your computer', so I never clicked, but now I'm wondering if it was acai berries too. Did anyone here ever click it? 86.164.58.119 (talk) 17:36, 4 February 2011 (UTC)[reply]

"WHY" does diamond form out of graphite?[edit]

It's a stupid question,I know.But I don't understand it "thermodynamically", because diamond has a higher enthalpy level than graphite and -seemingly- less entropy.Does diamond have a higher entropy than graphite or the Gibbs free energy equation does not work in conditions in which diamond is formed?--Sina-chemo (talk) 19:05, 3 February 2011 (UTC)[reply]

This "gem" of thermodynamic wisdom may help: "The nucleation and growth of a crystal are under kinetic, rather than thermodynamic, control." Here's a whole textbook on the subject, too: Crystals: growth, morphology, and perfection by Ichirō Sunagawa. From Chapter 3 of that book, excerpts of which are available online, it is evident that to thermodynamically consider crystal nucleation, you have to consider the kinetic processes as well - you can characterize these kinetic behaviors statistically (thermodynamically) by constructing an effective surface free energy potential for nucleation sites. Lattice thermodynamics is a huge ongoing area of research in materials and solid-state physics. Nimur (talk) 19:10, 3 February 2011 (UTC)[reply]

Note that enthalpy is equal to internal energy + pressure*volume. Now because diamond has density 3.52 whereas graphite has density 2.1-2.2, the volume will be a lot different for the two. The higher the pressure, the lower the enthalpy of diamond relative to graphite. Wnt (talk) 19:14, 3 February 2011 (UTC)[reply]

not a medical question, but if something smells very moldy and you're near it, is that comparable exposure to mold as the stuff people get sick from?[edit]

this isn't a medical question, but on a parts per million basis, if you are near something moldy (like a couch) that you can smell is very modly because it was from the basement and is just moldy, is smelling that single source of mold comparable to the level of exposure that people in moldy environemnts get sick from? In other words, is the nose very very sensitive, like the eye is to light in total darkness, or, on the other hand, is it the case that if it smells quite moldy then the level of exposure is on the same dangerous scale as being in a moldy environment itself is? This isn't a medical quesiton, just hypothetical... I'm really interested in the science of it. 217.136.93.63 (talk) 23:36, 3 February 2011 (UTC)[reply]

The answer is it depends. Some molds are benign, and others are rather toxic. Also, some people have a much higher sensitivity to molds than others, so any health effects you experience from exposure to a mold will depend on both what kind of mold it is, and what your body does with it. It is literally too complex to make any broad statements on your particular situation. See Mold health issues for a full list of the possible problems; again noting that you may or may not experience any of these with all of the caveats I listed above. --Jayron32 04:41, 4 February 2011 (UTC)[reply]

The Miasma theory of disease that considered that disease was spread only by foul air is largely obselete. Cuddlyable3 (talk) 10:30, 4 February 2011 (UTC)[reply]

So it's a good thing it has nothing to do with the question or with Jayron's answer, then. Matt Deres (talk) 14:18, 4 February 2011 (UTC)[reply]

@Cuddlyable3: It is, but considering that Airborne disease is a very real phenomenon, I don't see why your comment has anything to do with the discussion. --Jayron32 19:16, 4 February 2011 (UTC)[reply]

The point to be made is that a subjective impression of "smells quite mouldy" is not a reliable guide to the disease danger of an environment. Unlike most mammals' noses the human nose is poorly developed, its sense is more habituated than trained for objectivity, it is too insensitive to track prey, and it has no capability for "binostrillar" direction finding. The OP is interested in a scaling together of physical particle concentration in parts per million (of what?), mouldy smelliness, and a scale of danger. However one cannot by smell alone reliably detect microorganisms that cause diseases such as infections and allergic reactions. Examples of the ease of influencing or misleading the nose are perfumes that seem to freshen polluted air, and unpleasant odorants that are added to dangerous odorless cooking gas as a warning. Cuddlyable3 (talk) 03:03, 6 February 2011 (UTC)[reply]