Wikipedia:Reference desk/Archives/Science/2013 September 11

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September 11

Arrhenius base

To be defined as an Arrhenius base, does the molecule have to release hydroxide ions, or can it rather generate hydroxide ions?

Compare elemental sodium: 2 Na + 2 H₂O → 2 Na⁺ + 2 HO⁻ + H₂,
With sodium hydroxide: NaOH (s) → Na⁺ (aq) + HO⁻ (aq). Plasmic Physics (talk) 01:46, 11 September 2013 (UTC)

See Acid-base reaction: sodium metal would be defined as a Bronsted-Lowry base, but not an Arrhenius base. 24.23.196.85 (talk) 06:38, 11 September 2013 (UTC)

Really? I've never heard of sodium accepting a proton to become NaH⁺ under ordinary conditions. What about NH⁻
₂?

NH⁻
₂ + H
₂O → NH
₃ + HO⁻

Even though the amide accepts a proton, it also generates hydroxide; is just a BL base or is it also an A base? Plasmic Physics (talk) 07:08, 11 September 2013 (UTC)

It is also an Arrhenius base. The definition of an Arrhenius base is that it generates excess OH- ions in water solution. It can either generate those ions by adding them directly (i.e. NaOH) or by reacting with the water to extract a hydrogen ion from it, leaving the OH behind (i.e. NH3). What makes something an Arrehenius base is the generation of excess OH- when added to water. Or, if you prefer, if a water-based solution of it has a pH greater than 7 (which is functionally the same). --Jayron32 14:43, 11 September 2013 (UTC)

Once this all gets sorted out here, please make sure Acid-base reaction#Arrhenius definition and Acid#Arrhenius acids match it. One of those defines it as "dissociates in water to form hydroxide" and the other as "decreases the concentration of hydronium ions when dissolved in water". The first of those sounds restricted to only those that themselves contain hydroxide rather than also possibly causing water to dissociate to give it by some other means. This issue really needs to be resolved via high-quality cited references. Some leads (thanks IUPAC!) might be: Arrhenius, S., J. Am. Chem. Soc. XXXIV, 4 (1912)Bates, R.G. and Guggenheim, E.A., Pure Appl. Chem., 1, 163 (1960) DMacks (talk) 15:03, 11 September 2013 (UTC)

Evaporation before formation

If the time stops around event horizon, a black hole to us would never collapse into a singularity; not only, if we accept the Hawking's radiation a black hole would evaporate in a FINITE TIME, even to us; so any object falling into a black hole would evaporate with the black hole itself without ever crossing the event horizon; even if there were primordial black hole with a singularity would be for the same reason impossible for an infalling oblject (or people) cross the event horizon. So, the final point is: it is impossible to cross event horizon, because the infalling observer observer would evaporate before crossing it (as part of black hole). This would mean that is impossible, even for the infalling observer, to surpass the event horizon? 95.239.192.130 (talk) 11:29, 11 September 2013 (UTC)

From whose perspective?

The whole thing you are missing here is that all frames of reference are equally valid - there is no "right" or "true" or in any way "special" frame of reference.

Sure, from my perspective as an observer far from the black hole, an infalling object gets squished into an infinitely thin splat on the event horizon and stays there forever.

From the perspective of the infalling object, (aside from likely being shredded by tidal forces and irradiated into oblivion), they sail through the event horizon with graceful ease while the outside universe goes on fast-forwards to it's eventual demise.

From the perspective of the black hole itself...who knows? Things get weird because it's a singularity.

From the perspective of a different object that's falling into the same black hole a few seconds later...yet different things.

So it's all about which frame of reference you pick. What's worse is that every fundamental particle within the in-falling object has it's own frame of reference. Usually, it's a fair approximation to say that for something the size of a human being, the frames of reference for all of our atoms are about the same...but with the crazy tidal forces and steep gravitational gradients near to a black hole, every atom is running on it's own clock - so even more crazy stuff happens because of that. What does it mean for chemical reactions when the rate of time's passage is radically different between the atoms taking part in it? Predicting outcomes under such extreme conditions is exceedingly difficult.

SteveBaker (talk) 12:29, 11 September 2013 (UTC)

My thought was another. I simply said that if a black hole evaporates, even an infalling observer would never cross the event horizon because the black hole would disappear (by evaporation) before it happens. 95.239.192.130 (talk) 13:23, 11 September 2013 (UTC)

In which frame of reference? The question (and answer) is meaningless without specifying that. SteveBaker (talk) 13:37, 11 September 2013 (UTC)

For us (external observer) a black hole would evaporate in a FINITE amount of time (that varies with the size of a black hole) and on object falling into a black hole would (ever for us EXTERNAL OBSERVER) take an INFINITE amount of time to cross event horizon; so we would see black hole evaporating before any object could enter (and maybe ever reach) the event horizon. But even an infalling observer would see black hole evaporating before he crosses the event horizon (obviously for the gravitational time dilation for him the black hole would evaporate in a tiny fraction of second but howewer BEFORE CROSSING event horizon); so what i wanted to say was that it's impossible, if exists Hawking's radiation, to enter into a black hole; would be possible only if a black hole doesn't evaporate. This simply because a black hole would evaporate before any external thing could falling into (I think that the feeling for an observer falling into would be to cease to exist in the moment he reaches event horizon). 95.239.192.130 (talk) 14:25, 11 September 2013 (UTC)

You want to take into account the effects of black hole evaporation, but you can only do that that in a fully self-consistent way. This means that you need to write down the Schrodinger equation for the matter fields and space-time metric (the wavefunction is a "functional" of the fields, i.e. it assignes to each field configuration a probability amplitude). This is done in this article, and they reach the conclusion that there two possibilities, one is similar to what you claim here, the other is the conventional view. Count Iblis (talk) 13:57, 11 September 2013 (UTC)

I understood the first possibility (the one similar to my thought) but I didn't understand the other (the conventional view): what does it mean?95.239.192.130 (talk) 14:25, 11 September 2013 (UTC)

The conventional view (which by the way is the one that is most likely correct and that's why it is conventional) states that from the point of view of the infalling observer NOTHING special happens at the event horizon and the observer reaches the singularity after a finite amount of time. 64.56.89.67 (talk) 14:47, 11 September 2013 (UTC)

How the conventional view can conciliate with the Hawking's evaporation theory? 95.239.192.130 (talk) 15:02, 11 September 2013 (UTC)

Without too much difficulty (The only real sticking point is the information paradox. 64.56.89.67 (talk) 16:06, 11 September 2013 (UTC)

It's worth mentioning that from the point of view of an outside observer, black holes of any reasonable size take vastly longer than the life of the universe to evaporate...so this may not be a practical problem because from the perspective of the in-falling observer, the universe ends before they reach the event horizon anyway.

No, that's not right. from the point of view of the infalling observers they reach the singularity after a finite amount of time and NOTHING particularly interesting happens at the horizon. 64.56.89.67 (talk) 16:51, 11 September 2013 (UTC)

What's the idea of ​​the antigravity engine and how it works

What's the idea of ​​the antigravity engine and how it works — Preceding unsigned comment added by 37.238.81.178 (talk) 13:25, 11 September 2013 (UTC)

The idea is that it can generate anti-gravity, or be shielded from gravity. How it works is whatever you want, because it is fictional. 217.158.236.14 (talk) 13:34, 11 September 2013 (UTC)

There is no such thing as "an antigravity engine" (nor is there ever likely to be) - so it doesn't "work". If you're talking about some fictional device (E. E. "Doc" Smith used such a device in his "Skylark" series, for example) - then, um, the unobtainium flange remodulator disencabulates the phase differential of the graviton-muon interaction cycle which in turn de-mitigates the...ok...look, just make up your own B.S explanation and enjoy whichever book/movie/video-game/serious-theatrical-production you're interested in. SteveBaker (talk) 13:35, 11 September 2013 (UTC)

I can think of a couple theoretical ways to counter gravity, while stationary, on Earth:

1) Put a very large, dense mass directly above you. The attraction towards that mass will cancel the Earth's gravity. Ordinary matter wouldn't work, as you would need a mass equal to Earth if at the same density as earth. However, some super dense matter, like in a neutron star, would allow you to use far less, since it would be far closer, on average.

2) If something is found which repels mass, rather than attracting it, then you can gather a larger amount of that below the person, to counter the gravity from the Earth. Dark energy appears to do this, but is extremely diffuse. I'm not sure if it can be concentrated.

And of course, in either case, you'd have the problem of how to contain those materials.StuRat (talk) 13:57, 11 September 2013 (UTC)

So "No" then? SteveBaker (talk) 14:30, 11 September 2013 (UTC)

Yes, "no" then. 64.56.89.67 (talk) 14:38, 11 September 2013 (UTC)

No confirmed. 217.158.236.14 (talk) 14:58, 11 September 2013 (UTC)

Well, certainly no today. But in the distant future, maybe. StuRat (talk) 09:54, 12 September 2013 (UTC)

Negative mass is an interesting read. Two objects of equal but opposite mass would continually accelerate towards the one with positive mass - the gravitational force will have a negative sign, making it a repulsive force for the positive mass object. The negative mass reacts oppositely (pushing it causes it to accelerate towards you), so the "repulsive" force actually attracts it. They will have opposite mass, momentum and energy, so the mass, momentum and energy of the system is zero. If you could create something with negative mass and use it this way, I guess you could call it an antigravity engine. Katie R (talk) 15:09, 11 September 2013 (UTC)

Too bad negative mass objects don't seem to actually exist. 64.56.89.67 (talk) 18:13, 11 September 2013 (UTC)

The anti-gravity article is interesting. The idea seems unlikely, but we should never exclude the possibility that there are flaws in current theories. Remember, we're using a model of the universe in which there is no preferred frame, even though there is cosmic microwave background and practically everything in any given region of space is moving at 0.01c or less relative to it. And even within entirely legitimate, accepted relativity there are gravitomagnetic effects, such as the attraction of two wheels to one another depending on which way they are spinning, which potentially modify ordinary gravity as we know it. (Though making them big enough to be measurable, let alone usable, is not so simple) Wnt (talk) 18:50, 11 September 2013 (UTC)

There don't need to be significant flaws in current theories, just incompleteness. The Standard Model of quantum physics is not assumed to describe every particle that might exist in the universe, so there is always the possibility that something may yet be discovered with exotic properties. On the same note, there is no reason to suspect that we will discover a useful exotic particle any time soon. Someguy1221 (talk) 10:14, 12 September 2013 (UTC)

Power supply problems.

(Not 100% sure that the science desk is the right place for this - but people with big brains and electrical knowledge hang out here so...)

I have a small 24 volt, 3.2 amp, switched-mode power supply for my number-two laser cutter (The data sheet is here). The machine has been giving me trouble ever since I first built the it - and it turns out that the problem may have been that the power supply was not generating the 24 volts that my machine needs. The data sheet linked above says that the output can be adjusted between 22 and 27 volts - and there is a prominent blue potentiometer marked "ADJUST". So I assumed that I just needed to tweak the potentiometer and all would be well. I know that the under-load voltage may be different than the no-load volts on a switched mode supply - so I set the laser cutter off working and put a multimeter onto the terminals of the power supply. Lo and behold, the meter read 22 volts - so I slowly turned the pot and got it up to 25 volts (because I wanted a volt or two over 24 to allow for voltage drop though the circuitry - and the stepper motor controller that it's powering is happy up to 80 volts). But just as the multimeter reached 25 volts, there was a burning smell, the green LED on the power supply went out and my laser cutter stopped moving.

:-(

I have another power supply on order (they only cost $25) - but I don't want that one to undergo the same fate!

So here is the question: It seems like the power supply must have been faulty because it could not be adjusted up to the maximum 27 volts that the manufacturer claims. But what kind of failure mode would have caused this to happen - and could that explain why it only produced 22 volts in the first place? It's hard to imagine why it would do that. The two motor controllers that it's powering pull at most 1.5 amps each - so I wasn't exceeding the current limits - and it had been running for weeks (presumably at 22 volts) without any obvious problems other than that my motors were somewhat lacking in torque.

Any ideas?

SteveBaker (talk) 13:28, 11 September 2013 (UTC)

Lambda makes very nice power supplies, although we tend to use almost identical-looking Meanwell supplies here. In my experience they're usually very stable at the set voltage through all sorts of load conditions, and the Meanwell ones (and presumably the Lambdas) are factory-set to the advertised voltage. If it was running low in the first place, then I think it is likely that you simply had a faulty unit. I would run the new supply with no or light load to make sure it shipped at 24 volts, then run your full load. Tweak the potentiometer with the meter as close to the load where you need the 24V as possible to compensate for the voltage drop along the way rather than putting the meter on the supply and guessing at how much to push it up. The little bit you pushed it past 24V certainly shouldn't have been an issue - they expect you to push it a bit for exactly the reason you did. We've had another manufacturer's supplies get stuck in a low-voltage failure mode that requires a power cycle to clear, and could sometimes get spuriously triggered, but it would have dropped far lower than 22V if it had that "feature". Katie R (talk) 13:54, 11 September 2013 (UTC)

The power had been cycled many times - and the machine has had marginal motor power since we first built it - so I doubt that it had gotten stuck in some resettable failure mode. But I'll take your advice to measure the voltage up by the motor controller - I was indeed measuring at the terminals of the power supply just because it was easier to juggle two multimeter probes and a screwdriver in the cramped electronics bay while keeping all body parts well away from the 40,000 volt laser power supply!

What I'm curious about is the nature of a fault that could have the thing operating well at 22 volts and self-destructing at 24. SteveBaker (talk) 14:25, 11 September 2013 (UTC)

You need to get yourself some nice probes with clips! You should be able to hook up the probes, sit the meter somewhere convenient (or stick it somewhere if you get a case with a magnet on the back), and then have your hands free for adjusting the pot. Hopefully your post-mortem teaches you something - I don't know enough to be able to speculate on the failure. I am surprised it made it through their QA process. Katie R (talk) 14:48, 11 September 2013 (UTC)

Do you use the same supply on your other laser cutter? It would be interesting to see what voltage it's putting out under the same conditions. Katie R (talk) 13:59, 11 September 2013 (UTC)

Sadly, no. The other laser cutter is an older version with a completely different power supply setup. SteveBaker (talk) 14:25, 11 September 2013 (UTC)

I'd want to do a postmortem and find where the burned spot is in the power supply. That spot was probably also the reason for the low voltage to begin with. I'd also complain to the company. They may or may not replace it, but they should know they are making defective equipment, in any case. (Was it made in China, by any chance ?) StuRat (talk) 13:50, 11 September 2013 (UTC)

TDK (and TDK-lambda) is a Japanese company. I don't know where the power supply was made - but TDK have a fairly solid reputation. When I pull the old power supply out of the machine, I'll dismantle it and look to see where the Magic smoke got released from. It has a three year warranty from TDK - so we'll definitely send it back...be we needed a replacement in a hurry - and having a spare doesn't hurt. SteveBaker (talk) 14:25, 11 September 2013 (UTC)

A Japanese company might still farm out production to China, and get lower quality products as a result. I'd look specifically for a label saying where it was manufactured. StuRat (talk) 07:44, 12 September 2013 (UTC)

The datasheet says the supply has overcurrent protection. Nevertheless, there might be some value in setting up a meter to measure the output current, in case it isn't what you think it is. Jc3s5h (talk) 15:56, 11 September 2013 (UTC)

I'm not sure how it would cause your particular failure, but I suggest using a scope to check a questionable power supply, since a multimeter's reading does not typically disclose brief sags/surges/spikes which could be caused either by the power supply itself or by the load. In other words, there may be more to it than the average power level. Edison (talk) 16:07, 11 September 2013 (UTC)

Purely speculative, but your mains voltage might be a bit low. This could give you a current through the main rectifier / transformer primary / main switching transistors that's above the design maximum - the output current limit may be working OK, but there probably won't be anything more than a thermal fuse on the input, and the semiconducting elements will burn out before that trips. But finding the burnt out component will give you a better idea of where the problem was. Tevildo (talk) 19:55, 11 September 2013 (UTC)

Population before human bottleneck

According to the Toba catastrophe theory there was a bottleneck of human population, with numbers reaching as low as around 10,000 individuals. What was the human population just before this event? Staecker (talk) 16:27, 11 September 2013 (UTC)

I'd be surprised if such estimates don't exist, but any such number would be hugely speculative. It would depend on how you define human, since in adition to Homo spaiens sapiens there were other species and subspecies existing at that time. The 10,000 figure is derived from working backwards from genetic assumptions. More recent discoveries on Y-chromosome lineages that weren't known in 1998 would expand the number that passed through the bottleneck. Sorry if that doesn't strictly give the answer you are looking for. μηδείς (talk) 20:41, 11 September 2013 (UTC)

Obesity baloon treatment

I have heard that Gastroanterology offer a baloon treatment to Obese people. Could someone please elaborate on it's mechanism? Thanks guys. — Preceding unsigned comment added by 95.35.55.212 (talk) 17:03, 11 September 2013 (UTC)

I think you're interested in Adjustable gastric band. Wnt (talk) 17:13, 11 September 2013 (UTC)

Or gastric balloon. DMacks (talk) 20:57, 11 September 2013 (UTC)

A little more information at NHS - How weight loss surgery is performed - scroll down to "Intra-gastric balloon" at the bottom of the page. Alansplodge (talk) 08:27, 12 September 2013 (UTC)

Standing on a Vauxhall Corsa

I'd like to stand on the roof of a car to get a better vantage point for a photograph of a building. I'm a 5'2" female and I weight about 9 and a half stone. (You could say that makes me chubby but I prefer curvy with good muscle tone... :P ) The vehicle I have at my disposal for this purpose is a newish Vauxhall Corsa. Will it hold my weight? Will I dent it/damage it? Julia\^talk 17:23, 11 September 2013 (UTC)

This isn't answerable with the information given. What matters is the pressure. If you're wearing gigantic foam snowshoes, maybe. (I disclaim all responsibility...) If you're wearing cleats or stiletto heels, not so much. Wnt (talk) 18:38, 11 September 2013 (UTC)

I'll take my shoes off and just wear socks. My feet are a UK size 5. Does that help? ;) Julia\^talk 18:41, 11 September 2013 (UTC)

I'm pretty sure you'll dent it.

What usually works is to buy and install a suitable roof rack - then screw either planks or half-inch plywood across it. That produces a nice stable platform that you can walk on comfortably with no risk of damaging the car. This site suggests a limit of 75kg for a Vauxhall Corsa roof rack. Since 9.5 stone is 60kg - then even with the weight of the planks/plywood, you should be OK. You could probably get away with more than that if you're only doing this while the car is stationary because that 75kg number must include the dynamic load while cornering, etc. (Please tell me it's stationary!) SteveBaker (talk) 19:27, 11 September 2013 (UTC)

(ec) You don't want to stand on the roof of the car unless you know there is support under the part of the roof panel you're standing on. The edges are safe, but it's easy to lose your balance. Socks or shoes won't make a difference when it comes to denting it. Shoes may scuff up the top, but socks are a lot more slippery, especially on a waxed curved roof. If you can borrow a ladder it would be a much better idea. Something like the Little Giant Ladder System folds down enough to fit in the Corsa, but sets up much taller than the roof you would be standing on. Katie R (talk) 19:36, 11 September 2013 (UTC)

I'd be very surprised if I couldn't damage the roof of a modern car with mere hand pressure while standing on the ground. Don't risk it, buy a roofrack. Or hire a car. Greglocock (talk) 00:08, 12 September 2013 (UTC)

I used to use the roof of my (old) car instead of a ladder, but that was in the days when car steel was thicker. (It flexed a bit, but bounced back.) I wouldn't risk it on my present car. Dbfirs 12:01, 12 September 2013 (UTC)

Since you mentioned it and this is the science desk, with a height of 5 feet 2 inches (1.575m) and a weight 9.5 stone (60.3kg) this gives a body mass index of 24.3, which for someone of predominantly European descent (which from your user page I think you may be) is generally considered in the normal weight range (albeit at the upper end since the boundary is 25). BMI has it's problems as our article attests but given the figures you provided were height and weight, it's probably the best way to analyse them and it suggests people shouldn't consider you 'chubby' based on those figures anyway. Nil Einne (talk) 15:11, 12 September 2013 (UTC)

Diode as radioactive detector - detecting energies?

Will a diode setup in reverse bias indicate the difference in energy from ionizing radiation? ie not only detect events but also give precision on the energy of the same event? Electron9 (talk) 18:59, 11 September 2013 (UTC)

It certainly will - see Semiconductor detector. Tevildo (talk) 21:22, 11 September 2013 (UTC)

If you are considering some sort of home made project, see a series of articles on making a particle detector featured in the magazine Elektor recently. They ran a construction project and some follow up articles within about the last year or so. 120.145.207.247 (talk) 23:41, 11 September 2013 (UTC)

Is there a scientific definition of "shake well"?

Many canned or bottled bottled products inform the user to "shake well" before using. (I assume to blend ingredients that have a tendency to seperate when stored.) All things being equal, how much shaking is expected when a product is labeled as such? --208.185.21.102 (talk) 19:17, 11 September 2013 (UTC)

As for your header, check out chaotic mixing. Basically, what you want might be called a "homogenous" mixture, though that term can be misleading, what you really want is a uniform distribution of each material in the heterogeneous mixture. See also homogenization. For you final question, I can't imagine much serious scientific investigation has been done for consumer goods, but mixologists usually aim for 30 seconds, e.g. here [1] SemanticMantis (talk) 19:27, 11 September 2013 (UTC)

I think SemanticMantis possibly meant Homogenization (chemistry) when linking to homogenization which redirects to Homogeneity (disambiguation) --220 of ^Borg 11:10, 12 September 2013 (UTC)

"My salad dressing label said I should shake well before using, so I shook it up last week in anticipation of using it today." :-) StuRat (talk) 07:41, 12 September 2013 (UTC)

And when you went to get it, it cried, "Close the door! I'm dressing!" ←Baseball Bugs ^{What's up, Doc?} carrots→ 12:19, 12 September 2013 (UTC)

If it said "Shake well before using", then StuRat should have vigorously shaken himself before using it. Stu's getting on in years, perhaps he shakes a bit anyway. 120.145.207.247 (talk) 16:34, 12 September 2013 (UTC)

I would imagine that the viscosity (informally 'thickness') and perhaps particle size (for powders) of the material/s being mixed needs to be taken into account. A very thick liquid would need a longer time to re-mix properly (homogenously). Perhaps even the density of the various materials in a mixture, as some liquid mixtures may settle out rather quickly (oil and water for example), and require to be 'well shaken' for longer. Whereas if the various components are of similar density then they would likely remain in a 'mixed' state far longer, and be easier to re-mix quickly. (Martini? "Shaken, not stirred"?) A mixture of powders, like a cake mix for example could be expected to stay mixed, once 'homogenised', but may separate if subject to vibration during transport. In that case the larger particles would tend to rise to the top. I think that these would be somewhat harder/ take longer, to re-mix than a liquid.

• Looking at a jug of pancake mix it says "Shake bottle to loosen dry mix", so some foods may tend to stick in the container without a shake. (And after a shake the powder is indeed moving more freely. wp:OR). Hmm, pancakes!

Thus to the question, it depends to some extent on what is in the container you are 'shaking well'. Experience will tell if you've shaken it enough. --220 of ^Borg 11:10, 12 September 2013 (UTC)

If in doubt, get one of those machines they use at hardware stores to shake paint cans. That should do it. ←Baseball Bugs ^{What's up, Doc?} carrots→ 12:20, 12 September 2013 (UTC)

In case anyone is interested, this is similar to the question "How many times should I shuffle the cards to make the deck random?", dealt with at Shuffling#Randomization. Duoduoduo (talk) 20:11, 12 September 2013 (UTC)

Theoretically, they're "random" right out of the package. The probability of any particular arrangement of the cards is the same, right? ←Baseball Bugs ^{What's up, Doc?} carrots→ 22:39, 12 September 2013 (UTC)

Depends on how you define your terms. If you define the order out of the box as "perfectly ordered" and the next most ordered state as one card moved one place over in line, for example, you can define how "ordered" the deck of cards is, and then find the state of minimal order (or if you prefer, maximum disorder). The mathematics of this is actually relatively simple with a smallish set of objects, like a 52-card deck, but ultimately these sorts of questions have been worked out for any arbitrarily-sized system with any arbitrary number of states. You can find the basics of this at Boltzmann's entropy formula (known usually as "Boltzmann's equation" to most people, but there's another Boltzmann's equation article about a different calculation). Ludwig Boltzmann's best known work is on the concepts of order and disorder, aka entropy. But I'm probably going a bit afield. The simplest answer is that a maximally randomized deck of cards is one that is the farthest from (i.e. would take the most work to get back to) the perfect suited-and-numbered order you got it out of the deck. --Jayron32 22:58, 12 September 2013 (UTC)

The difficulty with cards is that human ideals of what is random are hard to reconcile. If you had a deck with all of the red cards (in random order) followed by all the black cards (in random order) then the math would likely say that the result was highly random...but people wouldn't accept that the deck had been shuffled at all - and a shuffle that produced that result would put a severe kink in the outcome of some card games! SteveBaker (talk) 16:01, 13 September 2013 (UTC)

Actually not. The card suit and the card number are both different Degrees of freedom. In something like statistical thermodynamics, one considers a "Degree of freedom" as a mode of motion of a particle; for example something like a helium atom (which is spherical) has less "degrees of freedom" than does an H₂ molecule, which is roughly barrel shaped, because H₂ has modes of motion (for example, tumbling end-over-end) not availible to a spherical object. These degrees of freedom are taken into account when dealing with calculations of entropy and order (which is all we're talking about here). So no, a deck of cards which are random in number, but ordered by color or suit, would be less randomized than one which was jumbled up in both, because one has to consider all the ways that a system could be organized (its degrees of freedom) when doing the analysis. --Jayron32 19:24, 13 September 2013 (UTC)

What is the radiation dose (Sv) from a CME when in commercial flight?

During a commercial flight at 10 000 m (30 000 ft) the annual radiation rate 20 mSv. But what is the radiation rate when a CME occurs? Electron9 (talk) 20:23, 11 September 2013 (UTC)

The question seems to relate to Coronal mass ejection (as opposed to Chicago Mercantile Exchange or [Continual Medical Education ). Dozens of news articles just refer to "increased" or "intense radiation" without even giving an approximation as to the extra dose. 22:50, 11 September 2013 (UTC) — Preceding unsigned comment added by Edison (talk • contribs)

I'll noticed that too :-) Electron9 (talk) 23:03, 11 September 2013 (UTC)

This NASA internal presentation might possibly be useful, particularly the example graphs on page 8 of the PDF: http://ccmc.gsfc.nasa.gov/support/SWREDI/Decision_Dashboard_YZheng.pdf -- The Anome (talk) 11:41, 12 September 2013 (UTC)

Bald Eagle flock...

How common is a scene like this? I was under the impression that the Bald Eagle is a fairly solitary species. --Kurt Shaped Box (talk) 23:04, 11 September 2013 (UTC)

The American Eagle Foundation writes: "The only time the bald eagle tolerates the presence of other bald eagles is during winter migrations when large groups of bald eagles gather in one place to share an available food source" ("Eagle Survival - Habitat or Environment"). Another reason for gathering in larger groups during winter migration might be "to scout out potential partners" ([2]) ---Sluzzelin talk 23:20, 11 September 2013 (UTC)

Perhaps relevant: the Bald Eagle population has been growing rapidly in the recent 10-15 years, see Bald_eagle#Population_decline_and_recovery. The species is now registered as "of least concern". SemanticMantis (talk) 01:55, 12 September 2013 (UTC)

Thanks for the answers, guys. Much appreciated. :) --Kurt Shaped Box (talk) 21:46, 14 September 2013 (UTC)