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March 10

Circular Motion

Imagine that you swing about your head a ball attached to the end of a string. The ball moves at a constant velocity in a horizontal circle. Can the string be exactly horizontal? Why? - Thanks —Preceding unsigned 00:23, 10 March 2008 (UTC)

Is this homework? Try setting up the problem with a horizontal string and working out all the forces involved. (There aren't too many—gravity, tension, centripetal/centrifugal.) -- BenRG (talk) 00:38, 10 March 2008 (UTC)[reply]

The answer is "no", and the presence of a downward gravitational force vector should make the reason why obvious. =Axlq 01:55, 10 March 2008 (UTC)[reply]

~~Sure it can. The swinger just has to provide enough force to counteract gravity.~~ Oops. Clarityfiend (talk) 02:25, 10 March 2008 (UTC)[reply]

No it can't as would be obvious if you draw and diagram and add the forces involved. Theresa Knott | The otter sank 05:21, 10 March 2008 (UTC)[reply]

Nitpick alert: the ball can't be moving at constant velocity either, though it may be moving at constant speed. Constant velocity requires travel in a straight line. AlmostReadytoFly (talk) 09:09, 10 March 2008 (UTC)[reply]

Energy Drink Death

First off, I know that drinking enough of anything can kill you.

My question is though, our article on Cocaine (drink) states that it contains 350% as much caffiene as redbull. That seems like a lot to me. How much caffeine can the human body tolerate? (I am very disinterested in killing myself, this is just curiosity) —Preceding unsigned comment added by 24.128.192.184 (talk) 00:35, 10 March 2008 (UTC)[reply]

Caffeine. Quote: "The LD50 of caffeine in humans is dependent on weight and individual sensitivity and estimated to be about 150 to 200 milligrams per kilogram of body mass, roughly 80 to 100 cups of coffee for an average adult taken within a limited timeframe that is dependent on half-life" Wisdom89 _{(_T / ^C)} 00:51, 10 March 2008 (UTC)[reply]

Wouldn't the LD_Lo or LD_min be more appropriate than the LD₅₀ in this case? —Keenan Pepper 20:15, 10 March 2008 (UTC)[reply]

The drink seems to have 280 mg of caffeine per serving. So that means you'd start getting into the dangerous range after 15 consecutive (and immediate) servings or so (when you start getting around 5 g of caffeine), depending on your weight. Which seems like a lot, but when compared to coffee, that's about an order of magnitude more toxic. --98.217.18.109 (talk) 15:21, 12 March 2008 (UTC)[reply]

Not quite an order of magnitude, more like less than twice (but yes, that's still quite a bit). Coffee contains an average of about 135mg of caffeine in 240mL, while Cocaine has 280mg per 240 mL (Red Bull has 80mg in the same volume). – Clockwork Soul 16:04, 12 March 2008 (UTC)[reply]

Note that most energy drinks in NZ recommend a 2 cans max daily intake. However the reason according to a number of sources (I believe one can said it and this source [1] also supports it) is not the caffeine level but the Vitamin B12 level. However Vitamin B12 doesn't mention anything about potentially lethal effects. But most energy drinks here have a level of caffein similar to a cup of coffee anyway (according to the can)... Nil Einne (talk) 16:49, 15 March 2008 (UTC)[reply]

Hair loss (non-balding)

Hi. This is not a request for medical advice. I think I've heard somewhere that the average adolescent loses roughly 120 hairs a day. Is this true? We really should have an article on hair loss that does not redirect to balness, then. Is that why when I drop food on the carpet it gets covered with hair even if it has been vacuumed? When does hair loss of more than one hair a day usually begin at, then? On average, when is the first white, gray, or silver hair occur in an average person (head)? Do people lose skin hairs too (not caused by accidental pulling)? This is not homework. On average, approximately how many hairs would have been lost (on a person that is not born will to be bald) and how many white/gray/silver hairs would have developed by the time an average person turns 18, for example? Do common mammalian pets (eg. cats, dogs, rabbits, hamsters, gerbils, mice, rats, etc) lose hair as rapidly as humans or more rapidly? What colour of hair loses hair most rapidly and which least rapidly? What about de-melanined hairs? Thanks. ~A H 1^(TCU) 00:49, 10 March 2008 (UTC)[reply]

These links should help you with your questions: [2],[3], [4]. Wisdom89 _{(_T / ^C)} 00:54, 10 March 2008 (UTC)[reply]

Anyone who has ever owned a cat will tell you how rapidly and in vast quantities they lose hair. As for the carpet, it's probably a mixture of hair which wasn't picked up by the vacuum, carpet fibres and other random dust and detritus. -mattbuck (Talk) 01:02, 10 March 2008 (UTC)[reply]

What is naphtha?

Is it really the same thing as petroleum ether? If so, why aren't the articles merged? If not, what is the difference? What is the main meaning of naphtha when the term is used in modern English? --Sonjaaa (talk) 00:55, 10 March 2008 (UTC)[reply]

Petroleum ether is obtained from petroleum refineries as the portion of the distillate which is intermediate between the lighter naphtha and the heavier kerosene says the PE article, while Naphtha says it is referred to as PE. My guess is that naphtha is not the same, but that there is a blurred boundary. -mattbuck (Talk) 01:00, 10 March 2008 (UTC)[reply]

All the terms used by the petro-chemical industry for the various fractions produced by oil refineries are ill-defined chemically and always include a large range of alkanes which overlap quite considerably with neighbouring fractions. It is more of a relative order of fractions up the distillation column which defines the terms. SpinningSpark 08:02, 10 March 2008 (UTC)[reply]

Question

Why do scientists use animals' Latin names? 58.168.209.250 (talk) 01:17, 10 March 2008 (UTC)[reply]

Biological_nomenclature#Value_of_binomial_nomenclature --Sonjaaa (talk) 01:20, 10 March 2008 (UTC)[reply]

That doesn't answer why Latin in particular, or the historical reasons for it. The simple answer is that Latin used to be the lingua franca of science—the "neutral" language that scientists of all countries used when they wanted to communicate with one another (though by the 19th century it was French and German; today it is English). It was expected that all educated men (and it was primarily men who were allowed to participate in the republic of letters for a good long time) would know the language, whereas whether they would know English, French, Russian, Italian, Swedish, etc. was less likely. Thus a large number of the early major scientific publications were originally written in Latin as well—De revolutionibus orbium coelestium, Philosophiæ Naturalis Principia Mathematica, etc.

The practice of using Latin specifically for binomial nomenclature (those animal names you are thinking of: genus plus species, Homo + sapiens) goes back to Linneaus, the father of modern taxonomy, and has stuck every since. Standardizing nomenclature in a neutral, non-changing language was important to making sure that a creature in one part of the world was really the same (or similar) creature in another, even though they had radically different names in the local languages. --98.217.18.109 (talk) 01:45, 10 March 2008 (UTC)[reply]

I've always enjoyed the fact that "Linnaeus" is itself a Latinized Swedish name. I suppose as "Carolus Linnaeus" it would be fully binomial. - Nunh-huh 01:42, 10 March 2008 (UTC)[reply]

Well, again, it's the language Linnaeus himself published his major works in. He wrote the whole thing in Latin, name included. He practiced what he preached! The custom of using only the Latinized last name for "major" writers is quite old (think also Copernicus). Scientists already drop the Albert from Einstein quite often; imagine how that'll progress in two hundred years. --98.217.18.109 (talk) 01:45, 10 March 2008 (UTC)[reply]

Yes, it's not that he was unusual in adopting a Latin name (Paracelsus, Helvetius, etc.) but that it was in his case an amusing coincidence. - Nunh-huh 02:05, 10 March 2008 (UTC)[reply]

There is a second reason Latin was used - it is a "dead" language. It is not changing due to modern usage. So, the definition of a Latin word 100 years ago is the same definition today and will be the same definition 100 years from now. English (or any other modern language) doesn't have that luxury. Consider "urchin". What is that? Easy, a nasty street kid. Or, perhaps it is one of those prickly things in the ocean. Not long ago, it was a prickly little animal in British gardens. Before that, it was simply a vulgarity. Who knows what it will mean in another 100 years. It is a good thing we don't use it for scientific purposes since it isn't nice enough to pick one definition and stick with it. -- kainaw ™ 17:04, 10 March 2008 (UTC)[reply]

Latin is mostly dead. But it is the official language of the Vatican, and that state's panel of Latin experts do formulate new words and constructions for things like "helicopter" and "hard disk" that the Romans knew not of. There are (or was) also available on CD the Elvis Presley classics, "Tenere me ama" ("Love Me Tender") and "Nunc hic aut humquaum" ("It's Now or Never"). - Nunh-huh 23:47, 10 March 2008 (UTC)[reply]

One big advantage of standarized bionomials: it's great for organization without the level of ambiguity and repetition of common names. Common names are often misleading (flying fox), used for a wide variety of different species (razor clam), or there may be several different names for a singe animal (cougar/puma/mountain lion/panther). A single, agreed-upon term helps everyone know just what animal we're talking about. — Scientizzle 20:19, 12 March 2008 (UTC)[reply]

Biosphere ?

Many people claim that houseplants "cleanse the air", by removing particulates and carbon dioxide and adding oxygen. While this is certainly true to some degree, I'm of the opinion that the quantity of plants needed to do a significant amount of this would be far more than would fit in the home, based on plants having much slower metabolisms than people. So, I'd expect that if the biomass of people and pets in your home is 1000 lb, that you would need maybe 10,000 or 100,000 lb of plants to clean the air of the carbon dioxide and other waste products the people create. How can I calculate the actual ratio ? StuRat (talk) 01:24, 10 March 2008 (UTC)[reply]

Well, this would take some research - if you can come up with the amount of CO₂ used per day by plants, we could do a quick and dirty calculation. On the human side: average CO₂ content in inhaled air is roughly 383 ppm by volume, which is 0.0383%.

Average CO₂ content in exhaled air is 4.5% by volume. The difference (4.5% minus 0.0383%) would be 4.1%. So the CO₂ output from a human is 4.1% of the volume of inhaled (or exhaled) air.

The tidal volume (volume of a normal breath) of a human is about 500 mL. The usual respiratory rate varies between 12 to 20 breaths per minute. For our purposes, let's say 18 breaths per minute.

There are 1440 minutes in a day, or 25920 breaths per day. 25920 breaths of 500 mL each = 12960000 mL = 12960 liters per day. 4.1% of 12960 liters is 531 liters of CO₂ produced per human per day. There are 22.4 liters/mole of gas at STP, so 531 liters is 23.7 moles. There are 44 g/mole of CO₂, so 23.7 moles = 1043 grams of CO2, or roughly one kilogram, produced per human per day.

I think :) - Nunh-huh 02:02, 10 March 2008 (UTC)[reply]

4.5% minus 0.0383% is 4.46%, not 4.1%. Oh, how you curse us, Mr. Decimal Point.-Running On Brains 21:13, 12 March 2008 (UTC)[reply]

Good work, now we need some info on the plant side. StuRat (talk) 03:15, 10 March 2008 (UTC)[reply]

You also need to account for the efficiency of the plant at converting CO2 to oxygen. I have read that plants are extremely efficient (95%[5]) at using light energy for this process. So even if plant metabolism is slower, it may be compensated by a more efficient CO2 to O2 conversion than a human's ability to produce CO2 from O2.

The amount of oxygen produced also depends on the light available. I had an aquarium containing hydrilla plants. When the sun shone on them, streams of oxygen bubbles would spew forth from the leaves. =Axlq 02:09, 10 March 2008 (UTC)[reply]

23.7 moles of carbon (C) is 284g, so the total carbon content of plant mass in your house must increase by 284g per day per human. If we assume that plants are reprented by the formula CH2O (sugars) it is 948g per person per day, but there are also phosphates and nitrogen that the plant takes in, so 1kg per day would be minimum...Which looks very unrealistic.--Shniken1 (talk) 02:31, 10 March 2008 (UTC)[reply]

Interesting approach. Aren't plants something like 95% water ? I wonder what percentage of their weight is carbon. If we figure 4% by weight, then one kilogram of carbon increase would mean 25 kilograms of plant mass increase. StuRat (talk) 03:15, 10 March 2008 (UTC)[reply]

Yeah water would make up significant amount of the mass of the plant so yeah the mass of the plant as a whole would have to increase a lot more than its carbon mass (depending on the plant), and as plants get bigger I would suppose that water would be harder to obtain (via transirpation) so it may be the limiting growth factor.--Shniken1 (talk) 03:52, 10 March 2008 (UTC)[reply]

I'd expect most of the water to be supplied by humans watering the plant, with the roots sucking it up from there. StuRat (talk) 23:55, 10 March 2008 (UTC)[reply]

Much of this depends on what you mean by cleanse the air. I would expect the impact on oxygen and carbon dioxide to be minimal unless you live in an air tight box. Transpiration is a likely reason for cleansing. Transpiration cools the air due to evaporation. It also adds moisture to the air. Both these could be beneficial inside a house. David D. (Talk) 18:55, 10 March 2008 (UTC)[reply]

Incidentally, the following link is in the "Biosphere 1, 2, 3, and J" section of our Biosphere article:

"Biosphere 3 (aka BIOS-3) - Experiment to be conducted starting Jan 2008 in the Gary C. Comer Geochemistry building at Columbia University's Lamont Campus in Palisades, New York."

However, when I follow the link, the article says:

"BIOS-3 was a closed ecosystem at the Institute of Biophysics in Krasnoyarsk, Siberia, in what was then the Soviet Union."

Which is correct ? StuRat (talk) 03:33, 10 March 2008 (UTC)[reply]

(after edit conflict) According to http://www.newton.dep.anl.gov/newton/askasci/1993/biology/bio027.htm, it takes around 353 plants to match the oxygen input rate that humans require. However, I have two concerns with the webpage:

(1) The webpage assumes that all oxygen breathed in that does not enter the inactive part of the lung (the "dead space") is absorbed. According to Breath#Composition_of_air, 15-18% of exhaled air is composed of oxygen by volume. For the 252 L/h of inhaled air that does not enter the dead space, and the corresponding 252 L/h that are exhaled, about 42 L is exhaled. Since 53 L/h of oxygen that does not enter the dead space is inhaled, this gives an hourly oxygen usage rate of 53-42=11 L/h. Assuming the plant calculations are correct, 76 plants would be needed.

(2) At the end of the webpage, the author states that "oxygen production decreases as carbon dioxide concentration increases". However, according to the photosynthesis article, the amount of CO2 plants take in is equal to the amount of O2 they output. Humans also necessarily expel as much CO2 as they breathe in oxygen, so if 76 plants provide enough oxygen to sustain a human, they also absorb enough carbon dioxide to keep the environment's CO2 level constant.

So 76 plants per person, and if we assume plants are 500 g on average, that's 38 kg/person. The plants would have to be properly maintained, of course, and cannot be allowed to grow; growth would mean a higher demand for CO2 than the humans can supply.

About BIOS-3: Most sources state that it was an experiment in the Soviet Union: [6][7][8]. Searching for "bios 3 'Gary C. Comer'" on Google yields only the Wikipedia biosphere article and mirrors of it. --Bowlhover 04:06, 10 March 2008 (UTC)

That seems like way too much oxygen production per plant kg. After all, a person weighs around 38 kg, and I have a hard time accepting that plants, with their far slower metabolisms, use CO2 at the same rate people use oxygen. Also, if the plant doesn't grow, where exactly does all the carbon go that it sucks out of the air ? StuRat (talk) 00:03, 11 March 2008 (UTC)[reply]

While it might matter for a biosphere, for your purposes as I understand them (the plant cleans up after the humans), it doesn't matter whether or not the plants grow. - Nunh-huh 00:44, 11 March 2008 (UTC)[reply]

But where does the absorbed carbon go if not to plant growth ? I suppose a fruit tree puts it into fruit and a nut tree into nuts (although this is a type of growth), but what about plants that don't produce food ? StuRat (talk) 17:29, 11 March 2008 (UTC)[reply]

List of air-filtering soil and plants may be of interest too. —Pengo 09:30, 11 March 2008 (UTC)[reply]

According to http://www.bio.net/bionet/mm/plantbio/2000-October/024096.html, the average plant produces 15.2 mL of oxygen per dm^2 of leaf area per hour. The "leaves" mentioned in the previous link I gave would then be 10 dm^2 each! Anyways, the total leaf area needed to support one person would be 750 dm^2, about 2.7 m x 2.7 m. I wonder how much that would weigh.

Also, the plants will grow, but I think they have to be trimmed so they don't use up CO2 at an excessive rate. --Bowlhover 15:18, 12 March 2008 (UTC)

Yes, that sounds more reasonable. Perhaps the leaves alone might weigh 38 kg, but including the stalks, branches, roots, etc., would make it much more. StuRat (talk) 17:01, 13 March 2008 (UTC)[reply]

We have had a similar question before. What if the person consumed plant material each day in the form of carbohydrates to total a significant portion of his daily caloric needs? Could the carbon in the plant material (carbohydrates) equal the carbon in the exhaled carbon dioxide the person emits? If the person exhaled more carbon than he consumed, wouldn't he become carbon deficient over time? Then there would be no net increase in the amount of carbon sequestered in the plant material. Ever see a demonstration of a sealed glass globe with water, plants, little fish and snails? The manage to keep recycling for quite a while sometimes until there is too much/too litttle sunlight or something else goes out of balance. Edison (talk) 14:04, 11 March 2008 (UTC)[reply]

I would expect that most of the carbon people eat ends up in their feces or urine. I'd expect animals would be needed to get this carbon back in circulation from there. Worms or insects can eat the feces, birds and frogs can eat them, and other animals can eat them, until we get to something people eat. StuRat (talk) 17:29, 11 March 2008 (UTC)[reply]

Some people might consume carbon in the forms of pencil graphite, diamonds, or charcoal and poop it out, but that is a small portion of the carbon consumed by humans.Human feces does contain a large proportion of undigested food which would still have much of its carbon content. Urine does incorporate some carbon from the carbon dioxide resulting from metabolism of food. Most carbon consumed is probably in the form of Carbohydrates, which are transformed along with some non-carbohydrates into Glucose by digestion, so that the cells of the body can use it as fuel. Glucose is (C₆H₁₂O₆). It yields energy and exits the body as water and carbon dioxide. Long term space travel will likely involve a closed system of recycling the products of human and animal metabolism into new food such as the Biosphere 2 experiment tried to test. Edison (talk) 16:47, 12 March 2008 (UTC)[reply]

How can you ever fall in a black hole?

Since time slows per the reference frame of an outside observer as an object approaches the event horizon, and 'freezes' at the horizon (thus the Russian term for black hole: 'frozen star'), how can a black hole ever increase in mass? This was once considered a flaw in the concept of black holes, but evidently was somehow resolved. However, I've never seen an account of how it was resolved. (And yes, I do understand that for an observer falling past the event horizon, supposedly nothing unusual would appear to happen - except that you'd think s/he'd witness the end of the universe.) — kwami (talk) 07:03, 10 March 2008 (UTC)[reply]

This question has been previously answered here [9] and probably many times before. However, I never felt satisfied with the answer given then, for the same reason given by Kwamikagami. If someone can elaborate, please do. SpinningSpark 08:15, 10 March 2008 (UTC)[reply]

I’m no physicist. But as the mass of the black hole increases, the event horizon will increase in radius and objects that previously were hovering on the edge will be enveloped. And objects don’t need to actually be at the center of the singularity to contribute to the black hole’s mass/gravitational effect. (This is just my speculation.) — Knowledge Seeker দ 08:33, 10 March 2008 (UTC)[reply]

But the BH can't increase in mass unless something crosses the event horizon in finite time in the external reference frame. — kwami (talk) 08:46, 10 March 2008 (UTC)[reply]

I’m not sure I agree with that statement. — Knowledge Seeker দ 08:51, 10 March 2008 (UTC)[reply]

How would it increase in mass, without mass being added to it? — kwami (talk) 08:54, 10 March 2008 (UTC)[reply]

Perhaps I should be more clear. I contend that a mass placed just outside the event horizon will increase the gravitational field/curvature of space (as it would anywhere) and that therefore a small volume of space which previously had almost, but not quite enough gravity to have an escape velocity greater than c (or equivalent formulation) now will; in effect, the combined event horizon around the two objects will be larger. (Again, speculation.) — Knowledge Seeker দ 08:57, 10 March 2008 (UTC)[reply]

A group of massive objects, act as if all their gravity were concentrated at thier center of mass. So from a distant observer's POV anything frozen at the edge of an event horizon would still contribute to the total mass of the black hole system. Does that resolve the problem? Theresa Knott | The otter sank 12:28, 10 March 2008 (UTC)[reply]

Does this mean that an unfortunate astronaut just outside the event horizon might be observed by a "local" observer (say, another astronaut also falling into the black hole) as not having quite reached the horizon, but that both might be "observed" by a distant observer to be within the event horizon (i.e. part of the black hole)? dbfirs 12:54, 10 March 2008 (UTC)[reply]

Hi. On a Discover magazine there's this guy named João Magueijo that has a theory called Varying Speed of Light (VSL). When he introduced this theory, one physicist said it really stood for "Very SiLly". However, according to this theory, time and the speed of light slow near a black hole and never allow anything to escape nor go in (after all, black holes are infinitely dense). Hope this helps. Thanks. ~A H 1^(TCU) 14:22, 10 March 2008 (UTC)[reply]

As far as I know Magueijo's VSL theory is supposed to be an alternative to cosmic inflation, so it only matters in the very early universe and doesn't have any bearing on (present-era) black holes. -- BenRG (talk) 21:33, 10 March 2008 (UTC)[reply]

Objects can and do fall in. They only appear to stop at the event horizon. 64.236.121.129 (talk) 16:43, 10 March 2008 (UTC)[reply]

The objects may see themselves falling in. An outside observer sees the objects dim and disappear as they approach the Schwartzchild radius. An alternate theory Gravastar suggests that matter falling into a compact stellar object will appear to cool and form a Bose-Einstein condensate. One consequence of general relativity is that perception is affected by perspective. Jehochman ^Talk 16:51, 10 March 2008 (UTC)[reply]

64 and Jehochman, it’s not that Kwami’s really disputing whether they fall in or not (the phrasing of the heading, unfortunately, is misleading). His question is (as I understand it), essentially: If, from the reference frame of an outside observer, an object never passes the event horizon [due to time dilation], how/when will he observe an increase in the black hole’s mass? — Knowledge Seeker দ 18:11, 10 March 2008 (UTC)[reply]

If matter has fallen inside the event horizon, how can you get back any information about mass inside? If light can't get out, neither can any other sort of information. If an outside observer can never see anything cross the event horizon, then presumably from their point of view, nothing ever has. A sphere or shell of mass around a point will be indistinguishable to an outside observer from all that mass being located at the center point. There would be no observable difference on the force of gravity affecting an outside observer. There are a lot of contradictions if we assume the existence of black holes. This may mean that our black hole model is wrong. Jehochman ^Talk 18:16, 10 March 2008 (UTC)[reply]

I don't know that much about black hole dynamics, but I think Knowledge Seeker is correct. Certainly it's a mistake to think that infalling matter has to fall through the event horizon to add to the mass, because nothing that happens inside the event horizon can be relevant to physics outside. Whatever it means for "the black hole to gain mass", it has to involve only the physics outside the event horizon.

The event horizon is acausal; it will happily expand faster than light to engulf infalling matter. Here's a special relativistic analogy. I'll use 2+1 dimensional Minkowski space because it's easier to visualize than 3+1 dimensions and 1+1 isn't enough for this example. At time t = 0 pick some region of the xy plane and magically destroy everything there. There's then a region of spacetime (a collection of events) with t < 0 which is unobservable to anyone who survives past t = 0, because any signal that might have originated from those events was destroyed. For example, say you destroy everything in the circular region x² + y² < 1 light year². Then the unobservable region is a cone in Minkowski space whose apex is x = 0, y = 0, t = −1 year and whose base is the destroyed disc. If you destroy a square (|x| < 1 ly, |y| < 1 ly) then the unobservable region is a square pyramid. The boundary of the unobservable region is the event horizon. The event horizon is always a null surface because of how it was defined, but it may have "creases" which are not null. For example, in the square pyramid case you have corners which go from x = y = 0 at time t = −1 year to x = ±1 ly, y = ±1 ly at time t = 0; they're effectively moving at c√2.

Let S be the destroyed set. Then the observable region at time −t is $\bigcup _{x\in S^{c}}B_{ct}(x)$ , where S^c is the complement of S and B_r(x) is the ball of radius r centered at x. Imagine S being eroded away from all sides at the speed of light as you go backward in time. Now say S is a sort of peanut shape, or an overlapping union of two circles. As you go back in time it will become a thinner peanut; then it will break into two teardrop shapes, with the pointy bits pointing toward one another; then the teardrops will shrink and become more circular until they disappear. Now run this forward in time: two expanding teardrops appear out of nowhere, extend toward each other faster than light, and merge. That's kind of what a black hole merger is like. The event horizon is not at all like a physical object obeying dynamical laws; it's a global property of the whole spacetime and it "knows about the future". In fact it's a theorem that the event horizons of two black holes that will eventually merge have the teardrop point on them from the beginning. That's for merging black holes, i.e. tossing one black hole into another, but I think tossing ordinary matter into a black hole works similarly. Whether any of this answers the original question I'm not entirely sure. This is a tricky subject. -- BenRG (talk) 21:33, 10 March 2008 (UTC)[reply]

I wonder if Knowledge Seeker might be on the right track: Once an object approaches a black hole close enough that their center of mass lies within the EH, does the EH then expand correspondingly? But this would happen while a small mass is still quite far from the EH, so it wouldn't be engulfed. (If you were to drop a sandwich into a Solar-mass black hole, it would hardly affect the EH at all, so we're left with my original question.)

It's not true that we can't know what's within the EH, only that no signal can be sent out. Fields extend beyond the EH, but any modulations of those fields would be red-shifted to a frequency of zero. However, with those fields we can still measure the total electronic charge, mass, and momentum (linear and angular). — kwami (talk) 22:23, 10 March 2008 (UTC)[reply]

Also, pace another comment, it isn't the case that the object just appears to slow down, due to some kind of optical illusion. It isn't light that slows down, but time itself. In the external reference frame, it really doesn't reach the EH. Thus the problem. — kwami (talk) 22:50, 10 March 2008 (UTC)[reply]

Intermolecular Forces

Can a molecule have hydrogen bonding, permanent dipole - permanent dipole and instantaneous dipole - induced dipole intermolecular forces acting upon it at any one time? 88.108.198.114 (talk) 17:38, 10 March 2008 (UTC) —Preceding unsigned comment added by 88.108.198.114 (talk) 17:37, 10 March 2008 (UTC)[reply]

Yes. something like liquid cholesterol would be a easy example to consider87.102.94.48 (talk) 17:45, 10 March 2008 (UTC)[reply]

True. Remember that molecules can be very very large (Protein), or even macroscopic (DNA), so different parts of one molecule can have completely different properties. --Bmk (talk) 17:59, 10 March 2008 (UTC)[reply]

Batteries in series vs. parallel arrangement

If one battery in a set is bad (low voltage), which arrangement is affected more? —Preceding unsigned comment added by 66.120.95.52 (talk) 19:22, 10 March 2008 (UTC)[reply]

Series. Voltages in series add, so a bad one will lower the voltage. In parallel, the remaining good batteries will try to supply whatever current it takes backwards through the bad battery to drop the supply voltage across it. If they can't supply enough current to do that, the voltage will fall, but if they can it won't, much. So a series arrangement is always immediately affected, and a parallel arrangement might not be affected very much at first. --Milkbreath (talk) 19:54, 10 March 2008 (UTC)[reply]

Could be either. If the one with low voltage is that way because the internal resistance has increased (low voltage under load, closer to normal voltage with no load, a common enough condition) then the parallel arrangement would work better, although the bad cell should be disconnected to avoid premature drainige of the others and because it is not contributing much to the output. If the bad cell has an internal short (happens sometimes) then it would drain the output of the others in parallel but would have less effect in series. In either arrangement the bad cell could explode. Edison (talk) 13:56, 11 March 2008 (UTC)[reply]

He asked about the case where one battery has low voltage, not where one battery is a dead short. "Could be either" is the answer to a different question, and is not a sensible answer to the question of "which" is affected "more". If it depended on the situation, "neither" or "it depends" would make sense. --Milkbreath (talk) 19:33, 11 March 2008 (UTC)[reply]

What bodily function is not possibel in space?

what bodily function is not possibel in space? —Preceding unsigned comment added by 89.243.182.3 (talk) 20:13, 10 March 2008 (UTC) - Please don't post in all caps, it comes off as yelling -- MacAddct  1984 ^{(talk  contribs)}[reply]

pushing the Caps Lock key a second time? Thomprod (talk) 20:14, 10 March 2008 (UTC)[reply]

If you mean in a spaceship, all bodily functions are possible, but can get a bit messy. That's why they have special devices for going to the bathroom. -- MacAddct  1984 ^{(talk  contribs)} 20:28, 10 March 2008 (UTC)[reply]

It hasn't been demonstrated that a child can grow in the womb in space. I'm not sure what people have thought about this though.Sancho 20:39, 10 March 2008 (UTC)[reply]

Larry Niven made a significant plot point out of this in his Known Space series. Pregnant women who lived in the Belt were required to live in a hollowed-out asteroid, called Confinement Asteroid, that was spun to provide centrifugal force in place of gravity. --Trovatore (talk) 20:53, 10 March 2008 (UTC)[reply]

Did they have another hollowed-out asteroid to send women with PMS ? :-) StuRat (talk) 17:35, 11 March 2008 (UTC)[reply]

Bodily functions remain intact, however, there can be serious muscular atrophy due to zero gravity unless the astronaut exercises regularly. Wisdom89 _{(_T / ^C)}

I think it's been demonstrated for mice though. IIRC, they were born a bit smaller than normal but otherwise okay. DMacks (talk) 02:49, 11 March 2008 (UTC)[reply]

According to Canadian astronaut Chris Hadfield you can't burp normally in a weightless environment - see Q2 here. Gandalf61 (talk) 09:49, 11 March 2008 (UTC)[reply]

Have anyone heard about sex in space? Considering that the blood does not flow normally in space and some parts of our body will need it, perhaps it is not possible... —Preceding unsigned comment added by 217.168.3.246 (talk) 03:03, 13 March 2008 (UTC)[reply]

Blood flows due to heartbeats and one-way valves, even in zero gravity. However, it does stop flowing at high g's, such as those experienced by a fighter pilot. A special suit (I forget the name) can help in this situation, by alternately compressing and releasing on the legs to push the blood forward. I believe sex would work just fine in space, although both parties would want to be contained in an absorbent bag, so that any fluids released don't go floating off down the hall. As a practical matter, though, the lack of privacy and not wanting to use up precious water to clean up afterwards make this a no-no for short space flights. Long-term or even multigenerational ships would need to find a way to make this work, however. I suspect that artificial gravity, via a rotating spaceship, would be used in such a situation, as some gravity seems important in maintaining the human body over the long-term. StuRat (talk) 15:16, 13 March 2008 (UTC)[reply]

Wikipedia knows all - see sex in space. Gandalf61 (talk) 15:24, 13 March 2008 (UTC)[reply]

One man's meat is another man's poison

Are there any spices with an unusually small difference between the standard amount used for flavoring and the minimum poisonous amount? Or with an unusually large difference between the LDLO and the LD50? --67.185.172.158 (talk) 21:11, 10 March 2008 (UTC)[reply]

I seem to rememeber that 2 nutmegs could be fatal...87.102.94.48 (talk) 22:21, 10 March 2008 (UTC)[reply]

I had to look these up: LD_Lo = Lowest published lethal dose and LD₅₀ = Median lethal dose. Mmmm, Eggnog with extra nutmeg! :-)) --hydnjo talk 01:51, 11 March 2008 (UTC)[reply]

Deadly meat - see fugu. 80.0.108.245 (talk) 19:32, 16 March 2008 (UTC)[reply]

Molecular Excitement

I know that molecular excitement (vibrations, oscillations - what is the correct term?) slows down as Temperature decreases - I'm thinking of a liquified gas in a tank - and increases as the Temperature rises, but what happens when the Pressure increases or decreases? What effect does Pressure have on Molecular Excitement?

P. Lamont —Preceding unsigned comment added by 72.39.249.249 (talk) 21:18, 10 March 2008 (UTC)[reply]

Temperature has the greatest effect on molecular rotation, pressure also influences this, through pressure broadening.--Shniken1 (talk) 22:54, 10 March 2008 (UTC)[reply]

Pressure can change the temperature, if you increase pressure, the substance is likely to contract, and work is done on it. This will add energy, and the temperature will increase. This head can be conducted out. Also if pressure increased so that the material is compressed to a white dwarf density, then the uncertanity principle will ensure a uncertain temperature and Degenerate matter exists. Graeme Bartlett (talk) 23:15, 10 March 2008 (UTC)[reply]