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April 2

911

is 911 a federal or local number — Preceding unsigned comment added by Wdk789 (talk • contribs) 02:54, 2 April 2011 (UTC)[reply]

Neither and both. See 9-1-1 for the history - it's the standard in Canada too. 9-1-1 call centers are locally administered, but the FCC and its Canadian equivalent set the standard. Acroterion (talk) 03:07, 2 April 2011 (UTC)[reply]

if u call from a cell phone without e 911 how does it route it to your city call center — Preceding unsigned comment added by Wdk789 (talk • contribs) 03:11, 2 April 2011 (UTC)[reply]

The very same article you were directed to above answers that question, if you read doen to the section titled "wireless telephones". --Jayron32 03:36, 2 April 2011 (UTC)[reply]

no it dont — Preceding unsigned comment added by Wdk789 (talk • contribs) 04:40, 2 April 2011 (UTC)[reply]

It did, but if you need more details, the article Enhanced 9-1-1 has more as well. --Jayron32 05:32, 2 April 2011 (UTC)[reply]

just answer me — Preceding unsigned comment added by Wdk789 (talk • contribs) 06:50, 2 April 2011 (UTC)[reply]

We did. We're not going to copy and paste the words from the articles here just because you don't wish to click the name... --Jayron32 06:52, 2 April 2011 (UTC)[reply]

it dont help i read it — Preceding unsigned comment added by Wdk789 (talk • contribs) 10:46, 2 April 2011 (UTC)[reply]

Wikipedia articles are written in the same English that we are speaking here. Some readers find that articles like these [1] [2] are easy to read. Cuddlyable3 (talk) 11:04, 2 April 2011 (UTC)[reply]

In summary, cell phones either transmit their geographical coordinates to the 911 station by a GPS system or location identified from the cell tower, or the caller has to tell the emergency reponders his/her location. ~A H 1^(TCU) 23:20, 3 April 2011 (UTC)[reply]

Energy

A 10 W Fluorescent lamp and a 10 W LED Tube light blows for one hour. Will they use same amount / quantity / level of energy? —Preceding unsigned comment added by 122.172.158.243 (talk) 07:09, 2 April 2011 (UTC)[reply]

Yes, but the LED is more efficient - more energy is converted into light. Plasmic Physics (talk) 08:33, 2 April 2011 (UTC)[reply]

To clarify. "10 W" is the electric power taken by the lamp. After one hour each lamp has dissipated 10 watt-hours, which is the quantity of energy one pays for. The LED tube gives more light from its energy consumption and has a longer life, but it costs more to buy in the first place. Cuddlyable3 (talk) 10:51, 2 April 2011 (UTC)[reply]

You meant "glows" instead of "blows", I assume ? There's also the issue that they don't seem to always count the electricity used in the electrical ballast in the base. So, is that 10W the amount used in the bulb only, or including the ballast ? StuRat (talk) 15:38, 2 April 2011 (UTC)[reply]

If the flourescent is a compact fluorescent bulb, then the energy lost in the ballast or electronic circuit should be included. If it is, say a 4 foot 32 watt straight fluorescent tube snapped into a light fixture, I would expect additional energy to be used in the ballast. Edison (talk) 19:41, 2 April 2011 (UTC)[reply]

What plant is that?

I noticed this in Budapest, Hungary.

Amongst the branches of a tree there was a very distinct bulb of leaves that seemed to be separate from the tree, an individual plant, different in colour (vividly green, unfortunately not seen on the photos). I don't think it's a birds nest. What could it be?

http://i53.tinypic.com/3586yp1.jpg

http://i52.tinypic.com/25ti6wz.jpg —Preceding unsigned comment added by 109.74.50.52 (talk) 08:27, 2 April 2011 (UTC)[reply]

Might be mistletoe. 77.3.139.229 (talk) 08:51, 2 April 2011 (UTC)[reply]

I think you've nailed it. Thank you!

109.74.50.52 (talk) —Preceding undated comment added 09:51, 2 April 2011 (UTC).[reply]

Eggs and earthquakes

Does an earthquake destroy eggs? Clearly they are in danger being hit by some crashing building or to fall out of the nest and hit the ground, but aside of that, is an earthquake strong enough to shake them so violently that they break? 77.3.139.229 (talk) 11:26, 2 April 2011 (UTC)[reply]

If you were asked to predict whether the egg falling from the nest will break, you would need quantitative information on the peak acceleration (deceleration) of the egg, the resilence of the surface that it contacts, and the geometry and material stiffness of the particular species of egg. The same difficult-to-supply information is needed to answer your question. Cuddlyable3 (talk) 12:41, 2 April 2011 (UTC)[reply]

I see you are right and it would be better to ask for information about acceleration magnitudes from earthquakes, preferably related to the common scales for earthquakes.Where can I find such information? 77.3.139.229 (talk) 15:32, 2 April 2011 (UTC)[reply]

Assuming the egg didn't fall or have something fall on it, it would depend on what surface the egg is in on. If it's on a soft surface, like the container they come in from the grocery store (or, presumably the nest), it should be fine. On a hard table, with a strong enough up-down ground motion, it might crack. A side-to-side motion could knock it off the table, but we already excluded that case. It might also run into something on the table. Again, if it's soft, like a cereal box, it should survive, but hitting a hard object like a cereal bowl might crack it. StuRat (talk) 15:33, 2 April 2011 (UTC)[reply]

Ah, I see this splits the question into three different aspects: 1) could the first acceleration damage the egg? 2) (I didn't think of it previously) could a vertical acceleration that lifts the egg without damaging it be strong enough that it cracks on falling back (hitting the same material that sent it off flying)? and 3) is it possible that eggs laying next to each other are accelerated horizontally in a way that they hit each other so that they get damaged? Note that I am not especially interested in eggs, it is just some kind of place-holder for something fragile that everyone has a common experience about. 77.3.139.229 (talk) 16:09, 2 April 2011 (UTC)[reply]

I see. Stability also comes into play here, as items in stable positions won't tend to fall. So, things on shelves are more likely to be damaged than things on floors. Of course, the stability of other things which could fall on top also matters. A long item (say a lamp) balanced on it's end will also be less stable than one on it's side. One interesting result of this is that it may make sense to leave everything where it lands after the initial quake (as those positions will tend to be more stable) until the aftershocks subside. You might also move things into more stable positions that survived the first quake (like putting lamps on their sides on the floor). StuRat (talk) 16:29, 2 April 2011 (UTC)[reply]

It is the momentary distortion of the"egg shell"shape during impact which causes the structural failure of the shells cohesive integrity. To protect the shell from damage from violent motion it's important to make certain that what ever energy absorbing packing material you use (eg.styrofoam) must make perfect contact with the whole of the shell surface so that kinetic energy induced by motion will be transfered to the packing material evenly at all points of the shell thereby eliminating shell ditortion and breakage. I'm not a mathematition but I'll bet one of our wikipedien freinds here can come up with an equation for that concept.190.56.107.186 (talk) 17:24, 2 April 2011 (UTC)[reply]

Simply divide the force exerted on the egg shell by the area of the shell where the force is applied, to get the average pressure. If that's high enough, the shell will break. If the force is unevenly applied, then you'd need to know the maximum pressure. Don't forget the force on the opposite side of the egg, too, where it pushes against the container. StuRat (talk) 04:39, 3 April 2011 (UTC)[reply]

See peak ground acceleration and dome. ~A H 1^(TCU) 23:17, 3 April 2011 (UTC)[reply]

Double-heading

Having just watched a double-header pass through my home town, I'm curious about the physics of double-heading steam locomotives. Presumably it is important that both engines exert roughly the same force, but how critical is this and how do the two crews arrange it, especially since communication between them is very limited?--Shantavira|^{feed me} 12:52, 2 April 2011 (UTC)[reply]

Why would communication be very limited? Wired telephony has existed since the 1800s, and presumably the crews today would have a wireless system, with in-ear receivers and noise cancellation systems in place to allow for ease of communication... --Jayron32 14:45, 2 April 2011 (UTC)[reply]

Or even more directly, it's hard to imagine that the operators in one locomotive don't have some more direct way to know what the speed/pressure/whatever is in the other locomotive. That kind of "communication" technology would not have to be very sophisticated, I wouldn't imagine. Though honestly I have no clue about trains. --Mr.98 (talk) 16:29, 2 April 2011 (UTC)[reply]

The simple fact is most steam locomotives didnt have wired telephones to other parts of of the train, and that since steam engines in the modern day are run for mostly the sake of nostalgia, they still dont. —Preceding unsigned comment added by 92.20.201.71 (talk) 00:26, 3 April 2011 (UTC)[reply]

I've wondered about the physics here, too. Here's how I've come to think about it:

One question is, what does the throttle setting on an engine actually do? Does it control how fast the engine tries to turn (i.e., in revolutions per minute), or how hard the engine is trying to work?

It's pretty clear that, for most engines, the answer is the latter. Think about a car: if you kept the accelerator pedal depressed with your foot to exactly the same position, you'd notice yourself going more slowly up hills, and more quickly down. If you jacked the car off the ground and kept the pedal depressed to the same position (or, more simply, put the transmission in neutral), the engine would rev to a very high speed indeed.

And in fact, typically what an engine's throttle does is control the rate that fuel is used. So by one of the most basic principles of physics there is, conservation of energy, the engine is going to try to do precisely enough work to transfer the amount of energy per unit time as is contained in the amount of fuel it's provided in that unit of time.

(Side note: some engines do have to move at a constant speed, but that's typically achieved by use of a governor, a basic mechanical feedback device which adjusts the engine's throttle so that the engine is always doing just the right amount of work to offset its -- possibly varying -- load, such that the speed stays constant.)

So, anyway, what happens if the two engines don't have their throttles set to precisely the same position, or if the two engines have significantly different power capacities, or for whatever reason aren't providing exactly the same force?

I used to think that the "stronger" engine would do all the work, and that the "weaker" would get a free ride. But no. If at any instant the stronger engine manages to do more work and pull ahead of the weaker one, the weaker one will see less resistance, and will speed up, until it is pushing, too.

The same physics end up applying in other situations. Suppose you've got a car that's stuck, and two of you are trying to push it, but you're not strong enough, and you hail a passerby, who agrees to help, and with the combined strength of three of you, you manage. Was it necessary that all three of you provided equal force? No. The force on the car is the simple algebraic sum of the forces provided by the three pushers. Every little bit helps.

I also wonder about those tag-along trailer bikes for kids. The adult on the bike in the front is clearly doing the lion's share (often all) of the work, but if the kid is actually pushing, would the adult feel the difference, that's he's having to do marginally less work to keep the train going? I think so. (It would also be instructive to interview people who ride ordinary tandem bicycles about their experiences when one rider is significantly stronger or is for whatever reason working harder than the other.)

Finally, I'm reminded of the story of two people who are trying to move a large, heavy piece of furniture through a doorway. They push and strain, but they just can't make it. A third person comes along and tries to help, but even with three people pushing and pulling, this way and that, they still can't get it to budge. Finally the newcomer says, "You know, I don't think we're ever going to get this thing into the room." "Into the room?", the other two exclaim. "We were trying to get it out!" —Steve Summit (talk) 17:24, 2 April 2011 (UTC)[reply]

Now, when I said, "the engine is going to try to do precisely enough work to transfer the amount of energy per unit time as is contained in the amount of fuel it's provided", that's a bit of a simplification. It can take a fair amount of engineering to realize a practical engine that can in fact transfer large amounts of power effectively and efficiently, even as the speed varies. It was probably simpler back in the days of steam, although even then, there were abstruse complexities having to do with, say, the cutoff. And the diesel-electric control section in our article on diesel locomotives goes into quite a bit of detail about how complicated the power control can be for a modern locomotive. —Steve Summit (talk) 17:37, 2 April 2011 (UTC)[reply]

In the days of the Midland Railway Company (who regularly used double header steam locomotives on the Settle-Carlisle Line where the practice was implicated in two accidents near where I live), the drivers were sufficiently experienced to know what sort of head of steam was required for different sections of the track, but I suspect that they just signaled to each other to approximately match power on setting off. If there was a big difference in power output, then the wheels would slip on the rails because they are linked to the pistons in the locomotive so they are much more likely to slip than those on the carriages. Dbfirs 18:01, 2 April 2011 (UTC)[reply]

It occurs to me that a considerably more interesting problem occurs when you have a helper engine at the rear, assisting one or more engines which are pulling conventionally from the front as a long train attempts to ascend a grade.

When two engines are both pushing (or pulling) from the same end, they're more or less rigidly coupled, so their speed stays the same, and the only thing that varies (that can vary) is the amount of force each applies.

But if some are pulling from the front, and some are pushing from the rear, the "slack" of the train comes into play. (The full train is far from rigid.) Some portion of the train will be being wholly pulled from the front, and some portion wholly pushed from the rear. If the rear engine(s) work a little harder than necessary, the demarcation point will move forward through the train (i.e. with more cars being pushed); if it/they slack off; the point will move back. Moreover, the feedback from this -- the point at which there's clanking as some car or another shifts from being pushed or pulled -- is invisible/inaudible to the engineers at both the front and the rear. So that must have taken some skill! —Steve Summit (talk) 20:29, 2 April 2011 (UTC)[reply]

Assuming the rear engine is not powerful enough alone to push the train, if he just goes at full throttle, the front engine controls simply adjusts their throttle until the required speed. The same principle occurs to both cases, one engine can just provide a constant force which adds to the force provided by the one which is regulating the speed. —Preceding unsigned comment added by 92.20.201.71 (talk) 00:23, 3 April 2011 (UTC)[reply]

Yes, there is some information about this problem at bank engine, but I suspect the physics of double-heading are different.--Shantavira|^{feed me} 08:26, 3 April 2011 (UTC)[reply]

In the days of steam it was not uncommon to see double headed trains being banked up the Lickey Incline, so two at the front and one at the back, or single headed trains being double banked, one at the front two at the back, and I'm almost sure (accepting that it was quite a long time ago) that some double headed trains also needed to be double banked, so two at the front and two at the back. All of which they seemed to manage without any problem. Mikenorton (talk) 09:04, 3 April 2011 (UTC)[reply]

Engine braking

When shifting down a gear to aid braking, where does all the cars kinetic energy go bearing in mind that the engine revs cannot increase instantaneously?--92.28.66.20 (talk) 16:11, 2 April 2011 (UTC)[reply]

We have an article on engine braking, though it could use some work.

I've always understood that you're essentially running the engine as an air compressor, and compressing air heats it up; therefore the kinetic energy is converted into heated gases that are released, either down the tailpipe or (I now learn from the engine braking article) in the case of a diesel engine, back out the intake manifold. —Steve Summit (talk) 16:23, 2 April 2011 (UTC)[reply]

Also if you have a petrol (US=gasoline) car, then the air intake valve will be closed, and the engine is trying to create a high vacuum in the air intake manifold. With a diesel car, there is no air intake valve, so the amount of engine breaking is somewhat less. Ronhjones ^(Talk) 20:24, 2 April 2011 (UTC)[reply]

Worth noting is that large diesel engines (the sort you would find in a tractor-trailer/articulated truck, not the sort you would have in your Volkswagen Jetta) are often equipped with compression release engine brakes, which pop open to vent the compressed air – and release the energy stored by its compression – as the piston reaches the top of the cylinder. This makes the engine braking much more effective in diesel engines that are so equipped, but can also make a terrific racket as the valves pop open and closed. TenOfAllTrades(talk) 16:31, 3 April 2011 (UTC)[reply]

We have an article on Regenerative brake (to make good use of the energy). Dbfirs 18:03, 2 April 2011 (UTC)[reply]

Kinetic energy is very readily converted to other forms of energy, such as heat and potential energy. When someone asks Where does all the kinetic energy go? it is highly likely someone has in mind the Principle of conservation of mechanical energy. Mechanical energy is conserved when the only forces operating are conservative forces, such as gravity and springs. When considering a motor vehicle, and the engine in a motor vehicle, virtually none of the forces operating are conservative forces, so the Principle of conservation of mechanical energy is not applicable. The kinetic energy of the vehicle will easily and rapidly turn into kinetic energy of exhaust gases, heat in the air, heat in the tires and on the road surface. Dolphin (t) 23:41, 5 April 2011 (UTC)[reply]

I'm confused by this as well. Most of the force in engine braking is due to pumping losses, with the cylinders trying to draw a vaccum against the closed throttle. Instead of the pistons pushing on the connecting rods, the connecting rods pull the pistons down. Braking force for each cylinder would be Patm-Pcyl * piston area. Drawing a vaccum actually cools the air though. The kinetic energy of the car must ultimately be converted to heat and be disappated into the air (presumably through the radiator and/or exhaust), but where in the process of creating a vaccum is the heat made and how is it rejected.--129.238.237.96 (talk) 18:22, 15 September 2011 (UTC)[reply]

why electrical arcs don't flow straight

When observing an electrical arc flow from one electrode to another or lightning between cloud and ground, I can't help noticing that (the stream of electrons)the spark does not flow in a streight line. the charge, presumably trying to reach the other node by the most direct route, should,it seems,take a direct streight line. Instead it obviously wiggles around taking many momentary directions, none of which aim directly at the target and constantly overcorrecting untill,like a shark seeking it's prey,it eventually homes in on it's target. I've tried to find an article that explains this but no luck.Can anybody explain why it doesn't travel in a direct streight line.Phalcor (talk) 18:10, 2 April 2011 (UTC)[reply]

It doesn't take the staightest line, it takes the path of least resistance. The electric field is a messy, complicated thing and does not exist as a clean, continuous gradient. It is constantly changing and in flux, not the least of which is changes created by the electric arc itself as it propagates. The two ends of the path are, of course, the extremes of the potential difference, but in between the electric field isn't a clean, regular gradient between those extremes. The result is that the electric field itself displays chaotic behavior; the electric spark will take the path of least resistance always, but that path is somewhat random and always changing. Sparks are actually highly fractal in nature, so their shape and behavior is explainable if not predictable. --Jayron32 19:20, 2 April 2011 (UTC)[reply]

(EC)With respect to an electrical arc flashing between two electrodes (assuming they are at the same height): the arc is at a temperature of many thousands of degrees, and the air and ionized gasses in and near the arc expand and become lighter than the surrounding air, so they rise in an arch shape (that is where the 'arc" got its name in the early 1800's). As for lightning bolts, much is still not known per the Lightning article. Speculating here: A bolt between cloud and cloud or cloud and ground might move in other than a straight line because it is following higher conductivity paths where there is more moisture in the air, or where ice crystals have caused a separation of charge in the atmosphere, creating an ionized path. There is no reason that the ionized path of lower resistance should be a straight line. Edison (talk) 19:36, 2 April 2011 (UTC)[reply]

A Jacob's ladder is an easily made demonstration of an electric arc being carried upwards by air convection. Cuddlyable3 (talk) 21:58, 2 April 2011 (UTC)[reply]

The path a river takes might be a simple analogy. You'd expect it to go straight, but it actually follows the terrain of the ground, which may seem flat at first look but isn't when you look at it finely enough. StuRat (talk) 04:34, 3 April 2011 (UTC)[reply]

Its actually an excellent analogy, StuRat, and we can carry it further: as the river carves the ground, it encounters differing types of soil and rock, so its course will be altered as it affects all of those soil and rock types differently. It both affects its own course through its own erosion, AND it is highly dependant on small and numerous differences in its environment on its course to the ocean. This is exactly like a lightning bolt, which has to navigate through air whose properties are different all along its course, and the lightning bolt affects its own environment by changing the electric field along its path. --Jayron32 04:56, 3 April 2011 (UTC)[reply]

Some atmospheric conditions, such as wind, can modify the lightning into ribbons. ~A H 1^(TCU) 22:57, 3 April 2011 (UTC)[reply]

Name this garden flower?

In south east England: http://img16.imageshack.us/i/unknownflower.jpg/ The plant is branching, so to look at there is a mass of flowers. The flower is wilting rather. Done on a scanner. What could it be please? Thanks 92.29.121.249 (talk) 18:21, 2 April 2011 (UTC)[reply]

Looks like a night-scented stock to me. --TammyMoet (talk) 18:51, 2 April 2011 (UTC)[reply]

Thanks very much. They could be night-scented stock but when searching for images of that, they look more like the illustration I found here http://www.suttons.co.uk/Shop/Flower+Seeds/Virginian+Stock+Confetti+Mix+Seeds+137036.htm so I think they are more likely to be Virginian stock.

Slightly off-topic but I think I scattered the seeds in the autumn, and they began flowering in late March. Another scatter-and forget seed I would recommend is Love-in-a-mist (latin name Nigella). The all-blue type looks better than the multi-coloured. They both self-seed so one seeding should give many years of trouble-free flowers.

Are there any other easy scatter-and-forget seeds anyone would recommend? Thanks 92.15.9.102 (talk) 08:34, 3 April 2011 (UTC)[reply]

If the flowers are on the smaller side, then Virginia stock is probably the one. There are a number of plants that, to quote my late father, "you only have to plant once", if you can get one of these plants then you won't need to replace it (or plant anything else!): Feverfew, mint, Leycesteria, any of the geranium family, aubrieta, herbaceous potentilla, Welsh poppy. My garden contains quite a few of these! --TammyMoet (talk) 09:14, 3 April 2011 (UTC)[reply]

Aluminum foil

Why is one side shiny and the other side dull? --70.244.234.128 (talk) 19:43, 2 April 2011 (UTC)[reply]

Aluminum_foil#Properties: "Aluminium foil has a shiny side and a matte side. The shiny side is produced when the aluminium is rolled during the final pass. It is difficult to produce rollers with a gap fine enough to cope with the foil gauge, therefore, for the final pass, two sheets are rolled at the same time, doubling the thickness of the gauge at entry to the rollers. When the sheets are later separated, the inside surface is dull, and the outside surface is shiny." --Mr.98 (talk) 19:46, 2 April 2011 (UTC)[reply]

Of course, it's not that difficult. (And I notice that our article doesn't supply a source for the claim that it is.) If manufacturing tolerances were the only concern, then the much-thicker sheets of heavy-duty foil could be manufactured in single thickness, and they would have two shiny sides. Of far more interest to the manufacturer, I suspect, is the fact that running two layers of foil through the rolling press simultaneously doubles their output. TenOfAllTrades(talk) 16:45, 3 April 2011 (UTC)[reply]

What has more hydrogen?

I read somewhere that a liter of petroleum (or gasoline can't remember) has more hydrogen in it than a liter of liquid hydrogen. What about water? Does a liter of water have more hydrogen in it than a liter of liquid hydrogen? ScienceApe (talk) 19:48, 2 April 2011 (UTC)[reply]

Liquid hydrogen has a density of 67.8 kg·m-3, which means that 1 cubic meter will have 67,800 grams of hydrogen molecules, or ~~135,600 hydrogen atoms~~. Petroleum is about 10-14% hydrogen by weight (according to our article). Petroleum is a mixture of a whole bunch of stuff, but according to List of crude oil products they seem to range in API gravity from 25-60, so lets take a nice middle of API gravity of 40, which translates to 141.5/171.5 = 0.82 g/cm3 or 820 kg/m3. 10-14% of 820 = 82.0 - 114.8 kg, which is 82,000 - 114,800 grams of hydrogen atoms per cubic meter. Divide those numbers by 1000 to get the per liter amounts. So no, on average petroleum does not seem to contain more hydrogen than pure liquid hydrogen, but those are large ranges we are working with; the densest (lowest API gravity) petroleum with the highest hydrogen mass (14%) may come close or even exceed the figure; this doesn't seem all that unusual as the types of intermolecular bonding that goes on in each substance is likely to be substantially different, accounting for differing numbers of atoms per unit volume. --Jayron32 20:13, 2 April 2011 (UTC)[reply]

Editing my response. The relevent comparisons are 67,800 grams of hydrogen per cubic meter for liquid hydrogen and 82,000 - 114,800 grams of hydrogen per cubic meter for petroleum. My math got confused for a second. So yes, clearly petroleum has mor hydrogen per cubic meter than liquid hydrogen. However, as I noted before this is unsurprising, as most of the hydrogen in petroleum is bonded via covalent bonds in large molecules, while the diatomic hydrogen in liquid hydrogen is in relatively small molecules. The intermolecular distances in a substance are somewhat larger than the intramolecular distances within a molecule; since petroleum has larger molecules, it can pack more hydrogen atoms in a smaller space. While by mass most of petroleum is carbon, by number of atoms it averages a little more than two hydrogens per carbon, so it is mostly hydrogen (by number of atoms), and the larger molecule size means that those hydrogens are, on average, packed closer together than in an equivalent volume of liquid hydrogen. --Jayron32 21:46, 2 April 2011 (UTC)[reply]

What about water? ScienceApe (talk) 02:21, 3 April 2011 (UTC)[reply]

Water has a density of 1000 kg/m3. 1/9 the weight of water is hydrogen. So 1000/9 = 111 kg or 111,000 grams per cubic meter. So, it too has more hydrogen per unit volume than liquid hydrogen. --Jayron32 02:27, 3 April 2011 (UTC)[reply]

This is truly a remarkable thing to point out. If you could "teleport" the oxygen out of a liter of water, and the two hydrogens bonded directly to one another instead, you'd be left with about 111 grams of liquid hydrogen. Which would want to expand about 65% under whatever conditions are given for the density figure above! In order to get it to fit, you'd have to increase its density to the brink of the transition to metallic hydrogen (1.08 to 1.33 g/cm³).^[3] I think that this emphasizes the difference between water and a typical nonpolar liquid loosely bound by Van der Waals forces. We recall things like methane, ammonia, hydrogen sulfide don't interact strongly enough with themselves to be liquid near room temperature. I think water is so dense because the network of hydrogen bonds holds it together almost like a solid, though with less stability. In a sense, if you think of the hydrogen bonds as a very weak covalent bond, you could almost think of a mass of water as a single molecule, like diamond. Wnt (talk) 04:01, 3 April 2011 (UTC)[reply]

I dunno. I don't find it all that remarkable, but then again my primary training and experience is as a chemist and chemistry teacher, so what I am likely to find commonplace and self-evident may be remarkable to others. Modern physics works roughly like magic for me, so I guess I understand your perspective here... --Jayron32 04:40, 3 April 2011 (UTC)[reply]

You're quite correct that water's propensity to form hydrogen bonds is what allows such an otherwise lightweight molecule to remain a liquid at room temperature, but hydrogen bonding is pretty much irrelevant to the fact that liquid water has a high number density of hydrogen atoms. Any solid- or liquid-phase hydrogen-rich molecule can be expected to exceed the hydrogen concentration of liquid hydrogen. If we consider completely nonpolar liquid methane (CH₄) at its boiling point (112 K), it has a density of 423 g/L (423 kg/m³). To a reasonable approximation one quarter of that mass is hydrogen, giving a hydrogen content of 106 kg/m³—within very close hailing distance to that of liquid water (and higher than the hydrogen content of water ice). If you want to compare with nonpolar molecules that exist at room temperature, hexane and cyclohexane contain 107 and 112 kg hydrogen per cubic meter. TenOfAllTrades(talk) 17:17, 3 April 2011 (UTC)[reply]

You make a good point; but those molecules link together more than two hydrogens into a single small unit. I can understand packing together hydrogen molecules by gluing them together with carbons. What amazed me about water is that it has the same number of molecules as liquid hydrogen, and so far as I know the molecules are bigger than diatomic hydrogen, yet they end up taking up less space. Wnt (talk) 17:32, 3 April 2011 (UTC)[reply]

That's because in chemistry, bond length and bond strength are inversely proportional; the intermolecular bonding in liquid hydrogen is the very weak london dispersion forces, so the distances between those individual H2 molecules will be quite long, comparatively speaking. For liquid water, the intermolecular bonding is via hydrogen bonding, a particularly strong type of dipole-dipole force, which being a stronger bond must also be a shorter bond; indeed much shorter to the point where the distance more than makes up for the increased H-H distance through that oxygen atom... --Jayron32 21:01, 3 April 2011 (UTC)[reply]

That's similar to what I said above ... but a lot clearer. ;) Wnt (talk) 22:04, 3 April 2011 (UTC)[reply]

Hissssss touch bringsssssss... decay!

In the fictional Judge Dredd comics, the touch of the evil Judge Mortis causes a person's body to instantly decay as if the person were already dead, causing said person to immediately die. If we assume such a thing is possible, is there a specific effect of the body decaying that causes immediate death? JIP | Talk 20:21, 2 April 2011 (UTC)[reply]

The scenario you mention as context is unscientific, and is therefore not within the scope of this desk. Regarding the specific question of what effects of decay could cause immediate death, this is interesting because it highlights key structures that are actively maintained in vivo. A few examples: (i) loss of integrity of major blood vessels, leading to massive internal bleeding; (ii) widespread loss of endothelial tight junctions, leading to multi-organ failure; (iii) loss of integrity of myelin sheaths on neurons; (iv) depolarization of membranes in crucial tissues like heart and brain. I am sure there are many others. -- Scray (talk) 20:40, 2 April 2011 (UTC)[reply]

You can rot while alive, see Gangrene and necrosis. Both can cause eventual death. --Jayron32 03:45, 3 April 2011 (UTC)[reply]

Of course, but those are piecemeal processes, generally. When systemic, they are rapidly fatal. -- Scray (talk) 04:49, 3 April 2011 (UTC)[reply]

Absoltely. Which is why I brought them up. Good to see we agree. --Jayron32 04:53, 3 April 2011 (UTC)[reply]

I don't really agree, because they don't answer the original question. Probably not worth further discussion, though. -- Scray (talk) 06:11, 3 April 2011 (UTC)[reply]

Human decomposition is a complicated process. Lots of individual parts of that process would be fatal were they to rapidly and systemically occur in a living human. Red Act (talk) 07:56, 3 April 2011 (UTC)[reply]

Spontaneous human combustion? ~A H 1^(TCU) 22:53, 3 April 2011 (UTC)[reply]

Don;'t some hemorrhagic fevers cause necrosis on internal organs? At least they seem to in popular media, but not having studied medicine, I expect that the truth is much less sensational then depicted. Googlemeister (talk) 14:00, 4 April 2011 (UTC)[reply]

Necrotic flesh is dead, so it's not performing its function as a collection of cells anymore. If, for instance, part of your heart necrotized (for whatever reason), the muscles would no longer be able to properly pump blood. — The Hand That Feeds You:^Bite 16:30, 4 April 2011 (UTC)[reply]

snowflakes

Why do snowflakes have hexagonal symmetry? I know that it has something to do with the hexagonal arrangement of the water molecules, but could someone be more specific? Thanks. 74.15.137.130 (talk) 20:54, 2 April 2011 (UTC)[reply]

The Wikipedia article Snowflake says the flake's six-fold radial symmetry is because the crystalline structure of ice is six-fold, and mentions several possible growth mechanisms whose details remain controversial. Cuddlyable3 (talk) 21:35, 2 April 2011 (UTC)[reply]

Thanks. 74.15.137.130 (talk) 03:21, 3 April 2011 (UTC)[reply]

Remember that not all flakes are hexagonal. Some come in the form of "columns" or "needles" and other dendrites. ~A H 1^(TCU) 22:51, 3 April 2011 (UTC)[reply]

Relativity paradox

Get a sample of a fissionable material that's just under the critical mass and accelerate it to near the speed of light. While it's moving, try to start a chain reaction. From a stationary view, it should have the critical mass and thus start a chain reaction, but an observer moving with the material would see that it doesn't have enough mass. What happens? --70.244.234.128 (talk) 21:46, 2 April 2011 (UTC)[reply]

Even the comoving observer will be able to detect the added energy from the acceleration, this added energy will, from his point of view, be the source of the criticality This isn't a paradox... The two will see the excursion happen at different times, but this is due to the relativity of simultaneity. Now, I am a physics retard, so someone next will come along and explain why my explanation is wrong, but that is how I read this situation. --Jayron32 21:50, 2 April 2011 (UTC)[reply]

To the observer who's moving with the sample, its mass will still seem to be less than critical because the relative velocity is zero. --70.244.234.128 (talk) 21:55, 2 April 2011 (UTC)[reply]

Yes, but the additional forces due to acceleration will be clearly evident. In accelerating the sample, the forces on the sample will create additional pressures which, to the comoving observer, will be the source of the criticality. If the comoving observer calculates the critical mass in conjunction with the forces introduced due to the acceleration, criticality will become mathamatically explainable. Criticallity is not solely mass-dependant. You can decrease the critical mass of a substance by, for example, placing it under pressure. Accelerating it rapidly can introduce that pressure. In other words, the comoving observer will see the mass remain constant, but will be able to detect a change in density due to forces introduced by acceleration; that change in density will be his reasoning for the criticality. Again, I am a physics retard, so please wait for corrections to my explanation... --Jayron32 22:00, 2 April 2011 (UTC)[reply]

If we don't take the acceleration literally, and just consider a lump of fissionable material, rest mass just under critical, moving relative to us with near speed of light, then the answer is: No, it won't go off. It doesn't go off in its rest frame, hence it doesn't go off in our frame. 70.244... is presumably thinking of the increase of "relative mass", a popular but ultimately useless concept, and the scenario proposed here is a fine example why it is useless. Bulk motion has no effect on processes that depend only on mass, i.e. it does not increase the mass. The effective mass of the lump can be increased if internal degrees of freedom are excited (e.g. heating it up), and this may indeed happen when forces due to a real acceleration act on it. --Wrongfilter (talk) 22:29, 2 April 2011 (UTC)[reply]

The critical mass is not a Lorentz invariant. Why would you expect it to be the same in any frame? —Preceding unsigned comment added by 92.20.201.71 (talk) 00:14, 3 April 2011 (UTC)[reply]

Sorry, my answer was incredibly ambiguous in a way which did not occur to me late at night , but that Dauto has well highlighted. To rephrase: The relativistic mass is not a Lorentz invariant (even when as the rest mass approaches critical mass). Why would you expect it to be the same in any frame? —Preceding unsigned comment added by 92.20.201.71 (talk) 11:48, 3 April 2011 (UTC)[reply]

It seems to me that the standing observer and the traveling observer are both totally irrelevant to action of the material. If the mass is increased to criticality it will just do it's thing and get really hot. not neccessarily go bang.Phalcor (talk) 01:40, 3 April 2011 (UTC)[reply]

Wrongfilter's answer above is the only correct one so far. Yes 92.20.201.71, your answer is also incorrect. Critical Mass, like any rest mass, is a Lorentz invariant and that's exactly the point here. 70.244.234.128's mistake is to confuse relativistic mass (That's the one that increase with speed) AKA energy with rest mass (That one doesn't change with speed and is a Lorentz invariant) AKA mass. Dauto (talk) 03:17, 3 April 2011 (UTC)[reply]

I agree with WrongFilter and Dauto.

I'm assuming in the following that the acceleration involved is small, and it's the object's speed that's what's actually of interest.

Unfortunately, the word "mass" by itself is actually ambiguous, since the word by itself is sometimes used to mean rest mass, and is sometimes used to mean relativistic mass. You can actually start a heated semantic argument about which of the two one should reserve the word "mass" by itself to mean, and a heated debate about the value or lack thereof of the concept of relativistic mass. But unlike the word "mass", the phrases "rest mass" and "relativistic mass" are at least unambiguous.

What's actually important in determining whether an object made of fissionable material will be critical is how many nuclei of the fissionable material there are. Mass is just used as a convenient easy-to-measure stand-in for the number of nuclei, which works because the two values are proportional (the element's atomic mass being the proportionality constant). The number of nuclei there are in an object clearly doesn't change depending on how fast the object is moving. So to remove the ambiguity of the word "mass" in the phrase "critical mass", you could think of the critical mass as being the "critical rest mass", although almost nobody uses that phrase. There is no paradox involved with critical mass, because the word "mass" in that phrase means rest mass, which doesn't change with the object's speed.

As explained in Critical mass, criticality depends on other things besides number of nuclei. And some of those things are indeed measured to change when an object is moving at great speed, due to length contraction. However, the Lorentz factors wind up balancing each other out, so the critical mass is again unaffected by an object's speed. But I won't get into that in detail, since you just brought up the potential paradox due to relativistic mass, and didn't bring up any length contraction issues at all. Red Act (talk) 03:49, 3 April 2011 (UTC)[reply]