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June 19

"Plant" Identification

I found these little white balls growing in my backyard. What are they? Baby mushrooms? Are they dangerous in any way? What is the best way to kill them for good? --Kainaw ^(talk) 00:05, 19 June 2007 (UTC)[reply]

They are definitely a fungus of some sort. --24.147.86.187 00:57, 19 June 2007 (UTC)[reply]

Yes, mushroom. As far as danger, they potentially could be harmful if eaten (don't try it) but beyond that prob not much danger. Complete eradication may be problematic. An application of fungicide would help with surface and shallow spores. Applying a good mulch would also help too especially if you plant some ground cover along with it. All that being said, they will probably disappear just as fast as they popped up. 161.222.160.8 01:05, 19 June 2007 (UTC)[reply]

I'm guessing those mushrooms came with the compost/manure that they're growing on. They're probably not harmful to the soil, although I am not suggesting you eat them. Since the spores are probably already in the soil, you're probably going to end up making the soil sterile getting rid of them. If they're unsightly, just turn the soil over, but you'll probably get more after a heavy rainfall. -- JSBillings 11:58, 19 June 2007 (UTC)[reply]

Yep - I raked them all up and turned the soil (as you can see from the pic, there's no grass there). We had a week of heavy rain and they popped right back up. I'll get a good fungicide on them. Then, it is back to the wisteria fight. --Kainaw ^(talk) 14:32, 19 June 2007 (UTC)[reply]

Those fungus may be benificial! I would think twice before eradicating them-most funguses are essential in the decomposition process.

Also, some fungus will appear quickly after heavy rain and disappear quickly when it doesn't rain in a few days. Give it a couple days and see if it is already gone (assuming it doesn't keep raining). Youth in Asia 22:15, 19 June 2007 (UTC)[reply]

Yep - they are already gone. That picture was from yesterday and today there is no trace of them. --Kainaw ^(talk) 00:31, 20 June 2007 (UTC)[reply]

Spring Constant Discrepancy

I'm sure I must be overlooking something simple here.

Take a hypothetical spring with a spring constant k. Hang it vertically and attach a weight of mass m to it. The mass stretches the spring by distance x. Using Hooke's Law, F = kx (removing negatives for simplicity, as far as I calculate it works the same).

However, in this case F = mg, as the spring is being stretched due to gravity, therefore F = kx = mg. Solving for k gives k = mg/x.

Now, the change in height of the mass, h will be the same as x. Therefore the change in gravitational potential energy can be found by E_p = mgh = mgx. Assuming no energy loss and that the mass started and finished stationary, this E_p is now stored as elastic potential energy in the spring, found by U_s = 0.5kx².

Therefore, mgh = mgx = 0.5kx². Solving for k gives k = 2mgx/x² = 2mg/x.

So, using Hooke's Law I get k = mg/x, using conservation of energy I get k = 2mg/x. Where's the mistake? --jjron 01:40, 19 June 2007 (UTC)[reply]

Good question. Think of it as an experiment? Hang an unstretched spring vertically, then add a mass and let it go. What happens? It accelerates downward, the spring stretches out, and eventually starts to slow down. At some point it turns around and comes back up. This cycle of simple harmonic motion repeats. The trick is that you don't only have gravitational potential energy and elastic potential energy, you also have kinetic energy. When the spring stretches to the equilibrium position, where there's no net force, you have just proven that half of the initial gravitational potential energy was converted to elastic potential energy, and the other half was converted to kinetic energy. Simple harmonic motion results as the energy keeps getting shifted back and forth between kinetic and potential energy. --Reuben 01:51, 19 June 2007 (UTC)[reply]

Yes, let's ignore friction in this 'ideal' world. I'm talking about the ideal situation here. If you regard the maximum stretch as zero height for simplicity, then at this point gravitational potential energy E_p is zero, and since velocity is zero, kinetic energy is also zero. Thus, all we've got is U_s stored in the spring, which is the problem I have with Reuben's answer, at maximum stretch you don't have kinetic energy. In my original question I was stopping it there, as Tenofalltrades suggests, but it doesn't really matter if it's oscillating (re above comment, 'yes', ignoring friction, but surely the mean extension is just the midpoint between zero extension and full extension). If it is oscillating, it will return to it's starting position, solely due to the energy stored in the spring. So this doesn't seem to be a solution. --jjron 03:00, 19 June 2007 (UTC)[reply]

But at "maximum stretch" (full extension), the stretch is 2x (if you take x to be the stretch at which gravity balances elastic force, i.e. x is the mean extension). That's what's causing your discrepancy. --Spoon! 03:25, 19 June 2007 (UTC)[reply]

You're right. Sorry, I hadn't really thought about the oscillation properly and I slightly misinterpreted what Tenofalltrades initially said. The oscillation does make a difference.

If you go back to the original question though I had the weight just hanging at stretch x, i.e., where 'gravity balances elastic force' (or vice versa) - E_p and E_k both at zero. It was Reuben that started it oscillating. And I was getting the discrepancy in this original situation, so I don't think that it does solve the discrepancy. --jjron 04:14, 19 June 2007 (UTC)[reply]

If the starting point is 0, and the point where gravity balances the spring's restoring force is x, then the point of maximum stretch is 2x. At 0, all the energy is in gravitational potential energy. There's no kinetic energy and no elastic potential energy. As the weight falls, it reaches maximum velocity at x. At that point, half of the initial energy is in kinetic energy. After that, it slows down again, until at 2x the weight is momentarily at rest. All the energy is now in elastic potential energy. If you're just sitting at the equilibrium point x (instead of oscillating around it), then you must have dissipated some of the initial energy through friction or some other loss mechanism. The energies don't balance, because some has gone into kinetic energy or been dissipated. --Reuben 04:52, 19 June 2007 (UTC)[reply]

Ah, now that makes sense, good explanation. You're right, if you think about that in a practical situation, to stop it at that midpoint you do have to dissipate energy, even in the theoretical ideal situation. Now it also makes more sense why for some situations you can only use Hooke's Law to find the spring constant. Thanks to all that made an attempt at explaining this, it's been troubling me for a couple of weeks. --jjron 07:40, 19 June 2007 (UTC)[reply]

Confusion with electrical units

Help me out, guys. I can't understand all these units that have to do with electricity. Okay, voltage is the potential, or how much power it can hold. If current is like how a river flows, then what in the world do coulombs do? --A Confused Student 03:13, 19 June 2007 (UTC)

A coulomb is a specific number of charges (usually electrons). Using your river analogy, it's a specific-sized bucket of water. An ampere (often shorted to amp) is a current flow of one coulomb per second: one bucket of water flows past a point on the shore of the river each second. More amps = wider river. TenOfAllTrades(talk) 03:27, 19 June 2007 (UTC)[reply]

And voltage is the water pressure. —Steve Summit (talk) 03:33, 19 June 2007 (UTC)[reply]

HHAHAHA THANK YOU SO MUCH! Wow, you guys answered it so quickly! Thanks Steve and Ten! --A Confused Student 03:46, 19 June 2007 (UTC)

And ohms are big boulders.--Shantavira|^{feed me} 06:55, 19 June 2007 (UTC)[reply]

Now someone try to fit henrys and teslas into this ;-) Someguy1221 07:08, 19 June 2007 (UTC)[reply]

Henrys (measuring inductance) are easy. An inductor is a spot where the river widens into a lake. Larger and deeper lakes represent greater inductance. (An inductor 'smooths out' high-frequency noise; a lake 'smooths out' short-term changes in river flow.) I leave teslas an as exercise for the next reader. TenOfAllTrades(talk) 13:15, 19 June 2007 (UTC)[reply]

That's clever, but inductors don't store charge; maybe an inductor is some kind of lock system instead. --Tardis 15:34, 19 June 2007 (UTC)[reply]

An inductor opposes any change in the flow of electrical current by generating an EMF which would tend to maintain the flow unchanged. The momentum or kinetic energy of the flowing water would appear to have this characteristic: if the channel is narrowed, the height (and pressure) of the water increase, and contrariwise if the water channel is broadened. A capacitor absorbs or releases charge in a way which tends to maintain a constant voltage (pressure) I picture the airfilled "anti-Water hammer" tube in the wall above a faucet acting in this way, to buffer changes in pressure. Edison 19:26, 19 June 2007 (UTC)[reply]

Capacitance is easy. It's like a rubber membrane across the river (or pipe, which is a better analogy), completely sealing off the two sides from eachother. We recognise a characteristic behaviour of a capacitance in that you can't pass a DC current through it. The membrane will bulge to one side if you try, but will soon resist any further flow. An inductance I imagine perhaps as a setup of paddles, that the water has to turn to pass through. The paddle system is made heavy, so it takes a while to get up to speed, and frictionless, so there is no limit to how fast the flow will go if you continue to apply a pressure difference across the paddle system. —Bromskloss 19:41, 19 June 2007 (UTC)[reply]

Is it just me or has this analogy become more confusing than what really happens?

We actually have an article on this subject: Hydraulic analogy. -Arch dude 01:39, 20 June 2007 (UTC)[reply]

Ah, nice. Didn't know about that one. I note, smugly, that my capacitance (rubber sheet) and inductor (heavy, frictionless paddle wheel) corresponds very well to the article! :) —Bromskloss 16:27, 22 June 2007 (UTC)[reply]

Of course real paddle wheels have some friction, and real inductors have some resistance.

How would a paddle wheel work in a completely closed system, anyway, though? — Omegatron 23:53, 22 June 2007 (UTC)[reply]

Parrot jaw strength

Can anyone here explain in layman's terms how parrots are capable of exerting so much pressure when biting down? For a (relatively) small creature, they have incredible jaw strength - it's an amazing 'design'. --Kurt Shaped Box 09:30, 19 June 2007 (UTC)[reply]

My guess would be that the muscle connection point on the lower jaw is rather far from the jaw pivot point (relatively to the jaw length). This results, for a given muscle mass, in larger force but smaller travel of the jaw. Pressure is determined as force divided by the area normal to which the force is applied. In parrots the lower (mandibular) beak is sharply pointed up, so the force is applied over a very small area, and thus a very high pressure is produced. I repeat, this is just a guess based on trivial observation (large force and small travel of the lower jaw). I couldn't find anything useful on the net, but if anyone does, a link would be great. Cheers, Dr_Dima.

That seems a pretty good explanation. Jaws basically operate as third-class levers - if it's a short lever like in the parrot jaw with the effort occuring relatively near the load and away from the fulcrum then you can generate a lot of force.

Muscles are incredibly strong, but we usually don't realise how strong because they're running via the bone levers in such a way that a small contraction results in a large movement, e.g., contracting your biceps to raise your hand. Consider for example how much easier it is to hold a big weight up at the bend in your arm than in your hand - at this point the mechanical advantage is much higher and you're getting more an indication of how strong your muscles really are.

You'll also notice that while parrots have a strong bite, they can't actually open their beaks very far. This is good indication that the muscles are inserting quite near the load (and away from the pivot) because muscles are limited in how far they can contract and elongate. Therefore a lot of the real muscle strength is being transferred in the parrot jaw, instead of being lost in causing a large range of motion. The pointy beak mentioned by Dr Dima is also relevant. --jjron 08:20, 20 June 2007 (UTC)[reply]

That makes sense. Thanks, guys. :) --Kurt Shaped Box 14:37, 20 June 2007 (UTC)[reply]

Carcinogens

In your information on essential oils you state "The smoke from burning essential oils may contain potential carcinogens, such as polycyclic aromatic hydrocarbons (PAHs)." I was wondering about a couple of things. Firstly, which exact essential oils are potential carcinogens. Secondly, is it only the burning of the essential oil that can make it carcinogenic, or can this still happen when the oil is simply heated to a temperature of no more than 80 degrees; that is warmed, not burned? If you don't know the answers, do you know where I could look to find them? Thanks for your helpDeb770 11:11, 19 June 2007 (UTC)[reply]

Warming or just volatalising the oil wont make the carcinogens - the oils would need to be burnt to do that.

Any burning carbonaceous matter has the potential to form this compounds - (even candles) - essential oils are more likely to form PAH's though as they already (often/sometimes) contain a ring.

NOTE if you just warm the oils ie 80-100C there won't be any SMOKE!

Did you ask this a few days ago, or am i déjà vecuing it? 213.48.15.234 11:58, 19 June 2007 (UTC)[reply]

She asked it on the Miscellaneous desk here and I suggested she ask it here instead. --jjron 08:00, 20 June 2007 (UTC)[reply]

Ah, thank you! 213.48.15.234 10:12, 20 June 2007 (UTC) [reply]

Note that a lot of this stuff is flying around - the combustion engines in cars burn oil and spew the carcinogenic stuff into the air we breathe. If you live in a city like Los Angeles or Bangkok, you might as well smoke tobacco. Well, maybe that is somewhat exaggerated (though not too much), but when in LA I saw a big anti-smoking sign by the side of a very busy road that rather made me laugh. DirkvdM 09:19, 20 June 2007 (UTC)[reply]

Actual complete burning doesn't create these things so much as heating in the absence of oxygen, via smoldering or such. Thus the problem with tobacco smoking, or with carcinogens from grilling food rather than broiling. Gzuckier 18:48, 20 June 2007 (UTC)[reply]

Broiling says that carcinogens are produced by both... — Omegatron 00:26, 23 June 2007 (UTC)[reply]

Most destructive particle

What particle, theoretical or real as of today, may be considered as having the most destructive force if in a sci-fi setting it was to be propelled towards matter? Ie in the form of a cannon against a human or spaceship or anything like that. Not necessarily restricted to only 1x particle, might as well bundle some few million or more of them together. It's not (all) about the gathered kinetic force, but also some interesting speculations as to how the particles exert different kinds of forces, magnetic properties, etc. So, I guess this half qualifies for the Entertainment desk, but I'd rather hear some fantastic suggestions here. :) Thanks in advance. 81.93.102.185 12:26, 19 June 2007 (UTC)[reply]

You are walking a tightrope of physics here. On one hand you would like your particle to penetrate the shields (whatever matter or field they may be "made of") without interaction, on the other hand you would like your particle to interact with the ship interior or a specific part thereof (living things, computers, engines, ammunition magazines, etc.). I am afraid that the lack of information on the shields / armor and on the inner structure of the ship makes any speculations on our part largely meaningless. Could you give us some information to work with, please? Cheers, and good luck with your writing. Dr_Dima.

A neutron fired at a ship made from uranium could do a bit of damage.

(Indent the above as I thought it might belong to Dima. Sorry if not.) Shields are of no concern, you might as well talk about common battleships or similar. Simply just a gun propelling some kind of substance that, in this amount, can be imagined to for some reason or another affect whatever is hit - say steel or something. For instance, some super-bullet or so of a tau lepton, so compressed and heavy due to its mass that it develops an amazing kinetic force. Or some other stuff. I'd really just fancy various sci-fi weapons with half-realistic roots in... well, reality. 81.93.102.185 13:12, 19 June 2007 (UTC)[reply]

The problem is that the 'bullet' you are shooting has to be accellerated somehow in the barrel of your 'gun'. The amount of energy it takes to accellerate it is precisely the same as the energy that's delivered to the target when it hits it. If you choose a very heavy bullet, for a particular amount of energy used to accellerate it, you get a lower top speed. A lower speed means it's easier to dodge. I think you want really light particles - accellerated to the fastest speed you can manage. The only reason we like heavy bullets here on earth is that they do better against air resistance. In deep space, shooting feathers at the enemy would be more effective than shooting depleted uranium - assuming you use the same amount of energy to propel them. SteveBaker 13:52, 20 June 2007 (UTC)[reply]

Antimatter is pretty destructive. Gandalf61 13:38, 19 June 2007 (UTC)[reply]

You could one-up the US military's gay bombs with some sort of homoerotic cruise missiles. --TotoBaggins 14:45, 19 June 2007 (UTC)[reply]

Antimatter en-masse like said above will cause problems. Antineutrons preferrably.--GTPoompt (talk) 15:50, 19 June 2007 (UTC)[reply]

If accelerated to high enough velocities, anything would do the job. Even sand.[1] --V. Szabolcs 06:52, 20 June 2007 (UTC)[reply]

At sufficiently ridiculous energies all impacts are the same because the forces/properties of the impactor are neglible compared to the effects of its kinetic energy. Even poorly interacting particles, like neutrons and neutrinos, will become strongly interacting if given enough energy. Dragons flight 08:40, 20 June 2007 (UTC)[reply]

The problem here is that the 'shields' present in most SciFi spaceships has no real world counterpart. We could imagine using a magnetic field to deflect charged particles - but if they fired a laser at you - or a big, heavy lead bullet - you really have no 'field' that's going to help you much. You can't use an electrical field because if won't affect neutrally charged bullets - a magnetic field isn't going to do much to non-ferrous metals like lead. If you had a chunk of neutron-star inside your ship, maybe you could have a gravitational field - but that would attract bullets - not make them bounce off! That leaves you with the strong and weak nuclear forces to make fields with - but you do that by having nice thick kevlar armor! So there isn't a known/possible way to construct StarTrek-like 'shields'. Given that - it's not really sensible to talk about how to defeat them. SteveBaker 13:45, 20 June 2007 (UTC)[reply]

So, a powerful magnetic field, combined with thick, mirrored armour would be the way to go? --Kurt Shaped Box 14:36, 20 June 2007 (UTC)[reply]

Yep - that would cover all the bases if you didn't know anything about your enemy. Solid armour might be a bad idea if he hits you with something really big - maybe I'd go with putting half a kilometer of aerogel around my ship - with mirrored space shuttle heat tiles on the outside. Keeping the ship light would allow much more accelleration - and "not being there when the shooting starts" is always the best defense. You'd definitely want 'stealth' capability too - so a very angular/factetted look with radar absorbing paint under your mirror finish. SteveBaker 19:55, 21 June 2007 (UTC)[reply]

If you want to get seriously kooky, you might try a stream of strange quarks [2] - just be sure to have some mechanism for disposing of the strangelet cum starship afterwards. Or if you have some mechanism for storing gargantuan quantities of energy but still want to keep your own starship maneuverable, bare quarks can rip additional matter out of the vacuum to satisfy color confinement. Conservation of energy is still satisfied, of course, but the cascades observed in modern particle accelerators can end up with dozens of times the original mass. This will also slow the initial beam down enough to have a reasonable interaction cross section with the target - no sense in shooting at something with a highly penetrating ray. -Eldereft 07:35, 21 June 2007 (UTC)[reply]

Hmmm - nasty! SteveBaker 19:55, 21 June 2007 (UTC)[reply]

Lost matter!

Hello! How much matter is lost from the earth's total matter everyday? I mean like, I've read that certain planets can lose their water into space through their atmosphere. So I was just wondering, of all the objects raining down on earth, plus all the matter that is lost through the atmosphere or sent out by space ships, how much is the mass of the earth changing every day/year/decade or whatever is a realistic way of measuring it. ► Adriaan90 ( Talk ♥ Contribs ) ♪♫ 13:54, 19 June 2007 (UTC)[reply]

I do not know the answer to your question, but I think you should note that the Earth's magnetic field protects it from much of the atmosphere loss that other planets suffer. If the Earth were to lose its magnetic field, the atmosphere would be stripped away rather quickly - which means that we would eventually lose a lot of our water. Of course, we wouldn't notice because the oxygen would be gone first. --Kainaw ^(talk) 14:47, 19 June 2007 (UTC)[reply]

The Earth gains much more from debris hitting it than it loses to stuff in the high atmosphere getting pinged away at escape velocity. Figures for the mass increase vary widely; one number prominently batted around is 40,000 metric tons per year. [3] [4] --TotoBaggins 14:54, 19 June 2007 (UTC)[reply]

Remember the suggestion a few years ago that Earth's atmosphere is hit every day by numerous tiny comets? Was that finally laughed down? —Tamfang 06:13, 20 June 2007 (UTC)[reply]

I seem to remember NASA research or video indicating that little objects do constantly shower down on the earth's atmosphere. Edison 17:41, 20 June 2007 (UTC)[reply]

Retractable claws

Can animals with retractable claws retract them individually, or not? And what if a beast with opposable thumbs and retractable claws arose? Would it need indivivdually retractable claws too?

The claw article doesn't really say, but to the best of my knowledge they cannot individually control them. Animals with retractable claws such as cats cannot individually control their toes (unlike say we can with our fingers), they tend to all do the same thing, e.g., scrunch up one toe on a given foot, they all scrunch up. Going on this, I can see no reason that they would have individual control over their claws.

The second/third question is an interesting hypothetical. I can't see why it would need individually retractable claws, but could it evolve them? Perhaps. It may come down to the evolutionary history of the beasts involved. For example, is their anything preventing a primate with already opposable thumbs from evolving retractable claws? I don't know.

Or if you go the other way, can a beast with retractable claws evolve opposable thumbs? One interesting 'experiment' in this is the giant panda (I'm not sure whether it has retractable claws, but both they and cats are part of the Carnivora Order, so not that far separated in evolutionary terms). In order to 'grip' the bamboo it eats it has evolved a 'false thumb' out of one of its wrist bones, rather than having one of its digits become opposable (basically for the reasons I outlined above about all its toes doing the same thing). The red panda article actually seems to give a better description of this, and also says that its claws are semi-retractable. Whether you can draw any conclusions about retractible claws from this I don't know, but it may shed some light on the issue. --jjron 03:09, 20 June 2007 (UTC)[reply]

Who in their right mind would answer this question?

Why is the human brain split down the middle? Lateralization of brain function answereth not. Is this true of other animals? Clarityfiend 19:20, 19 June 2007 (UTC)[reply]

To answer the second part of your question, the article on the Medial longitudinal fissure seems to indicate it is a feature of vertebrate brains. There is a good picture on this page that shows comparative brain anatomy for fish, birds, amphibians, and reptiles, all of which have a medial division of the cerebrum. If you look at the brain of the lamprey, which is a primitive vertebrate, it almost looks like the cerebrum is a glorified olfactory bulb. My guess is that the cerebrum has two separate halves because in more primitive animals it solely consisted of the olfactory bulbs, which of course are separate. Over time the two halves came together, and eventually in mammalian brains the corpus callosum developed to connect the two halves (according to that article, monotremes and marsupials don't have a corpus callosum). So the answer to your first question comes out of the second. The brain in higher animals like humans has compensated for having a split forebrain by developing the corpus callosum and lateralization of brain function. --Joelmills 21:53, 19 June 2007 (UTC)[reply]

I think it's because God likes dual-core processors. --Steve Summit (talk) 22:50, 19 June 2007 (UTC)[reply]

It actually basically comes down to the fact that we are bilaterally symmetrical organisms. We could then go into the evolutionary history of this, the embryology, etc, but as to why it's 'split down the middle', well that's why. It is true of other bilaterally symmetrical animals (not just vertebrates), though may not be so obvious as their brains don't have the complexity. --jjron 02:33, 20 June 2007 (UTC)[reply]

I think that's more of an answer of what, not why. Evolution describes what very well. Even fills in the blanks. But why can be very elusive. Since evolution is responsible for brains that are both medially split and not medially split, nature has selected for both. Why are mammalian vertebrates split? is a more general question and I submit that it is not answerable with our current understanding. As noted above, perfectly successful organisms have survived and evolved without the split. So while it is notable to understand what the split is, how it functions, how it evolved, etc, the answer as to why will be elusive because it cannot be tested as a hypothesis. --Tbeatty 14:32, 20 June 2007 (UTC)[reply]

But only vertebrates have a telencephalon (I think), which is the only part of the brain that is completely split in half, except for the corpus callosum in mammals and the variouscommissures. I think that is what Clarityfiend was asking about, but I could be wrong. --Joelmills 03:42, 20 June 2007 (UTC)[reply]

Your interpretation of my question is right, Joelmills. Thanks for the info. Clarityfiend 06:43, 20 June 2007 (UTC)[reply]

Nevertheless, bilateral symmetry is the constraint evolution has worked under, and structures may have further evolved / expanded in a way that created a fissure. "Why" questions are dangerous to ask, it can imply functionality in a system when the truth might merely be "pre-existing constraints made this path the easiest to evolve". Perhaps it is easy to expand regions of the brain in size / layers, but more difficult or less important to evolve neural circuits that "connect" the halves (after all, severing the corpus callosum isn't exactly lethal), leading to the formation of a fissure. This is merely my own speculation here. Madeleine 07:02, 20 June 2007 (UTC)[reply]

PS - a perhaps more interesting question is why brain asymmetry evolved making right and left brains different ("lateralization"), since symmetry is the default. Maybe helps with multi-tasking. [5] [6] Madeleine 07:19, 20 June 2007 (UTC)[reply]

Yeah, but we're not very symmetric on the inside. Show up at X-ray with a heart on each side of your chest, or even one that's more central than usual, and watch the excitement.Gzuckier 18:45, 20 June 2007 (UTC)[reply]

Yeah, look at the spinal cord in cross section: [7] It's already split lengthwise down the middle, and the brain is just a sort of mushrooming of the end of it. That's becausein development, the spinal cord forms from the neural fold, in which a stripe down the center of the back sinks down and the edges curl over it to make a tube, which has a seam running up the back. Gzuckier 18:45, 20 June 2007 (UTC)[reply]

to random questions.

Don't blame me, someone else asked me to ask you. 1Why doesn't vodka freeze when you put it in a freezer? 2can you get different types of lightning?

If it doesn't (I don't know if it does or not), it's because its freezing point is lower than the temperature of the freezer. The freezing point of a solution is generally below that of its constituent parts.
Sure. See lightning. — Lomn 20:34, 19 June 2007 (UTC)[reply]

Vodka contains water, Ethanol, and almost nothing else. The freezing point of ethanol is -114.3 degrees C. 100 proof vodka is half water, half ethanol. I have determined empirically that it does not freeze at -20 Degrees C, neither does 80 proof gin, but 80 proof gin mixed as 4:1 with 24 proof vermouth begins to freeze at about that temperature. see also freeze distillation. -Arch dude 23:18, 19 June 2007 (UTC)[reply]

There is the well known phenomenon Ball lightning, where a glowing ball floats through the air (often during thunderstorms), but scientists are so far unsure of whether this is a form of lightning. Anvil-to-ground lightning is pretty interesting too: the lightning can travel horizontally for miles, often hitting a site so far away that those it hits cannot even see the thundercloud! Lightning doesn't have to be produced by the weather either: spacecraft, nuclear bombs and volcanoes have all triggered lightning strikes (as this particularly apocalyptic image shows!) Laïka 19:24, 20 June 2007 (UTC)[reply]

Two of my 9v adapters got fried

Hello, I play guitar and in the electric guitar world there are effects pedals that act as a footswitch to turn a desired effect on or off. These need to be powered in some way, and I prefer using adapters. I started using a universal AC adapter for one of my pedals and set it to 9v (these kind of adapters have the 3 tips on the end). After a few months, it stopped working one day. I looked at it and noticed that the LED light did not light up at all and the little sticker on the top had become warped; the adhesive attaching it to the surface of the adapter was pretty much worn away too. So I tried another adapter (the exact same kind) on the same pedal, and within about a month, the same thing happened. The sticker is warped and everything. It seems like it melted off or something, I'm assuming the adapter got too hot and just fried. This hasn't happened to any of my other adapters - incidentally, none of them are universal. Anybody have a clue what's going on here? Thanks! NIRVANA2764 20:15, 19 June 2007 (UTC)[reply]

It may be that the universal adapter you used couldn't supply as much current as that particular effects pedal was using. For example, even though you correctly set the adapter for 9 volts, if the pedal was trying to draw 300 milliamps, and the adapter was only capable of supplying 120 mA, and if the adapter was cheaply made (without any internal overcurrent protection), it could have burned itself out.

Any time you're matching power supplies, you want to pay attention to both the voltage and the current. For conventional arrangments (i.e. all you're likely to see), the voltage should match exactly, and the current supplied by the adapter should be equal to or greater than the current required by the device. (If the adapter can supply more current than the device requires, this is no problem, and gives you a margin of safety. The adapter won't blow out the pedal with "too much current" or anything; the pedal will take only as much current as it needs.)

The current required by the effects pedal you've been having problems with might be printed on its nameplate, or is probably printed on the original power adapter that came with it (if by chance you still have that available). --Steve Summit (talk) 22:14, 19 June 2007 (UTC)[reply]

Those guitar effects pedals will run for an hour or so on a regular 9v PP3 battery (at least the ones that my son uses do) - so we can make a guess at how much current they demand. Most 9v batteries produce about 500mAh - so if they run down and die after an hour, they are providing ~500mA for one hour. So I'd make sure that whatever adapter you are using is able to provide at least (say) 1 Amp at 9 volts. If you bought an adaptor that can only provide 500mA, it'll be running at maximum capacity and it will tend to overheat and die prematurely. So go with a 1A adaptor or higher so it'll run cooler. Note also that some of those universal adaptors produce different amounts of current depending on the voltage you demand from them - check carefully on the box. SteveBaker 13:29, 20 June 2007 (UTC)[reply]

Plants that tend to produce hollow seed pods?

Hello, So, weird plant question:

Does anyone know of varities or strains of plants that produce mostly hollow, or all hollow plant seeds or husks? I am picturing a rice grain, which is core less. Also, any methods of inducing this in other plants?

Thank you.

Gourd Loofah end up with hollow fruits. Even the coconut ends up hollow after the milk has drained out. I cant help you with tiny hollow seeds though. GB 07:10, 20 June 2007 (UTC)[reply]

Try the pitaya fruit. Its seeds are hollow. And there are many. Cheers!!! -Zacharycrimsonwolf 12:03, 22 June 2007 (UTC)[reply]

Calculating required force for gym lifts

Originally posted this question in the math section but was advised to post it here instead. I'd be glad to receive any answers that you can give. The question was this: How do you calculate the force required to perform classic gym lifts like the deadlift? If it was only a matter of E = mgh it would be easy, but the fact that joints are working in different directions makes it difficult. I'm unable to make use of the information in the torque article, which I suppose should be of use. Could anyone give me a few guidelines how to calculate this kind of movements? Thanks in advance, Jack Daw 21:57, 19 June 2007 (UTC)[reply]

In the most general case, you will experience axial force, shear force, moments in two different planes and a torque (6 quantities in total). You really need to calculate all these things, because all of these need to be separately balanced by your effort. Also remember that when you lift a weight, it accelerates, so the total force includes not only the gravitational force but also inertia force. The way to find all these quantities is to draw a good sketch of the process in 3D showing all the dimensions, and use it to draw free body diagrams. Let me know if this makes sense, or else I can keep elaborating on it. deeptrivia (talk) 00:39, 20 June 2007 (UTC)[reply]

Hmm ok... I'm completely unable to do any of that. Please elaborate more. I don't need an equation that is 100% accurate, only one that gives you an estimate. Jack Daw 13:32, 20 June 2007 (UTC)[reply]

Do you really need the forces and moments (harder to calculate) or just the energy/power required? If lifted slowly, and under some assumptions, the energy needed will approximately be mgh. I'll have to draw figures to show how to calculate forces, which I'll do at a later time. Some information required are: dimensions of the person's arm, speed at which the load is lifted as a function of time (or maybe just the highest acceleration that the load would ever experience), the type of lift, the shape of the object being lifted (if it's not "small") and any significant friction/damping in the joints (can ignore these for a rough estimate). deeptrivia (talk) 14:47, 20 June 2007 (UTC)[reply]

Well I suppose I don't need the forces and moments, I'm just interested in Watts, basically. What dimensions do you need? And can't one just use variables for that (length l, diameter d etc) and include that in the final equation? I don't know the speed either, that varies from lift to lift. Friction of the joints can be ignored yes, that would be too much to ask, as well as any resistance from the air upon the person and weight. (who knew this was so complicated) Jack Daw 17:24, 20 June 2007 (UTC)[reply]

If you just want a total number for the power in Watts - then E=mgh is just fine. If you're ignoring friction and stuff like the inefficiency of muscles - then you calculate the energy required (m=mass in kg, g=gravity in ms^-2 and h=height in m) and divide it by the time taken to perform the lift to get the average wattage. So if a champion power lifter can hoist 100kg to 2m in maybe 2 seconds, they are producing 100*9.8*2/2 which is nearly 2kJoules over 2 seconds - almost a kiloWatt!...wow! In reality, it's more than that because at the start of the lift, their body is closer to the ground than at the end - so you need to figure out where their center of gravity is at the start and end of the lift and push that through the same equation. Our article on muscles says that the quadricep muscle can produce 100 Watts for several minutes - so with arms and legs working together, 400 Watts is believable - and for a well-trained individual with really huge cross-section muscles, a kilowatt doesn't seem that unlikely. It also says that human muscles have been measured to be between 14% and 27% efficient. Since 2kJoules is about half a kcalorie, our power lifter will need to burn (say) 4kcal to do the work...about what you'd get from eating a single sugar cube! SteveBaker 01:42, 22 June 2007 (UTC)[reply]