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November 4

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emissivity

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physical meaning of emissivity —Preceding unsigned comment added by Quiet.mp (talkcontribs) 02:15, 4 November 2010 (UTC)[reply]

See Emissivity. -- kainaw 02:18, 4 November 2010 (UTC)[reply]

Chemistry

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What substance melts at Negitive 97 degrees and boils at 76 degrees? —Preceding unsigned comment added by 67.182.212.131 (talk) 03:34, 4 November 2010 (UTC)[reply]

Seems like that might depend on whether you mean degrees Celsius, Fahrenheit, etc. DMacks (talk) 03:36, 4 November 2010 (UTC)[reply]
Regardless of which temperature scale you use, there are likely to be hundreds of substances which broadly meet these requirements. Given the millions and millions of possible substances in existance, it is quite impossible to identify a substance given only those two bits of info. --Jayron32 03:38, 4 November 2010 (UTC)[reply]
See acetyl bromide for a common substance with those specific melting/boiling points (within 1 degree). -- kainaw 03:43, 4 November 2010 (UTC)[reply]
Why isn't the acetyl bromide boiling point a sharp point? Is it because there are actually multiple species upon boiling, or that there are different microphases? John Riemann Soong (talk) 04:01, 4 November 2010 (UTC)[reply]
Probably because no one has bothered to measure the boiling point any more specifically. If you want to get an exact boiling point (where vapor pressure = 1 atm), you're going to need to control both the temperature and pressure pretty well. Anyway, my CRC handbook gives 76 C. Buddy431 (talk) 04:32, 4 November 2010 (UTC)[reply]
Just to explain in a bit more detail, while melting point is usually vital information for identifing a compound (melting point analysis is usually a standard analytical technique for identifying both identity and purity of a compound), boiling point usually is not, being that boiling point tends to be had to control for, as it will vary slightly due to a LOT of different factors. So, while someone could nail down a more exact boiling point (equipment and standard techniques do exist for this) they may not have, at least in the literature used as a reference for the Wikipedia article. --Jayron32 05:06, 4 November 2010 (UTC)[reply]
Is this slightly harder (than for say, water) because of the tendency of acyl chlorides to be impure? John Riemann Soong (talk) 06:34, 4 November 2010 (UTC)[reply]

Autoclave naming

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Autoclave means self-locking device. But why did they get this name? It seem to me they got named after an insignificant feature compared to their general purpose. —Preceding unsigned comment added by Jib-boom (talkcontribs) 04:39, 4 November 2010 (UTC)[reply]

I guess that the self locking devices to prevent blow-out on the early pressurised autoclaves were unusual enough for it to be included in the name of the machine. Perhaps something like "Modern Autoclave Sterilising Machine. Like all new words 'autoclave' became the significant part of the name and possibly it became 'The Autoclave', with the purpose now well known. The English language is a strange and infinitely mutatable medium. There are many examples of names that have left their original meanings back in the mists of obscurity. ?Chauffeur,[1] for example. Caesar's Daddy (talk) 07:48, 4 November 2010 (UTC)[reply]
See google books: The London literary gazette and journal of belles lettres.. the name was invented by fr:Pierre-Alexandre Lemare, ironically a linguist.213.249.225.56 (talk) 11:52, 4 November 2010 (UTC)[reply]
His linguistic skills may have been a bit week as "auto" is Greek and "clavis" is Latin. Further, "self-lock" (using his Greek-Latin system) would be something more like "autoclaus", which my personal limited Latin knowledge tells me is close to meaning "self-contained". -- kainaw 12:18, 4 November 2010 (UTC)[reply]
Just like the French and German mix in the "YesYes" board game -- Sjschen (talk) 19:45, 4 November 2010 (UTC)[reply]
So if it sanitizes would it be a "saniticlaus?" Edison (talk) 18:31, 4 November 2010 (UTC)[reply]
There ain't no saniticlause! APL (talk) 13:04, 5 November 2010 (UTC)[reply]

Escape Trajectory

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I understand an orbiting mass around a larger mass can be pulled into a escape trajectory by a passing mass. But what would be the simplest trajectory that would illustrate this interaction?

Just to clarify with an example: what would the trajectory of the Moon following its last orbit around the Earth look like if a planet size mass passed by and pulled the moon into an escape trajectory, avoiding a collision with both bodies? TheFutureAwaits (talk) 09:16, 4 November 2010 (UTC)[reply]

Even if we ignore the gravitation of the Sun, what you ask is a Three-body problem. They can be difficult. Cuddlyable3 (talk) 10:48, 4 November 2010 (UTC)[reply]
Sure but we know there are solutions that involve putting a mass into an escape trajectory...I'm just asking what the simplest case would be? It's really just the opposite of a capture trajectory which involves a moving mass passing near another mass at hyberbolic speeds when a 3rd body removes enough velocity through its own gravity to reduce the eccentricity <1. For seem reason I have a much harder time visualizing the reverse... TheFutureAwaits (talk) 11:34, 4 November 2010 (UTC)[reply]
Newtonian gravity is time symmetric. So, if you can visualise the geometry of a capture trajectory, just run the film backwards to visualise an escape trajectory. Gandalf61 (talk) 11:40, 4 November 2010 (UTC)[reply]
But the starting condition in this case has one body already "captured" by another. I think TFA is imagining eg. the moon being "captured" from the earth by a third body? So running that backwards gives you another "capture", not the case where all three bodies gravitationally escape each other. WikiDao(talk) 16:12, 4 November 2010 (UTC)[reply]
Simulations of three-body problems and passing rogue stars are available from this astronomy site. ~AH1(TCU) 18:05, 7 November 2010 (UTC)[reply]

Tropical Laserbeam

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If I were to take a green laster pointer and point it at the night sky will some of its light make it out of the atmosphere? What about if I use a red or blue laser pointer? Or a flashlight? TheFutureAwaits (talk) 12:38, 4 November 2010 (UTC)[reply]

Unless it is cloudy at least a few of the photons should be able to make it out into space, but unless it is a very powerful light, you would not really be able to differentiate those photons from other light "clutter". The atmosphere is pretty transparent to the visible portion of the EM spectrum. Heavy airborne pollution or dust would limit the amount escaping, but without anything to absorb the light or bend it back to earth, it will go into space. Googlemeister (talk) 13:15, 4 November 2010 (UTC)[reply]
Lasers are used to precisely measure the Earth-Moon distance, so clearly they can make it out of the atmosphere (and back); see Lunar Laser Ranging experiment. Sure these are not just your garden variety laser pointers though, but as Googlemeister says, undoubtedly some the photons from those would escape as well. The article linked above states that "...out of 1017 photons aimed at the reflector [on the moon], only one will be received back on Earth every few seconds, even under good conditions (they can be identified as originating from the laser because the laser is highly monochromatic)." --jjron (talk) 13:37, 4 November 2010 (UTC)[reply]
(ec)To avoid confusion, it should be noted that the reason so few photons return is not because so few make it out of the atmosphere, but because the the diameter of the laser beam will be very large by the time it hits the Moon (kilometers across). Most of the photons which reach the Moon strike the lunar surface instead of the retroreflector; similarly, most of the photons that do get reflected back end up spread over the Earth's surface rather than entering the (comparatively) small aperture of the receiving telescope. I wouldn't be surprised if the answer to the original poster's question is actually "most of the light makes it out of the atmosphere" — provided that it isn't cloudy. TenOfAllTrades(talk) 14:14, 4 November 2010 (UTC)[reply]
Here is a simple thought experiment: The laser pointer, or flashlight, or torch in your hand appears much brighter than Sirius. Light from Sirius easily makes it through the atmosphere, so light from your local device will make it up, too. In numbers, Sirius puts out about 1028W. Your local light source puts out ~1W. Sirius is about 80228200000000000 times farther away that the flashlight (2.6 parsec vs. 1 m), so it appears 6.4e+33 times fainter than it would at one meter. So Sirius is ~60000 times fainter than you local light, and still is clearly visible through the atmosphere. --Stephan Schulz (talk) 14:03, 4 November 2010 (UTC)[reply]
Sirius is spreading light in all directions, while the laser sends all it's light down a narrow path, so I don't think your math works out that easily. Googlemeister (talk) 14:32, 4 November 2010 (UTC)[reply]
You are right, but the laser light is just as attenuated by 1/r2 as Sirius. So yes, we lose a small constant factor because we do not see the backside of Sirius, but that's a minor effect. --Stephan Schulz (talk) 14:48, 4 November 2010 (UTC)[reply]
I am not sure that you mean attenuation because (a) that only happens in a medium, not in a vacuum and (b) it follows an exponential law. Are you saying that the intensity of a laser beam in a vacuum decreases as 1/r2 ? I find that hard to understand. I can see there would be a some fall-off in intensity with distance due to a small amount of beam spreading, but I don't see how it could be anything near 1/r2. Gandalf61 (talk) 15:52, 4 November 2010 (UTC)[reply]
Well, there's nothing out there focusing the beam, so the sides of the beam form straight lines, so the diameter of the beam is proportional to r, so the area of the beam is proportional to r2, so the intensity of the beam is proportional to 1/r2. Red Act (talk) 16:12, 4 November 2010 (UTC)[reply]
But the beam has already been focussed - it doesn't need continual refocussing. If the sides of the beam are almost parallel then your r is not the distance from the source - in effect, r is the distance from a "virtual" source, a long way behind the real source, where the sides of the beam would meet if projected backwards. Gandalf61 (talk) 16:48, 4 November 2010 (UTC)[reply]
That is true, but only important if r is of the same order of magnitude as (or smaller than) the distance back to the virtual source. On astronomical scales, the virtual source of the laser beam and the physical location of the laser aperture are in essentially the same place. TenOfAllTrades(talk) 18:04, 4 November 2010 (UTC)[reply]
A beam with a wavelength of 600nm and an original diameter of 2mm has a divergence of 0.0003 radians (even larger if the beam is partial spatial coherent).[2] Still assuming a 2mm aperture, that means that the "virtual" source is only about 6m behind the "real" source, which is completely negligible compared to the 100km thickness of the atmosphere. Over the 100km thickness of the atmosphere, the beam will widen to a diameter of about 30m, compared to which the original 2mm diameter is negligible. Red Act (talk) 18:10, 4 November 2010 (UTC)[reply]
I follow your argument, but something is askew in your calculations or model or assumptions, because a beam width that increased by a factor of 15,000 (30m/2mm) over 100km would increase 6 x 107 times over 400,000 km, and then a 2mm aperture would give a beam width of 120km at the Moon's surface, whereas the actual beam width at the Moon's surface in the Lunar Laser Ranging experiment was only 6.5 km - see below. That's different by a factor of 20. Gandalf61 (talk) 11:54, 5 November 2010 (UTC)[reply]
The diffraction limited divergence angle of a laser beam is inversely proportional to the initial diameter of the beam; see the link I provided in my earlier post. The OP, and Stephan Schulz's first post, refer to the laser as being a laser pointer, so for my calculation I assumed an initial beam diameter that would be reasonable for an inexpensive battery-powered handheld laser pointer. The observatories participating in the Lunar Laser Ranging Experiment use much more expensive, much higher power lasers, with a much wider initial beam diameter. My point would still stand for the LLRE lasers, though. Even if the distance from the virtual source to the real source is 20 times bigger than in my calculation (in the ballpark of 120m), that's still negligible compared to the 100km thickness of the atmosphere. Red Act (talk) 17:39, 5 November 2010 (UTC)[reply]
Actually, I just found out that the divergence of the laser beam used by the Apache Point Observatory Lunar Laser-ranging Operation (APOLLO) isn't even diffraction limited, because they use a beam expander to expand the beam to a huge 3.5m initial diameter. Instead, atmospheric divergence dominates in their case.[3] Red Act (talk) 18:48, 5 November 2010 (UTC)[reply]

In light of this discussion, here is another question: If I were sat atop the retroreflector used for the lunar range finding experiment, would I be able to see the beam with the naked eye? HappymulletukHappymulletuk (talk) 22:44, 4 November 2010 (UTC)[reply]

Well Lunar Laser Ranging experiment says "the reflected light is too weak to be seen with the human eye", but nothing about your question. However it also says "at the Moon's surface, the beam is...about 6.5 kilometers wide", and as stated above "out of 1017 photons...only one will be received back on Earth every few seconds". Now allowing for a pupil diameter of 8mm and assuming the 6.5km wide spread of the laser is circular, my back of the envelope calculation says that some 6,059 photons would enter your pupil 'every few seconds' (which is horribly imprecise). However, that many photons would most likely be detectable (technically you could 'see' one photon), but I doubt that it would be intense enough that you would determine its source to be that laser. (Willing to be corrected here...). --jjron (talk) 07:24, 5 November 2010 (UTC)[reply]
Could you be a bit more specific about your assumptions here? Your number feels like it's a bit high. (Here's my envelope; feel free to compare with the back of yours.) Given 1e17 photons spread uniformly (er...) over a 6.5 km diameter circle on the moon, and assuming a 1-square-meter retroreflector (I think the actual exposed reflective surface is smaller, but we'll go with that), then we're down to 3e8 photons striking the retroreflector. If we then assume that the reflected beam spreads no more than the original beam (a generous assumption, I suspect) and reaches a detecting telescope with an aperture of 1 meter diameter, then the scope picks up just 70 photons. Meanwhile, the astronomer next to the telescope is looking in vain — with his tiny 8 millimeter pupils, he gets 0.005 photons per shot.
In practice, the actual return rate is much worse; the Apache Point Observatory Lunar Laser-ranging Operation (APOLLO) project boasts a 3.5-meter telescope, and is very happy to see on the order of one photon returned per pulse. I suspect that this is largely due to divergence of the beam after it strikes the retroreflector, rather than inefficiency in the ground-based detector. TenOfAllTrades(talk) 18:39, 5 November 2010 (UTC)[reply]
The question asked by Happymelletuk is about someone sitting on the retroreflector, not about someone standing next to the telescope.
The divergence of the reflected beam is indeed somewhat worse than the incident beam, but not enormously so. Due to thermal deformations of the retroreflector, the reflected beam has a divergence of about 7 to 10 arcseconds.[4] Red Act (talk) 19:08, 5 November 2010 (UTC)[reply]
Yeah, my calc was done based on someone sitting on the retroreflector on the moon as was asked. FWIW I only calculated it for one eye, so open both eyes and you could double those photons! :) (Mind you it could still be out - just quickly retried and got a different answer, but still in the magnitude of ≈103). --jjron (talk) 16:38, 6 November 2010 (UTC)[reply]

You can easily compute the apparent magnitude of a laser as a function of distance. For a black body radiator at a temperature of roughly 5500 K (e.g. the Sun) the formula is:

where F is the flux produced by the black body source. Now our eyes are about 7 times more sensitive to green light of the wavelength produced by green laserpointers than to sunlight. Our eyes are most sensitive to light of these wavelengths, while any black body radiator will always emit a lot of energy to either longer and shorter wavelengths. So, in case of a green laserpointer, we have:

Taking the beam divergence to be 0.0003 radians (as Red Act explains above, this is the ideal beam divergence that can be obtained by a laserpointer with a 2 mm initial beam diameter), the flux of a 5 milli-Watt laserpointer a distance d away is:

Inserting this in the above formula for visible magnitude gives:

So, even at a distance of 1000 km, it will still appear to be magnitude -3.1, approximately similar to Jupiter in the sky right now. Count Iblis (talk) 20:10, 5 November 2010 (UTC)[reply]

The laser ranging beam will appear to be magnitude -5.9 for someone looking into it from the Moon. Count Iblis (talk) 21:40, 5 November 2010 (UTC)[reply]

Although the light from hand-held laser pointers may not escape the atmosphere, they can still be bright enough to cause temporary blindness in the cockpits of aircraft kilometers above the surface, so much that shining a laser at a plane is illegal. The laser beam usually expands as the distance increases, so the light may be less focused. However some non-laser lights such as the xenon lamps used to light IMAX screens would apparently be visible from Earth, were it placed on the Moon. ~AH1(TCU) 18:04, 7 November 2010 (UTC)[reply]

algae and plankton

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Would destruction of all the algae and plankton on the Earth destroy human civilization as we know it, and if so how? --96.252.213.127 (talk) 14:04, 4 November 2010 (UTC)[reply]

I can't find a reliable source quickly, but it's usually said that the ocean's algae is responsible for some large percentage of Earth's oxygen. Numbers between 50% and 80% are often mentioned in this context. Even assuming it's only 50%, that kind of oxygen loss would probably be devastating. Especially if it happened suddenly. APL (talk) 14:22, 4 November 2010 (UTC)[reply]
Not to mention that the whole food web would be shot to pieces. --Stephan Schulz (talk) 14:25, 4 November 2010 (UTC)[reply]
Based on the previous two answers: Yes. -- Sjschen (talk) 19:43, 4 November 2010 (UTC)[reply]
I'm not so sure. Most of Earth's population gets its food from land-based food sources (i.e. Maize), so the collapse of the marine ecosystem wouldn't really matter so much. The oxygen's a bigger deal. This document shows what happens to people in low-oxygen environments. If algae do really contribute this significantly to oxygen levels, things could get bad. Humans are pretty inventive though, so I bet enough people would create oxygen collecting and breathing devices (we already have them for Mountain Climbing) to prevent the total extinction of the human species. Buddy431 (talk) 00:58, 5 November 2010 (UTC)[reply]
Wouldn't there also be a massive upshot in atmospheric CO2? That could be bad on its own. (Though, I suppose it would help the maize in the short term.) APL (talk) 13:03, 5 November 2010 (UTC)[reply]
The Earth system's two main sources of oxygen are the oceanic plankton (and cyanobacteria) and the tropical rainforests. The oxygen synthesis and carbon dioxide removal capacity of both are declining. The rise of Carbon dioxide in Earth's atmosphere would also simultaneously cause a depletion of atmospheric oxygen at twice the rate that the carbon dioxide is rising. ~AH1(TCU) 17:59, 7 November 2010 (UTC)[reply]

Superinfections?

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How Many HIV Superinfections are there. I understand this is a second strain of the virus. and I also understand that the second stage of the virus may cause more rapid disease progression or carry resistance to medicines. How long is the life expectancy if someone is infected with a superinfection.

Questions are: How many Hiv Superinfections are there? How long is the life expectancy is someone is infected with a Superinfection? —Preceding unsigned comment added by 91.111.85.243 (talk) 16:16, 4 November 2010 (UTC)[reply]

According to HIV superinfection and the reference it cites, HIV superinfection is when an infected person becomes infected with a second strain of the virus, making the disease harder to treat. This seems to be different to what you understand of the term, at least from what you've written anyway. Because of this, there aren't a certain number of superinfections that could be counted. There are however, many Subtypes of HIV, but I think that it is impossible to say how many strains there are since it mutates so frequently. Life expectancy will vary depending on many factors, the HIV article has a section on prognosis and superinfection isn't going to make it any better. SmartSE (talk) 16:35, 4 November 2010 (UTC)[reply]
Also take a look at SuperAIDS. ~AH1(TCU) 17:57, 7 November 2010 (UTC)[reply]

Acceleration at event horizon

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In the wikipedia's article on gravity acceleration is written that on event horizon the gravity acceleration is infinite. But isn't infinite only at singularity? The event horizon is a few kilometres from singularity, so shouldn't be infinite. Francesco —Preceding unsigned comment added by 95.234.212.148 (talk) 16:52, 4 November 2010 (UTC)[reply]

Wikipedia does not have an article called gravity acceleration. There is an article on gravitational acceleration but this says nothing about any event horizon. Please could you state exactly which article you are talking about?--Shantavira|feed me 18:06, 4 November 2010 (UTC)[reply]
Yeah, I thought maybe there was just a missing comma between "gravity" and "acceleration", meaning the article in question was Gravity. But that article also doesn't talk about acceleration near black holes. Neither do General relativity or Black hole, for that matter. At any rate, near black holes, the Newtonian gravity approximation completely breaks down, and you have to use general relativity. And in general relativity, gravity isn't even viewed as being an acceleration. Instead, objects travel in a straight line (a geodesic) along a curved spacetime. Red Act (talk) 19:23, 4 November 2010 (UTC)[reply]
The event horizon is typically the edge where the acceleration or speed of matter is moving toward/around the black hole or away from the observer at faster than the speed of light, and it is not necessarily infinite but may be in certain equations. ~AH1(TCU) 17:56, 7 November 2010 (UTC)[reply]

Capillary DNA sequencers on eBay?

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How much do sequencers cost? The department I've joined (as a PhD student) sends out samples to a company within another university for sequencing and it just makes me wonder why most large universities that carry out such research don't have their own internal sequencer for internal use. Wouldn't that work out cheaper than outsourcing, even with the cost of a technician to operate it? ----Seans Potato Business 18:03, 4 November 2010 (UTC)[reply]

Many financial decisions in research are very opaque to a graduate student. It is very possible, for example, that your research group or department has received funding grants with specific line item requirements - for example, $ X to be spent on outsourced technical work; $ X to be spent on capital equipment purchases. (If the funding comes from the government, this may be in support of a wide-reaching policy of using research to foster certain economic activity, or other totally opaque objectives from your point of view). Maybe the best resource to ask is your PI or Ph.D. advisor; depending on your research conditions, they may be willing (and even happy) to discuss research-funding details with the students. At my school, one detail was that money granted to individual professors was "taxed" (by the University!) at a different rate depending on its purpose: capital equipment investments were taxed differently than money granted for student labor, research expenses, and so forth - so the grant-writers, knowing this, played all kinds of mind-bending games to make sure they maximized the impact of their money. Nimur (talk) 18:13, 4 November 2010 (UTC)[reply]
(ec) I'm not sure what they cost, but remember that fully-operational sequencing services can't just spring out of nowhere. Someone has to write the grant application, find space, hire the staff, and supervise the facility. Unless there exist a few principal investigators who each do a lot of sequencing (and therefore see sequencing as big line items in their budgets) and who get along well enough with each other to write such a grant proposal, it's unlikely that it will happen at your university. It may also be that unless utilization of the sequencers remains high (50% of the maximum throughput? 80%?) and there are enough tasks to fully occupy a full-time technician (how many sequencers can one full-time tech operate?) then any theoretical per-sequence savings will be smothered in operating overhead costs, and then you've gone to a tremendous amount of time and effort to build a facility that's less cost-effective than the original outsourcing option.
As well, outsourcing 'future-proofs' your sequencing needs. If someone introduces a new technology then it's up to your contractors to figure out how to pay for it — you don't have to write a new grant. If the output from one service isn't up to your requirements, then you can change to another lab (or send piecework off to other labs in cases where specialized treatment or tools are required). TenOfAllTrades(talk) 18:35, 4 November 2010 (UTC)[reply]
Oh, right, I see. Well, I'll put £50 in the pot to buy Edinburgh Uni's own sequencer. :) I'll ask my supervisor what it's all about. ----Seans Potato Business 18:48, 4 November 2010 (UTC)[reply]
Keep in mind that the cost of the machine itself is a *very* small part of doing sequencing. There's disposables/reagents associated with each run. (You replace the acrylamide in the capillary after each run, and the capillary itself is only good for a limited number of runs.) Then there's the maintenance costs associated with the equipment (usually 5-10% of the list price of the equipment *per year*). Finally, there's the time cost for the the technician to run it, and personnel costs tend to be the most expensive part of any business. Given that there are a number of companies who have brought assembly-line efficiency with razor-thin profits to DNA sequencing, you'd have to be doing a lot of sequencing to make it cost effective: think enough work for several full time technicians. You say you are using "a company within another university" currently, which makes me guess that you're probably already getting the same near-at-cost price the other university gets. It's highly unlikely that your university would be able to do it cheaper by itself. Indeed - after evaluating the price, quality, and turn-around time, for us it's actually better for to go with a private, commercial provider for sequencing, rather than use either of the two on-campus sequencing services we have. -- 140.142.20.229 (talk) 21:59, 4 November 2010 (UTC)[reply]

Are we very tiny in the universe?

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I don't mean this as a silly question, but I can't get over how tiny the earth is compared to the sun and how tiny the sun is compared to Arcturus. So, are we like the tiny people living in the locker of the movie Men in Black, where we don't even know how small we are compared to the universe around us? Are there other planets as big as Arcturus, and is it possible there are giant lifeforms and we are just a speck of dust compared to them? (If I sound insane, please let me know. Thanks.) AdbMonkey (talk) 18:31, 4 November 2010 (UTC)[reply]

I think I get what you're saying. You're asking whether we are to some ET lifeform as, say, small insects are to us? ----Seans Potato Business 18:42, 4 November 2010 (UTC)[reply]
I think the OP is referring to the film closing, which zooms out on our planet/solar-system/etc. to reveal that it's just the inner molecular structure of a giant alien. While it's tempting to look at solar-system models and compare them to bohr atom models, they are not the same. Different physical laws govern the behavior of our solar system (mostly, gravity), as opposed to the electromagnetic and nuclear effects that govern atomic behaviors. So, the very large structures in the universe are probably not just bigger versions of atoms. What we can see in the universe around us is large-scale structure of the cosmos - things like galaxies, and then at larger scales, clusters and filaments, and so on. We are not capable of "zooming out" to arbitrary limits - we are bound by our experimental capabilities. But we have some pretty good, albeit incomplete, ideas about the physics that governs the extremely large processes in our universe. It would be unlikely that any "macro-world" built out of the larger structures of our universe could exist and be compatible with what we do observe. Nimur (talk) 18:48, 4 November 2010 (UTC)[reply]

Well, thank you for answering, but I didn't think that the universe was a atom, or that our tiny planet was an atom, but I know what you're talking about from that ending scene. I was actually wondering if there could be gigantic life and if we would be super, super tiny compared to them. I was wondering if there were giant planets as big as Antares, because all I hear of are gas giant stars. So, to reframe my question, are planets capable of getting to be the size of Arcturus? Thanks for explaining that we're not likely a tiny component in a "macro-world." AdbMonkey (talk) 19:00, 4 November 2010 (UTC)[reply]

No, you couldn't get planets as big as a star, because once you get a collection of that much matter in that small a space, the pressures in the core are enough to get nuclear fusion going, which is what makes a star a star in the first place. (Hmm, actually, thinking about it I don't know whether you could theoretically in a very metal-rich environment have ridiculously big planets, but you'd have to artificially create the conditions or wait a very long time for enrichment of the ISM to get you to that point ...) --86.130.152.0 (talk) 19:37, 4 November 2010 (UTC)[reply]
So does that mean there is an upper limit of how big a matter based planet can become before undergoing "auto-fusion"? -- Sjschen (talk) 19:42, 4 November 2010 (UTC)[reply]
Would depend very much on the exact elemental composition, I would think. Some elements are much easier to fuse than others, and many need energy putting in for it to happen. But the (baryonic) matter in the universe is overwhelmingly hydrogen, so you do get stars. --86.130.152.0 (talk) 20:53, 4 November 2010 (UTC)[reply]
So if I were to construct a planet "brick-by-brick" with some kind of baryonic matter, say clay bricks, is there a way to calculate the critical mass (in the way they discovered the critical mass for U-based nuclear fusion) in which the pile would start crushing itself so much to start undergoing fusion? Or directly collapse into some kind of neutron star, black hole, or "weird"? -- Sjschen (talk) 05:38, 5 November 2010 (UTC)[reply]
The size of a biological creature is related to 3 things: Gravity, Heat, and Energy. A larger creature will retain heat better (good in the cold), but will also have a harder time getting rid of heat created by exercise. Gravity is obvious, but the strength of biological matter is unlikely to change much, so if gravity was lower you would have taller (since it's easier to support), but not heavier (what would be the point of over-muscling the creature, i.e. the strength of muscles are likely to be about the same). (So it's up to you if that counts as bigger.) And energy from various redox (oxygen burning) reactions is more of less the same, but a bigger creature will need more energy which could be hard. Of course what I write here could just be a failure of imagination, but I suspect that our size is likely to be typical (for internally supported creatures anyway - if you have water dwelling, or gas filled creatures things could be very different). Ariel. (talk) 20:01, 4 November 2010 (UTC)[reply]

So what's the maximum size a lifeform can get? Since there's an upper limit on planet size? And life can't exist in a star, right? And the gravity is why roaches on this planet aren't big, right? AdbMonkey (talk) 01:46, 5 November 2010 (UTC)[reply]

The answer is that we don't know. We're pretty sure that the maximum size for earth-like life is represented by animals like the blue whale and Apatosaurus, which are among the largest animals ever to have lived on earth. However, consider that Earth-based life does some pretty outrageous stuff, chemically speaking. We seem to have this sense that aliens on other planets will have a chemistry that resembles ours, a body plan that resembles earth-based animals, etc, etc. The truth is, it is much more likely that alien life will bear absolutely no resemblence to what is on earth. Considering the rather exotic and convoluted ways that life on Earth has come up with to solve Earth-type problems, there's no telling what sorts of exotic ways that life on other planets may arrive at. For example, the heat-dissipation problems and musculature issues that Ariel discusses still asumes a roughly Earth-like body plan and chemistry, if you untether your mind from those limitations, it becomes almost boundless the sorts of life that may exist. My favorite statement along these lines was made by Douglas Adams, who once postulated on a "hyper-intelligent shade of the color blue". While that is plainly farcical, it does demonstrate that life could really be just about anything. --Jayron32 04:38, 5 November 2010 (UTC)[reply]
Solaris describes this very nicely.--131.188.3.20 (talk) 11:52, 5 November 2010 (UTC)[reply]
I thought the biggest living thing was some fungus somewhere. Sequoiadendron giganteum trees could be bigger than blue whales. If you are floating rather than being affected by gravity, and are in a medium of the right temperature or are 'cold-blooded', then weight and heat might not matter to much. 92.15.10.141 (talk) 13:17, 5 November 2010 (UTC)[reply]
A better candidate for 'largest' organism on earth is Pando_(tree). Also probably oldest. All hail Pando! SemanticMantis (talk) 15:54, 5 November 2010 (UTC)[reply]
Another thing to think about, which Jaryon alludes to above. How are you defining life? Depending on your definition, the Great Read Spot might qualify (homeostasis, reproduction, growth etc). Along these lines, some sci-fi have speculated about things like 'living' nebulae. What Nimur and Ariel say above is generally correct though, if you're looking for carbon-based biology basically similar to life on earth, there are some pretty solid limits at large scales. SemanticMantis (talk) 14:41, 5 November 2010 (UTC)[reply]

I am defining life as something capable of thought. So probably like something rarely seen on this planet. AdbMonkey (talk) 22:38, 5 November 2010 (UTC)[reply]

Gas giant planets may get up to 13 Jupiter masses; beyond that they condense into the approximate size of Jupiter and become a brown dwarf. That being said, it's certainly possible that large bodies of life live on the suborganic gases of the atmosphere of those planets, getting their energy from chemical reactions similar to photosynthesis while allowing their size to be comparable to those of entire countries on Earth. It's all theoretical. ~AH1(TCU) 17:53, 7 November 2010 (UTC)[reply]

THANK YOU AH1! That is exactly what I was looking for 13x Jupiter. Is mass and size the same? Or could it be heavier and smaller that 13 put together Jupiters? And is that just for our solar system, or for the entire universe? AdbMonkey (talk) 14:36, 8 November 2010 (UTC)[reply]

Definition of planet. Brown dwarfs are usually roughly Jupiter-size, regardless of mass. ~AH1(TCU) 03:15, 13 November 2010 (UTC)[reply]

Why car radiators are called so.?

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I know because it radiates heat..... but i think the prime mode of removal of heat is convection not radiation in a car. —Preceding unsigned comment added by 59.95.101.211 (talk) 18:41, 4 November 2010 (UTC)[reply]

Many thermal loss processes are happening simultaneously to cool the engine. The radiator is radiating heat directly (as infrared emission); and it is also undergoing a (very complicated) conduction/convection process, wherein it conducts heat to nearby air, and then the air flows away, both laminarly and convectively (usually with the assistance of an air intake duct and/or a fan). This doesn't even begin to address the complex cooling mechanism of the fluid that is circulating inside the radiator - again, a mix of conduction, convection, and radiation, all rolled into one. If you needed to calculate very accurately, you could measure geometry and material properties to determine how many watts are attributable to each process.
Regarding the etymology: I'd rack it up to "loose use of terminology." Radiators existed (for various applications) long before thermodynamic theory developed and formalized specific definitions. Nimur (talk) 19:09, 4 November 2010 (UTC)[reply]
The importance of conduction/convection versus radiation is shown by the speed with which the engine overheats and stalls if the fan quits blowing. That interferes with convection more than with radiation. Edison (talk) 17:11, 5 November 2010 (UTC)[reply]
Note that domestic radiators also heat mostly by convection, while electric fires that do substantially radiate are not called radiators! --ColinFine (talk) 17:31, 7 November 2010 (UTC)[reply]

Entirely different forms of power

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I've been watching the BBC series on the National Grid and it got me thinking. Is there any conceivable power delivery mechanism that's better than electricity? I don't mean like nuclear power or hydro or whatever where power is generated and turned into electricity (or wind), I mean the actual end thing. I.e. where we no longer have electrical powered devices we have 'x' powered devices. I don't want to get into whether it's economical or realistically going to replace electricity, purely thinking - is there research going on to look at different non-electricity based ways of powering devices. ny156uk (talk) 18:42, 4 November 2010 (UTC)[reply]

There is some interest in both extra-orbital and surface-to-surface microwave power transmission. -- Finlay McWalterTalk 18:46, 4 November 2010 (UTC)[reply]
(EC)Please provide some metrics of how "better" is defined. There are doubtless special cases where superior systems might be proposed. , or provide a test case, such as "Transmit 1 mw of power to a spaceship 1 million km from Earth for 1 month."Some sort of laser might get the power there better than a pair of copper wires. "Transmit 10000 watts of power from the gasoline engine of a car to the rear axle." A hydraulic and mechanical transmission might do a good job at a lower cost. or "Transmit 80,000 BTU of energy from a furnace to the living space of a house." A steam or hot water boiler might be cheaper and lower maintenance than a steam turbine and generator with electric radiators. "Transmit energy from a solar power satellite to the ground." Microwave might work, copper wires are doubtful. Getting 600 megawatts of power 100 miles from a power station to a city? A drive belt, or a rotating shaft would be expensive and have very high friction losses. A hydraulic or compressed air system would be subject to leaks and friction losses. A laser or microwave or hydraulic system would probably be more expensive than electric wires and transformers. Edison (talk) 18:53, 4 November 2010 (UTC)[reply]
Purely of historical interest, many factories were powered by a central power source (stream, or sometimes a waterwheel) and this power was transmitted to many users inside the factory mechanically, using belts and shafts (info). Even then the mechanical losses were severe, but I guess that (if electrical power just didn't work for some science-fictiony reason) we could all get power delivered to our homes by means of miles of rotating shafts. -- Finlay McWalterTalk 18:55, 4 November 2010 (UTC)[reply]
  • I believe Finlay meant to say "steam, or sometimes a waterwheel". --Anon, 03:59 UTC, November 5, 2010.
(ec) Energy is never really "produced" - it's moved around, and it changes form. Pumped storage uses gravitational potential energy, in the form of mechanically conveying (pumping) water - moving energy from the (electric) pump into the water. Many large institutions use a steam plant, instead of delivering heating via electric distribution; and large pipes full of steam deliver hot gas for use in water-heaters, building heating, and other energy-needs - they move thermal energy around. Natural gas lighting delivers energy in the form of chemical (delivering gas by pipeline), instead of running copper wire to power an electric light bulbs. Compressed air systems exist, and are in wide use to deliver enormous quantities of energy to industrial equipment, dentist-drills, construction sites, and so on. All of these schemes convey energy from a point of production to a point of use. But ultimately, what do you want to use energy for? If you want to power electric devices (like transistors, computers, and mp3 players), you need electricity at some point (whether you convey the energy in the form of compressed air or an electromagnetic wave, transistors can't run on compressed air. You can think about on-site electric generation: but there are few options that make things easier than electric distribution. There might be a future in filling up your iPod with alcohol so that it can run a fuel cell and produce electricity internally. Barring truly revolutionary changes in the way we understand physics, I don't think there is much foreseeable future in replacing transistors with non-electronic equivalents. They are simply the best, smallest, most energy-efficient, easiest-to-manufacture, devices, and we have an entire technology infrastructure set up to use them for all kinds of neat applications. Nimur (talk) 18:59, 4 November 2010 (UTC)[reply]
London had a high-pressure water system for powering small sites from 1872 to 1977. The UK grid and local distribution network is over 92% efficient at getting electricity from power station to the customers. 62.56.53.12 (talk) 19:51, 4 November 2010 (UTC)[reply]
A hydraulic power station, note the water tank on top, for more info see the image description
An alternative is Hydraulic power network, systems of which were quite common in cities during the early electricity age - a good method of tranmitting torque, etc, but not for lighting.94.72.205.11 (talk) 19:57, 4 November 2010 (UTC)[reply]
In Ada, Vladimir Nabokov presents an alternative history in which electricity has been banned and telephones and certain other devices operate by means of hydraulics. He doesn't explain how airplanes and automobiles (which exist in the fictional world) function without electricity, though. Deor (talk) 00:37, 5 November 2010 (UTC)[reply]
You don't technically need electricity for airplanes or automobiles, do you? I don't know how you'd use hydraulics for them, though, but if good-old-fashioned internal combustion engines are still viable then I think you're OK. --Mr.98 (talk) 00:57, 5 November 2010 (UTC)[reply]
Compressed air could be used see this article. Strong laser light in fiber optic "wires" could be a possibility. I'm not sure if those are "better" though. Ariel. (talk) 20:06, 4 November 2010 (UTC)[reply]
In Micropower are described some present day research projects aimed at using fuel to generate electricity at the point of use, such as for powering "Land Warrior" electronic devices used by soldiers in high tech wars, rather than batteries which have short lifetimes and are heavy. It would be easier to deliver liquid fuel to refill the fuel reservoir than to deliver enough batteries to supply the same watt hours for computers and communications devices on the battlefield. Some research has looked into using hydrocarbon fuel to power laptop computers via fuel cells [5], [6]. Just as in cars, chemical fuel provides greater power density or greater runtime than batteries. Natural gas is being used to power fuel cells or microturbines for "distributed generation" at the customer's building rather than at a power plant. The waste heat, perhaps 2/3 of the energy in the fuel, can be used for space heating, air conditioning, or hot water rather than being a waste product with negative value at the power plant. So a pipeline or fuel can can be a form of delivering energy to the point of use rather than by a power line. Amish farmers, at least in the movie Witness (1985 film), have used a windmill to power a mechanical power transmission system consisting of a pair of cords alternately pulled back and forth to transfer mechanical power to the house of barn to perform useful work. It would work much like alternating current. Such back and forth motion could be mechanically converted to rotary motion. It's unclear how the transmission efficiency would compare to a simple drive belt. [7] describes how a 3/4 inch hydraulic line can transmit 100 horsepower on a farm, via the "PowerCube". The PowerCube has a "Lovejoy Coupling" which sounds like fun. It should be noted that natural gas, liquid fuel, rotating mechanical shafts, tractor powertakeoff shafts, hydraulic lines, and compressed air have killed many workers, and are not intrinsically safer than electricity. Edison (talk) 17:02, 5 November 2010 (UTC)[reply]
It's worth mentioning that steam or hot water produced by cogeneration plants really is a "better" method of distributing power, in the sense that a Carnot engine inevitably leaves waste heat behind. If you don't distribute the heat in the form of hot water used to heat homes, you have to build a cooling tower or cooling pond or kill the fish in a heated river instead.
I have speculated that in the future fissile isotopes will be broken down through a managed cascade of nuclear isomers eventually converting much of their nuclear energy into usable chemical, mechanical or electrical forms. Wnt (talk) 11:28, 8 November 2010 (UTC)[reply]

Flavours and colours of quarks

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I was wondering how many different quarks there are, which led me to three questions. First, can any flavour of quark be any colour (excluding anticolours)? This would allow six flavours times three colours times two for regular and anti for a total of 36 different types. However, I noticed in our article that quarks can change flavours and change colours. Can they also change from regular to anti, meaning that any quark can be any of the 36 types, or are regular and anti completely seperate? As an unrelated question, could I make a regular-matter proton using three anti-down quarks? It wolud have a charge of +1, but I assume that something would be wrong with it. Thank you. —Arctic Gnome (talkcontribs) 19:24, 4 November 2010 (UTC)[reply]

Quarks are in fact constantly changing colour -- every time gluons are exchanged (which mediate the strong force between the quarks, so ALL THE TIME), they're transmitting colour between the quarks. They can't change directly from "regular" to anti, though of course in pair production you get quarks and antiquarks created at the same time. And three anti-down quarks wouldn't be a proton, they'd be an anti-Delta-baryon (you may find list of baryons an entertaining article -- see also list of mesons). Note that you can't have three quarks all of the same flavour in a baryon of angular momentum 1/2 due to exclusion principle effects -- you have to be in the J=3/2 mode. HTH. --86.130.152.0 (talk) 19:35, 4 November 2010 (UTC)[reply]

supernova

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How far away would a supernova be if it had an apparent magnitude on earth of -10? Googlemeister (talk) 19:52, 4 November 2010 (UTC)[reply]

That depends on the type of supernova. Type Ia supernovae are considered to be fairly uniform (with some caveats), with an absolute magnitude of -19.3. (That last article, incidentally, shows how to calculate distance given absolute and apparent magnitudes.) Type II supernovae are more varied, but all are dimmer than Type Ia. TenOfAllTrades(talk) 20:05, 4 November 2010 (UTC)[reply]
(edit conflict) That of course depends on the supernova. According to supernova, all type Ia supernovae have an absolute magnitude close to -19.3. Thus they would have to be about 724 parsecs away to have apparent magnitude -10. Type IIs are dimmer than that, though. Our articles don't seem to say much about typical magnitudes of type Ibs and Ics. Algebraist 20:06, 4 November 2010 (UTC)[reply]
The eventual supernova of Spica could have roughly that magnitude. ~AH1(TCU) 17:49, 7 November 2010 (UTC)[reply]

nuclear war

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Would a nuclear war kill all the algae and plankton on the Earth? --96.252.213.127 (talk) 20:51, 4 November 2010 (UTC)[reply]

No. To achieve that, you'd need many orders of magnitude more energy than the total kilotonnage of the world's nuclear stockpiles. --86.130.152.0 (talk) 20:58, 4 November 2010 (UTC)[reply]
Direct energy released matters very little for this question. What is more relevant are the radioactive and meteorological effects — nuclear fallout, nuclear winter. But it seems fantastically unlikely that even under the worse conditions it would kill off all algae and plankton — I'm not sure what level of blocked sunlight you'd have to get to do that, but it must be fantastically high, probably worse than anything humans could actively do in the short term. --Mr.98 (talk) 00:34, 5 November 2010 (UTC)[reply]
Forget about algae - a nuclear war would have a hard time even killing all the humans. Despite fears, we don't actually have enough nuclear weapons to kill everyone, and the earth is very very big. Ariel. (talk) 05:58, 5 November 2010 (UTC)[reply]
Furthermore, life persists in some of the most harsh places, even if we could exterminate all humans, or even all higher animals, extremophiles show that life would probably persist in some form or another. Though, as Ariel notes, despite alarmist views, even the sum total of nuclear weapons couldn't directly kill all humans. --Jayron32 06:27, 5 November 2010 (UTC)[reply]
We could make life pretty miserable for people, though. Radioactive climate change... it's not a good thing. I think we could probably say with confidence that it would kill human civilization "as we know it." It's not really an "alarmist view" to say that a full nuclear exchange (which fortunately today is a lot less probable than it was) would be devastating for the species on par with an asteroid impact. --Mr.98 (talk) 14:09, 5 November 2010 (UTC)[reply]

Note that the latest thinking is that life on Earth existed during the late heavy bombardment:

The Late Heavy Bombardment and the "re-melting" of the crust that it suggests provides a timeline under which this would be possible; life either formed immediately after the Late Heavy Bombardment, or more likely survived it, having arisen earlier during the Hadean. Recent studies suggest that the rocks Schidlowski found are indeed from the older end of the possible age range at about 3850 Ma, suggesting the latter possibility is the most likely answer.[11] Schidlowski's argument remains a topic of heated debate.

More recently, a similar study of Jack Hills rocks shows traces of the same sort of potential organic indicators. Thorsten Geisler of the Institute for Mineralogy at the University of Münster studied traces of carbon trapped in small pieces of diamond and graphite within zircons dating to 4250 Ma. The ratio of carbon-12 to carbon-13 was unusually high, normally a sign of "processing" by life.[12]

Three-dimensional computer models developed in May 2009 by a team at the University of Colorado at Boulder postulate that much of Earth's crust, and the microbes living in it, could have survived the Bombardment. Their models suggest that although the surface of the Earth would have been sterilized, hydrothermal vents below the Earth's surface could have incubated life by providing a sanctuary for heat-loving microbes.[13]

Count Iblis (talk) 17:34, 5 November 2010 (UTC)[reply]

Radio waves

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I read the article on crystal radio (radios powered entirely from radio waves, with no need for a power source) but it doesn't explain the process? How much electricity is there in radio waves? Could they power a light bulb for example? 82.44.55.25 (talk) 21:01, 4 November 2010 (UTC)[reply]

According to the article, the power received by crystal radio antennas is at most measured in microwatts (millionths of a watt). To get an understanding of how much power a microwatt is, that’s 100,000,000 times less power than a standard 100W incandescent light bulb uses, and about 30,000 times less power than is used for a rather dim LED. Red Act (talk) 22:20, 4 November 2010 (UTC)[reply]
And if you've ever used a crystal radio, you know that the volume is very low. Immediately following the discovery of radio waves, a lot of inventors put a lot of effort into inventing a quality amplifier for use in loudspeakers. Nimur (talk) 23:24, 4 November 2010 (UTC)[reply]
If you were right next door to the transmitter,or even a kilometer or more away, a crystal radio could certainly drive a loud speaker. I have heard such demonstrations. A lightbulb of the right sort next to the transmitting antenna could be made to light up. 19th century scientists from Joseph Henry [8] in the 1830's to Thomas Edison (Etheric force)in 1875's and Heinrich Hertz in 1886 demonstrated that radio waves could produce a visible spark many meters from the transmitter in a receiving loop. Tesla did demonstrations of "wireless power" over short distances. The power drops off dramatically as you move farther away. Edison (talk) 16:33, 5 November 2010 (UTC)[reply]

Electricity

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We're currently studying electricity, and while I can use the equations (Ohm's law and all that) just fine, i'm having some trouble intuitively 'getting' how electricity really works. Although analogies like the Hydraulic analogy are useful, they can break down when looked at closely.

My question is: What is voltage? It's the amount of energy that each coulomb of energy has, yes, but how is that energy stored? it's not like each electron has a liitle pouch they can store energy in, and the energy can't be stored as kinetic energy of the electrons, because that is current! And how does the application of a voltage make a current flow?

Thanks for helping sort my confusion, --HarmoniousMembrane (talk) 21:35, 4 November 2010 (UTC)[reply]

You can get offtrack thinking of electricity as something belonging to electrons. Applying a voltage to a circuit and the current flowing through it is NOT well represented by a mental picture of little "electron men" running into one end of a tunnel with some amount of momentum and eventually emerging from the other, and a resistance is not well modelled by the little electron creatures climbing a hill. The hydraulic model is not bad if you visualize the pipe as always filled with fluid, whether there is a flow or not. Think of voltage as pressure: in general the more voltage applied to an ohmic circuit path, the more current will flow. A higher resistance is like a longer or skinnier fluid filed tube. The tube is always filled with fluid, so it starts flowing out the distant end as soon as the valve is opened or the pressure is applied. Picture the voltage source or battery or generator as a pump which receives back into its input fluid as fluid is pumped out the outlet, for current electricity. (Such a model breaks down if carried too far, but helps with many circuit problems). Do lots of problems and get so you find the right answer easily, and it will all become more intuitive. Edison (talk) 00:15, 5 November 2010 (UTC)[reply]
I guess that no one else has answered this, because they are flabbergasted that the educational system has collapsed to the point that no one understands anything any more; rather they just have to memorises stuff. What happened to education and the long-winded pons asinorum. Never mind, the Chines are ready to take over world trade and economic leadership. --Aspro (talk) 00:22, 5 November 2010 (UTC)[reply]
Voltage is pressure, not just in the hydraulic analogy but literally. Mobile electrons repel each other and, given the choice, will space themselves out evenly in a wire. If there's a local excess/deficit in one part of the wire, that part will have a negative/positive charge that will repel/attract electrons to restore the equilibrium. Liquid water behaves in the same way, though for more complicated reasons, which is why the hydraulic analogy works. Imagine a closed loop of pipe with a pump in one location and a water wheel in another. The water downstream from the pump and upstream from the waterwheel is under pressure and its density is slightly higher than the average, while the rest of the water is under tension (negative pressure) and its density is slightly lower. Replace the water with a sea of electrons and the pump and wheel with a battery and electric motor respectively, and the same thing happens. The part of the wire that's under pressure has a slight net negative charge and the other part a net positive charge, so there's a macroscopic EM field (outside the wire), and that's where the energy is stored. -- BenRG (talk) 02:56, 5 November 2010 (UTC)[reply]
(The following assumes you're a GCSE-ish level student, apologies if that is a faulty assumption.)
Take heart, OP. There are studies that show that people's ability on the electricity part of school Physics courses is completely uncorrelated with the rest of the subject, so if you're struggling with this one particular bit it doesn't mean you've suddenly got much worse! I know that personally speaking, all this stuff only really clicked for me when I got to university and did Maxwell's equations -- once I understood those (and Gauss's Law and so on) I was able to go back to the "simple" circuit stuff and it made a lot more sense to me, but I wouldn't advocate trying to learn lots of calculus just on the off chance your brain happens to work the same as mine.
I would suggest that you look at all the different analogies out there (hydraulic, the "electron men"/"buckets of charge" carrying certain amounts of energy ideas -- which latter does come with several important caveats!) whilst always bearing in mind that they are just analogies and none of them is going to give the complete picture -- different ones are helpful for understanding different situations (simplistic presentations of the hydraulic analogy, for example, have a hard time with series circuits because your pipes are suddenly changing length all the time when you add new components). One thing you may find helpful is to read up on drift velocity, which is often left of Physics at lower levels these days, but makes a strong connection between the overall flow of current and the behaviour of the individual electrons in the electric field across the wire, and might help you disentangle some of the things you're confused about. --86.130.152.0 (talk) 06:53, 5 November 2010 (UTC)[reply]

Thanks For all the great ansers, I definately feel like I've got a handle on the Voltage issue now. And yes, that assumption was spot on, I am a GCSE student. Incidentally, Aspro, it's not as if my teacher has not tried to help us understand electricity, with analogies like the aforementioned 'Buckets of Charge', it's just that a full understanding of it takes quite advanced maths, and given the fact that definately at least half of us, if not more, will stop physics after GCSE, it's frankly much more constructive to inforce simple understandable analogies and simple ideas about power and resistance, which will help in everyday life, than try and go the whole hog and cover all of the subject in excruciating detail, so we can 'understand'; which will be pretty much useless unless you are going into a highly specialised field, in which case it can be assumed that you will be taking A-level Physics, where you will learn the 'truth' anyway. Again, thanks for all the very illuminating answers!--92.10.28.38 (talk) 16:10, 6 November 2010 (UTC) Oops, that was me: I forgot to sign in.--HarmoniousMembrane (talk) 16:12, 6 November 2010 (UTC)[reply]

Yes, think of voltage as a pressure or as a potential energy, and think of current as a kinetic energy, and compare the whole situation to Bernoulli's principle in relation to fluid flow. David Tombe (talk) 17:27, 7 November 2010 (UTC)[reply]
It may also be helpful to think of voltage as electric potential difference. ~AH1(TCU) 17:48, 7 November 2010 (UTC)[reply]

Entropy

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Hi! I'm in a 1st year physics course and now we're looking at entropy. In my text it says "If the temperature changes in the process of the addition of heat to a system, the change in entropy can be calculated using advanced mathematics. This chapter deals primarily with changes in entropy in isothermal systems. For a discussion of how a change in entropy is calculated when the temperature changes, see the Supplemental Text on the next page" I looked on the next page(s) and.... no supplemental texts :( I think they got lost somehow when the book was edited. But how is entropy calculated when the temperature changes? I tried your Entropy article but as I'm only in first-year physics that was not very helpful :( —Preceding unsigned comment added by 24.92.78.167 (talk) 22:03, 4 November 2010 (UTC)[reply]

Taking the heat to be added slowly (and reversibly), so that the system remains all at one temperature throughout, . So, you can integrate (let's say over time) . The trick is that you have to know how the temperature changes; if you know the heat capacity C of the system, it becomes nicer to integrate over the added heat: , where . That last equation is the tricky one: is on both sides, so unless C is a constant you actually have to treat it as an autonomous differential equation . Solve that with the "advanced mathematics", then you have , and you can do the S integral. If C is constant, then and you can directly write . If you then try to start from absolute zero () you get nonsense; as the third law of thermodynamics says, C can't be constant at absolute zero. --Tardis (talk) 00:32, 5 November 2010 (UTC)[reply]
It's a tricky question. As a chemist, I start by telling my students that, for an ideal gas, dS = CV. That puts them into an adequate state of confusion, because they can't believe it's as simple as that. Of course it's not, not for a real system! But practical measurements of entropy on pure substances still rely on heat capacity measurements to very low temperatures, and on finding the various phase changes. Physchim62 (talk) 01:21, 5 November 2010 (UTC)[reply]
You must have left something out: one and only one side of your equation is a differential. For constant C I get , but that's not very much like . --Tardis (talk) 13:53, 5 November 2010 (UTC)[reply]

How to control this old oscilloscope

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I have this old oscilloscope but it didn't come with a manual. I hope it might be a simple enough machine that someone with relevant experience might be able to tell me what I'd need to know to operate it. I'd like to be able to connect a circuit driven by rectified but fluctuating current from a dynamo and determine the voltages present. What's that "vert" socket, bottom left for? --90.209.7.130 (talk) 23:52, 4 November 2010 (UTC)[reply]

You lucky man..
The good news is that all Oscilloscopes operate the same. This might get you going.Oscilloscope--Aspro (talk) 00:01, 5 November 2010 (UTC)[reply]
That article doesn't say anything about the 'vert' socket... when plugged in and switched one, should the 'pull on' light come on or is that only under certain conditions? Why does the on/off switch have a 1 and a !0 instead of a 1 and a 0? --90.209.7.130 (talk) 00:25, 5 November 2010 (UTC)[reply]
The "Vert" connector is just where you connect the signal. The switch with 1 and 0 looks like a custom add-on. The light you speak of is just a power-on indicator. Good luck. (Did the scope come with probes?) PhGustaf (talk) 00:40, 5 November 2010 (UTC)[reply]
If you have $32 plus my substantial finder's fee that you may donate directly to the Wikipedia Foundation then you can have the handbook you lack. Cuddlyable3 (talk) 14:46, 5 November 2010 (UTC)[reply]
One thing to watch out for in an oscilloscope is whether the input "ground" terminal is actually an earth ground, or whether it is floating. If it is truly a ground then you cannot connect the scope across a component which has some voltage to ground on each side, or something will make smoke. Similarly it is important to note if the grounds for each channel are independent of each other or connected together internally. Edison (talk) 16:18, 5 November 2010 (UTC)[reply]