:{{small|My great-uncle, who was a farmer, had an effective treatment for moles: his shotgun. Your results might vary. ←[[User:Baseball Bugs|Baseball Bugs]] <sup>''[[User talk:Baseball Bugs|What's up, Doc?]]''</sup> [[Special:Contributions/Baseball_Bugs|carrots]]→ 22:23, 10 June 2012 (UTC)}}
:{{small|My great-uncle, who was a farmer, had an effective treatment for moles: his shotgun. Your results might vary. ←[[User:Baseball Bugs|Baseball Bugs]] <sup>''[[User talk:Baseball Bugs|What's up, Doc?]]''</sup> [[Special:Contributions/Baseball_Bugs|carrots]]→ 22:23, 10 June 2012 (UTC)}}
:I use a pharmacy solution of salicylic acid and lactic acid. It works wonders for both warts and moles. In Australia it is sold as Duofilm, made in Ireland. [[User:Plasmic Physics|Plasmic Physics]] ([[User talk:Plasmic Physics|talk]]) 22:33, 10 June 2012 (UTC)
== Photoelectric threshold frequency ==
== Photoelectric threshold frequency ==
Revision as of 22:33, 10 June 2012
Welcome to the science section of the Wikipedia reference desk.
The best answers address the question directly, and back up facts with wikilinks and links to sources. Do not edit others' comments and do not give any medical or legal advice.
Here are two Herschel documents about the planet,[1][2] neither including a list of names, but the latter includes in passing the names of both Jupiter and Saturn. Thincat (talk) 09:27, 6 June 2012 (UTC)[reply]
Interesting to see that Hershel called it TheGeorgium Sidus in writing. I guess what I'm most curious to find is the kind of list that would be given to school children to memorize. As in, here an the eight planets: Mercury... Saturn and Georgium Sidus. --Anonimasimio (talk) 09:31, 6 June 2012 (UTC)[reply]
Here is the Nautical Almanac for 1820 where the name is in a list and it is called "Georgian" (page 34, for example, of the PDF). The preface refers to "the Planet Herschel, called the Georgian Planet by us" (page 25). Hardly appropriate for school children though. Thincat (talk) 10:55, 6 June 2012 (UTC)[reply]
Thanks, Thincat. That's the first contemporary, non-Hershel reference I've seen. I'm digging this up because there is a modern story told by many of how Uranus was almost named 'George' but I've never been able to find anything from the time that actually uses that name. If anyone can find a that reference, or any other contemporary usage of the planets' name before we settled on Uranus, I'd be grateful. --Anonimasimio (talk) 11:56, 6 June 2012 (UTC)[reply]
I have seen (in the library of a National Trust property, but I won't say which one, because I don't think we were supposed to be looking at the books) a school atlas that lists the planet as "Hershall" (It was a long time ago, but I'm pretty sure of the spelling). --ColinFine (talk) 15:38, 6 June 2012 (UTC)[reply]
Anonimasimio, I've found a publication on Google Books by Tiberius Cavallo in 1803 that refers to Uranus as "the Georgian Planet" in several places -- [3] He does, in a footnote, mention that the planet is also called Uranus or Herschel by some. There are a number of other titles from the 19th Century in Google Books that also refer to it as "the Georgian Planet", but I thought Cavallo was of sufficient stature to be an important one to mention. I didn't hunt for "Georgium Sidus", your original question, but you may have luck searching for it as well. Best of luck with it! Jwrosenzweig (talk) 06:02, 7 June 2012 (UTC)[reply]
Another citation perhaps worth mentioning is this book by Jacques Ozanam[4], which I point out only because it's an example of a non-British scientist who is clearly familiar with the name (and appears to be using it as the planet's standard name, although I don't know if this book is translated from a French original) -- anyway, most of the citations I'm finding are from scholars born in or working in England and Scotland, and I thought it was worth pointing out that "Georgium Sidus" had traveled across the Channel as at least one of the planet's names, if not its standard name. I'll also note that I'm seeing citations as late as the 1850s that seem to be using "the Georgian Planet" as Uranus's standard name, though it might be worth investigating whether these are just reprints of earlier editions, or if the name persisted in use that long. Jwrosenzweig (talk) 06:10, 7 June 2012 (UTC)[reply]
To clarify, Sidus is merely Latin for "star" (which in the old sense included planets). And Georgium is the accusative singular of Georgius, a proper name from the ancient Greek Γεώργιος, a name based on γεωργός (farmer) = γῆ (earth) + ἔργον (work). We recognize Γῆ to this day as Gaea, and thus, oddly enough, Uranus was very nearly named after the Earth! Wnt (talk) 17:20, 7 June 2012 (UTC)[reply]
Sidus gives us words like 'sidereal' and 'consider'. Stella gives us 'stellar', 'constellation' etc. And the words 'star', 'Astarte' (= Venus), 'Ishtar' (= Venus), 'asterisk', 'asteroid', 'astro-' words (like 'astrolabe', 'astrology', 'astronaut', 'astronomy', 'astrophysics', etc), 'disaster' and 'catastrophe' are all related. It's written in the stars. -- ♬ Jack of Oz ♬ [your turn]05:26, 9 June 2012 (UTC)[reply]
Battery charging time
I have installed a custom made 12 V 750W UPS with 170Ah battery. The charging current is to 14A, so how long will it take to charge the battery.
To properly answer this question, you need to tell us more about the battery and the load - I'll say more about this but first the simple, and very inaccurate answer:
The battery capacity rating Ampere-hours (AH) is defined as the product of current and hours from full charge to a defined flat condition under conditions meant to be typical for the battery in question. The charge capacity is roughly about the same as the discharge capacity. Thus, in your first case, time to battery "flat" = 170 Ah / 14 A = 12 hours. Since the maximum output of a 750 W 12 V power supply is 750/12 ie 62.5 A, I assume your load is equal to output - charge ie 62.5 - 14 = 48.5A. Your second question cannot realy be answered as you have not given the charging current. If it is assumed to remain 14 A then the charge time is unaltered.
In real batteries, the product of discharge (or charge) and current is a function of current. The capacity increases as current is reduced. Eg for typical lead acid batteries, halving the current will not double the discharge (or charge) time, it will as much as treble it.
In real batteries, the ampere-hour capacity is strongly temperature dependent. Capacity increases with temperature.
How long a charge takes or lasts is a function of how flat you can tolerate. There are industry standard cutoff voltages for each type of battery, so that you can make comparisons, but in practice your load may cease to work properly at a different voltage. Lead acid batteries in particular are damaged if flattened. Types intended for standby service will give a short life if routinely flattened more than 50 to 70 % of nominal capacity.
All this means that in order to meaningfully calculate a charge or discharge time, you need more comprehensive data about your battery, and you need to know more about your load. Battery manufactuers usually provide graphs of performance over a range of charge and discharge currents.
You mention a "custom made 12V UPS". Note that if you are using a nominal 12V battery, and your power supply has a constant output voltage of exactly 12V, it will not fully charge the battery. For instance typical lead-acid batteries require to be charged to 2.3V per cell (13.8V for a nominal 12V battery) at 25 C. Other types are similar.
If you are installing your own system, you should ensure you are thoroughly conversant with battery safety - particularly with ventilation requirments if using lead acid, and charging requirements if using lithium-iron types. Get it wrong and you can have explosions, posionous gas, or both.
Something's not right somewhere, or information is missing. Your stated load totals 305 W. Allowing for typical invertor efficiency that means a DC current of ~28A on a 13.8V bus. For a 750W system, 54 A is available. For a continuous bus system, charging current will equal 54 - 28 ie 26 A. But you siad charging current is 14 A. This is possible if the UPS is of the switched bus type (battery on a separate charger circuit, and switched to feed the invertor when required), but you haven't told us about the 1000W UPS. Is it continuous bus or switched bus? If switched bus, what is the charging current? Possibly the link you gave was meant to tell us more, but it is not functional. Wickwack58.164.238.58 (talk) 14:52, 6 June 2012 (UTC)[reply]
When I had the job of testing some early UPS units a few years ago, I noticed that the discharge rate was a brutal one in UPS mode, and much higher than the charge rate. This made sense, because the device is not really designed to operate on ,say, a daily basis, but rather to operate in the occasional emergency. The battery is called upon to discharge at an extremely high rate, at the cost of decreasing the battery life, in order to keep initial cost down, but still provide a relatively large amount of AC power to the protected computer or device. Protecting the continued operation of the computer (don't lose the term paper you have just finished typing, or don't lose all the pizza orders held in the computer, or the financial transactions, or the control of the elevators in a highrise, or the air traffic control, or the power system operations) is judged more important than the well-being of the battery. Fewer ampere-hours could be obtained from the battery in this mode than if the discharge rate was lower. Lead acid batteries have very low internal resistance, so they can be called on to do heroic discharge rates for a short while, like cranking a car engine (hundreds of amps from a 70 amp-hour battery for a few seconds). Power interruptions are often a few seconds if automatic reclosing of distribution feeders clears the fault, or a few minutes if supervisory switching can solve the problem. If a PC is operating on UPS with no generator backup, the typical practice is to save your work and do an orderly shutdown rather than keeping on for an hour or whatever until the battery dies abruptly. Ten volts as the battery voltage when the discharge is ended,, as 182.178.210.45 stated, seem an extreme state of discharge, and likely detrimental to the battery life (number of charge discharge cycles) which can be expected. Wasn't a ten hour rate a classic rule of thumb (capacity divided by ten = charge rate and discharge rate?) Even so, a typical charger tapers off the charge rather than continuing at the initial high rate, which would require raising the charging voltage over time. The website of the battery manufacturer might offer specific recommendations for charging protocal. Edison (talk) 14:31, 6 June 2012 (UTC)[reply]
You have it pretty right, Edison. However, the 10 hour rule used to be commonly quoted for small batteries intended for powering portable equipment such as cameras, tape recorders, and the like. It has largely gone by the wayside now, and was never used for larger sizes, nor used for vehicle, industrial, or UPS batteries. Also, constant-current chargers became common when switchmode AC/DC conversion became economic. In continuous bus UPS designs (common because they're cheap, reliable, and simple), the current available for changing the battery is simply the AC/DC convertor output, which must be capable of sustaining the full rated UPS system load, minus the current drawn by the DC/AC convertor in supplying the load. Hence if for some reason the load is minimal, the full AC/DC convertor output is available to charge the battery. Wickwack58.164.238.58 (talk) 14:52, 6 June 2012 (UTC)[reply]
Do you have a reliable source for charge rates way higher than capacity/10 being the present day common practice? I agree that for occasional emergency use (UPS, emergency stairway lighting) a rate which discharges the battery in an hour or so at the cost of battery health and at the cost of ampere hours is common, but I question charging it really fast, since I expect damage to result. The AC/DC converter is ONLY used to charge the battery in many UPS systems, which run the load off the mains until mains failure, then quickly switch to inverter from the battery, fast enough that the PC stays online. Better and more expensive UPS systems indeed run from the inverter off the battery all the time, perhaps with a generator offline to start when the mains fail. Edison (talk) 04:29, 7 June 2012 (UTC)[reply]
For the medium to large scale UPS's (10's to 100's of kW) I had experience with, the battery charging rate was roughly of the same order as the discharge by application design. This is because they were continuous buss systems, so the charge current is essentially the rated output minus the actual load, as I said. This should be taken into account when choosing the battery. Switched bus systems can deliver a fixed charge current independent of actual load and are thus kinder to the battery as you have said. And since the switching can be from raw AC input to invertor output, the AC/DC convertor can be much smaller and cheaper. However, continous bus systems can be prefered for their reliability and inherently stable and spike-free output. Lead acid batteries in particular will not take kindly to rapid charge. Generally, the slower the charging, the longer the life and the greater the capacity. As far as I am aware, batteries designed for rapid charge are used for small portable applications like personal electronics, phones, small tools, and the like. For performance, and service life, for any charge rate, consult manufacturer's data. Batteries are made for a wide range of cycle conditions - even within the same basic chemistry (gell lead acid, nickel-iron, whatever). So I don't think a source with one simple rule of thumb or formula, beyond what I gave above, can be cited. Incidentally, where diesel backup is employed, it usually because an outage cannot be tolerated, such as telecomms, business critical server farms, or hospital operating theaters. This is quite different to a PC or minicomputer application, where you only need enough time to shut down in an orderly way (a few minutes). Diesel backup is usually provided with a UPS battery capacity for 3 hours or so. This is so that, should the mains not came back on in that time (an unlikely occurance) and the diesel fail to start, there is enough time to call out the mechanic, plus time for him to fix it immediately if he can, or request delivery of a portable genset. In such cases, the UPS battery will be very large and inherently tolerate a lot of charging current. Wickwack120.145.54.86 (talk) 05:28, 7 June 2012 (UTC)[reply]
Which will system will be better, 24V or 12 V ?
It depends on many factors. But, as a general rule, if the load power is the same, and the battery capacity in Ampere-Hours (Ah) is kept the same, then 24V will be better as the discharge current (and potentially the charge current) will/can be halved. I stress that this is only a very rough guide. You would normally expect to halve the battery Ah size as well on 24V. Of perhaps more reliable to state is that AC/DC conversion and especially DC/AC conversion is slightly more efficient at 24V. Wickwack58.167.249.78 (talk) 12:05, 7 June 2012 (UTC)[reply]
I've already answered that to the limitted general degree possible given that you have not supplied sufficient information. Is the 24V sytem to be switched bus or continuous bus? If switched bus what is the charging current? 2 x 170 Ah batteries, or 2 x 85 Ah batteties, or what?? And, as I siad, battery capacity varies with operating conditions (cutoff voltage, charge current, temperature, etc etc). You need to look at manufactuers data for batteries you are considering. Your questions are like asking us "If I now eat 2kg of food each day and propose to eat 2.5 kg, how much will I weigh?" How should I know, it depends on so many factors that haven't been given. About all I can say is you will get fatter, if all other factors are kept constant. Similarly, I can't say what the charging time will be, beyond the rough guide I've already given. Wickwack121.215.41.248 (talk) 15:31, 8 June 2012 (UTC)[reply]
Charging current is 14A, Battery 2 x 170 Ah, cutoff voltage 10V, room temperature is around 36C . Manufactures usually don't provide such extensive data. — Preceding unsigned comment added by 182.185.171.47 (talk) 09:04, 9 June 2012 (UTC)[reply]
In that case, since you have neither changed the ampere-hour capacity, nor the chanrge current, then the charge time is unchanged. However the discharge time for the same load power will increase somewhere around 2.5 to 3 times, ignoring the 35 C temperature. Battery manufacturers certainly DO provide such "extensive data" - it is essential for the reasons I have explained. You just haven't asked in the right place. I have used such data myself. A 35 C temperature is excessive for an office environment. Wickwack124.178.139.104 (talk) 12:47, 9 June 2012 (UTC)[reply]
Source of uncertainty of the Earth's age ?
On the page Age of the Earth it is stated that the age is 4.54 ± 0.05 billion years. What are the main sources of the 1% uncertainty, and what is most dominant?
Is it the accretion time (100 million years is indeed 2% of the 5 billion years age) itself?
From the first citation on that page, the uncertainty seems to be from the dating methods. The time appears to refer to the end of the accretion of solid bodies such as Earth, which should have occurred at roughly the same time throughout the inner solar system.-RunningOnBrains(talk)13:39, 6 June 2012 (UTC)[reply]
There was an interesting article a few weeks back that one of the longer lasting isoptopes had been found to have a different half-life than believed, and that the earth's age might need to be recalibrated . Anyone recall the article or isotope? μηδείς (talk)
Yes, that kind of radioactive dating tells you when that particular rock formed. What we know is that the Earth must be at least as old as its oldest rock (everything melted during the formation, so any older rocks won't have survived intact). It is unlikely that the Earth is significantly older than its oldest surviving rocks. Our estimates of the age of the Earth are consistent with the ages of meteorites, which theory tells us are left over from the formation of the planets. It's also consistent with our estimates of the age of the Sun. It's very unlikely that our age estimates are completely wrong. --Tango (talk) 16:12, 7 June 2012 (UTC)[reply]
To expound further, I realize my original point was unclear: I'm not saying there couldn't be new evidence (or new theories consistent with the current evidence) that pointed to an older earth, but the idea that we have a half-life wrong is about as close to impossible as I'm comfortable to say. -RunningOnBrains(talk)17:12, 7 June 2012 (UTC)[reply]
bond energy of ionic compounds
how will we define the bond dissociation energy or bond dissociation enthalpy for ionic compounds ?
i have read the definition of bond dissociation energy as " The bond dissociation enthalpy is the change in enthalpy when one mole of covalent bonds of a gaseous covalent compound is broken to form product (gaseous atoms) in gas phase ." the problem is that ---firstly the ionic compounds don't occur in gas phase , secondly if we heat it we will get ions (& the energy involved is lattice energy) & not the atoms, moreover the definition tells only about the covalent compounds .ncert class 11 chemistry text book part one pg 171 http://ncert.nic.in/NCERTS/textbook/textbook.htm?kech1=0-7117.225.240.240 (talk) 14:03, 6 June 2012 (UTC)[reply]
As a side note: boiling sodium chloride, produces discrete molecular sodium chloride or 'chloridosodium'. Further heating to the point of ionization leads to diassociation into a gas plasma. This consists of a menagerie of ions, including sodium(1+) and chloride(1-), as well as polyatomic ions. Plasmic Physics (talk) 23:30, 6 June 2012 (UTC)[reply]
Most of the angular momentum of the Solar System, I seem to remember reading, is that of the orbit of Jupiter. But of course every rotating or revolving body has its own bit of a.m., and their vectors are not all parallel. Is the variance of these vectors (mass-weighted, of course) a meaningful concept? If so, is it known (or estimated)? If so, how big is it in radians?
It does seem like an interesting concept, although I doubt if it would be measured in radians. If we assume that the initial rotating cloud of gas was uniform in it's rotation (not sure if this is true), then any variance must be caused by outside influences (passing comets, etc.). However, once there is even the slightest variation, it's possible for two objects to give each other more variance by passing near each other and knocking each other out of position (perhaps a bit more with each orbit). Pluto seems to be the (dwarf) planet with the most deviation of the inner planets, so something interesting must have happened there. Objects out in the scattered disc and Oort cloud are far more random in their vectors than those in the inner solar system and, to a lesser extent, the Kuiper belt (with exceptions in the inner solar system allowed on for smaller objects, easily knocked out of the plane of the ecliptic, like the centaurs). The following chart may also be relevant. StuRat (talk) 19:07, 6 June 2012 (UTC)[reply]
Distribution of trans-Neptunian objects, with vertical axis showing inclination from the plane of the ecliptic, and horizontal axis showing distance from the Sun.
"Inner planet" usually refers to Mercury, Venus, the Earth and Mars. Pluto isn't an inner planet. It's orbital inclination is normal for Kuiper belt objects, as your chart shows. --Tango (talk) 19:39, 6 June 2012 (UTC)[reply]
Cis-Plutonian. But as the orbit of Neptune, rather than of eccentric Pluto, is taken as the delimiter between the outer solar system and the next broad category of solar system objects, people would be more apt to talk about Cis-Neptunian objects. List of trans-Neptunian objects shows there are about 40 known TNOs with perihelions closer than Neptune's, but none with an average orbital radius smaller than Neptune's Tautologically; they wouldn't be TNOs if they were. -- Finlay McWalterჷTalk09:19, 7 June 2012 (UTC)[reply]
Let's start by checking if what you remember reading is correct, since it's a simple enough calculation. I would guess the major contributors are going to be Jupiter's orbit and the Sun's rotation, since the Sun and Jupiter make up essentially the entire mass of the solar system. Jupiter's orbit gives an angular momentum of 779,000,000 km (orbital radius) * 13 km/s (orbital speed) * 1.9*1027 kg (mass) = 1.9*1037 km2kg/s. The Sun (assuming uniform density, which will overstate its angular momentum since a lot of the mass is concentrate in the core) has an angular momentum of 2/5*m*r2*2*pi/T (see List of moments of inertia, m, r and T are the mass, radius and rotation period of the sun, respectively). That gives us 2/5*2.0*1030kg*(0.7*106km)2*2*3.1/25 days=1.1*1036 km2kg/s. So yes, Jupiter is the main contributor. I'll just quickly check Neptune, since it is so far away it might be higher than I expect: 4.5*109 km * 5.4km/s * 1.0*1026 kg = 2.4*1036 km2kg/s. That is higher than I expected, but not as high as Jupiter's. --Tango (talk) 19:39, 6 June 2012 (UTC)[reply]
Awesome Science things in Europe?
So, not exactly a science question, but i'm hoping you can help. It's my dad's 60th soon and we want to take him somewhere in western europe to see something awesomely sciencey. CERN is an obvious one but it seems like most of their stuff for tourists is museumy rather than seeing anything real. Trying to pull some strings to get into ATLAS but i'm not hopeful. Something like the JET Tokamak comes to mind but we live in the UK and it's not exactly an exotic location to spend more than a day. Any thoughts would be appreciated or reprimands for putting this here also welcome! 137.108.145.21 (talk) 18:50, 6 June 2012 (UTC)[reply]
The Deutsches Museum in Munich has a massive collection of all manner of science and industry, but it's not an active site of research, which seems more like what you're going for. --Mr.98 (talk) 01:00, 7 June 2012 (UTC)[reply]
I know you say you want to visit western europe but a few locations in England stick out too much not to mention, even though they are probably more of a day trip rather then a nice stay away somewhere. Maybe if not this time, you can at least put them on a list for "next time". I'm in Australia and just over a year ago I stayed for a month with my brother who was living in London at the time. I did a day trip to Down House, i'm a big fan of Darwin so that was very high on my list of things to do, I thought it was set up really well. The other two places I'd love to see would be blatchley park and the cavendish laboratories, those places obviously hold legendary status in the history of science, not having seen them I can't say how "interesting" their exhibits would be to people not already interested in their history. Vespine (talk) 04:09, 7 June 2012 (UTC)[reply]
Have you considered Iceland? There are various geothermal energy experiences to appeal to the sciencey-minded, the opportunity to see active geysers, lava fields and other awesome volcanic things, and of course the possibility either of 24-hour daylight or the aurora, depending on when you travel. We're approaching solar maximum, so aurora-spotting has an improved chance of success next winter, particularly in higher latitudes. Karenjc18:16, 7 June 2012 (UTC)[reply]
In the days of Scott and Amundsen, how would an explorer identify his position as being exactly at the South Pole, and how accurate would this be? --rossb (talk) 22:52, 6 June 2012 (UTC).[reply]
Not so much "motions" as "positions" -- we note at history of longitude that "determining latitude was relatively easy in that it could be found from the altitude of the sun at noon with the aid of a table giving the sun's declination for the day." For the poles, that's enough right there -- latitude 90, longitude irrelevant (also, time of day functionally irrelevant). This site confirms that a sextant reading of the sun was used, and notes that Amundsen was within 200m of the true South Pole. — Lomn23:25, 6 June 2012 (UTC)[reply]
If I recall Huntford's book The Last Place on Earth correctly, in those few days of at-pole peregrination, Helmer Hanssen was probably the one who got closest to the geographic pole. It was my recollection that Scott carried a theodolite, which would have allowed him to make much more accurate measurements of the position of the Sun than Amundsen could with his sailor's instruments (and that the additional weight thereof was one of many contributory factors to Scott's failure). But, while there was a theodolite on the expedition, it seems it wasn't taken to the pole (picture and info). -- Finlay McWalterჷTalk20:10, 7 June 2012 (UTC)[reply]
At night, they could track the celestial South pole to give a rough direction. Somehow I don't think that option was available, you'd be pretty foolish to attempt such an expidition during the polar winter. Plasmic Physics (talk) 23:34, 6 June 2012 (UTC)[reply]
For the next three days the men worked to fix the exact position of the pole; after the conflicting and disputed claims of Cook and Peary in the north, Amundsen wanted to leave unmistakable markers for Scott. After taking several sextant readings at different times of day, Bjaaland, Wisting and Hassel skied out in different directions to "box" the pole; Amundsen reasoned that at least one of them would cross the exact point, from Amundsen's South Pole expedition. If my visualisation of the trigonometry is right, when you're at the South Pole the sun stays at the same altitude all through the day (+- a little depending on if it's rising or setting). These days of course it's much easier; you know you're there when you trip over this thing. FiggyBee (talk) 23:31, 6 June 20126 (UTC)
Not really, that thing is not really at the pole, the pole drifts around. Yes, incase you're wondering, both the magnetic and actual poles drift. That is just an arbitrary point that someone chose when the pole actually used to be in the vicinity. LOL, it never even passed through that particular point either. Plasmic Physics (talk) 23:38, 6 June 2012 (UTC)[reply]
I'd think we could tell that much change all over the world, if it happened during the quake. If not, by what mechanism does the Earth's axis slowly change as a result of the quake ? StuRat (talk) 06:25, 8 June 2012 (UTC)[reply]
Can anyone identify or suggest an identification for this plant? It is one foot tall. I will be able to recognize the flowers if someone suggests the correct identification. A friend had one such plant in his yard in South Jersey last year, and now it has taken over the entire property. The blooms, when they come, are white and look like bluebells Hyacinthoides non-scripta, although the stalk is more erec,t or wild orchids of some type. Thanks. μηδείς (talk) 23:29, 6 June 2012 (UTC)[reply]
Looking at the links and pictures I can find on google that is close but I haven't seen anything that looks like an exact match. I'll have to see if I can get a picture sent to me after the flowers mature. μηδείς (talk) 15:52, 8 June 2012 (UTC)[reply]
I guess the question title doesn't really need much expansion; I live on the Isle of Man but oddly haven't actually bothered taking the Horse Tram at any point (I don't live in Douglas and there's no real point in it for normal transport anyway). I have been held up by it when driving along the promenade many times but never seen it traversing points.
So... the question is fairly simple. How on earth does it USE these points? Do they use "ramps" in the tracks? Do they force the horses to swerve violently with the coaches? Something else?
P.S. As a secondary point, the original google maps link affected the spam filter. That's cause I chose short url which shortened it using goo.gl... anyway it was a nuisance wiki complaining about this! EggCentric00:34, 7 June 2012 (UTC)[reply]
It seems most likely that those points are no longer used, and they've simply removed the blades and tarmacked in the gaps. Alternatively, the only option I can see is lifting the tram at each end and moving the wheels across to the point rail, which I can see being a reasonable alternative to mechanical points (which would need to be quite complex if they're set in a road surface safe for pedestrians and other vehicles) if the trams are light and those points aren't used often or by in-service trams. I had a zoom up to the end of the line on Google Earth and the crossover at the terminus (the roundabout near the ferry terminal) seems to have normal switchblades.FiggyBee (talk) 00:57, 7 June 2012 (UTC)[reply]
That makes considerable sense, thank you. I also wonder hwo they do change the working points. Tell ya what, I'll try to get on one during the weekend and ask the folk involved what they're doing. And then incorporate it into the wiki article.
I can't answer that medical question, but chlorhexidine and eugenol have been used in conjunction in research, e.g. [5][6] Note that, as explained at [7], eugenol is not seen as entirely benign in some situations all by itself, and eugenol-free periodontal dressings have been developed. The real problem with any answer is that "harmful" depends on the context. A bottle of cyanide is not harmful... provided it stays in the bottle or under a fume hood. While water toxicity really does occur. The dose (and circumstances) makes the poison. Wnt (talk) 18:19, 7 June 2012 (UTC)[reply]
Drug Delivery
I have had this doubt for a long time. Suppose I have a pain in certain area of my body - say X. I take a pain killer tablet. So the chemicals in the tablet dissolve into my blood stream. Now how does the drug get absorbed exactly in the painful area and relieve the pain ? What is the mechanism that makes the drug to get absorbed at the exact painful spot ? A even more localized example is when you have a sore throat and a tablet works wonders ! — Preceding unsigned comment added by 117.193.139.99 (talk) 08:14, 7 June 2012 (UTC)[reply]
A systemic painkiller (taken orally and then circulating in your blood) doesn't have to target the physical location or cause/origin of the pain for which you are taking it, it could just dull "your sense of pain". DMacks (talk) 08:17, 7 June 2012 (UTC)[reply]
Yes, I expected that reply. But say, there is some problem with the liver/ spleen and tablets need to be taken ( I mean some localized disease), how does it work. It would be a real waste to have the drug in the entire blood stream rather than localize its concentration at the target site. So how does it happen. — Preceding unsigned comment added by 117.193.139.99 (talk) 08:29, 7 June 2012 (UTC)[reply]
Your typical non-steroidal anti-inflammatory painkiller (tylenol, ibuprofen, aspirin) functions by broadly inhibiting your body's ability to transmit localized pain signals. In this sense, it is acting locally, but it's acting locally everywhere. Opioids, on the other hand (as well as GABA analogues, although less dramatically) block your brain and spinal cord's ability to register pain signals. In that sense, the drugs are acting non-locally from one place, but are still present throughout your entire body assuming you took a pill. As you are probably aware, some painkillers can simply be injected directly to where they need to work, such as for dental procedures. In this case, since the drug stays and acts locally, much higher local concentrations of the drug can be achieved with minimal side effects.
As for liver effects, it turns out that most drugs you consume actually wind up in your liver, given its purpose of breaking down most unusual chemicals that you consume. But everything else, unless it is injected locally, pretty much ends up everywhere. It is a goal of medical science to to develop drugs that target to specific tissues, actually. But that's not to make the drugs cheaper. Rather, the hope is that since the drug will only be going where it is needed, there will be fewer side effects from what the drug would do to tissues that don't need it. Someguy1221 (talk) 08:36, 7 June 2012 (UTC)[reply]
(edit conflict) Some drugs are designed to bind selectively to certain types of tissues or cells. For example, they could have a high affinity for chemicals (and hence cells with these chemicals on their surface or vicinity) that are known to be produced in the case of a specific biochemical situation (inflammation, cancer, etc). In this situation, the drug does localize and concentrate to a certain site where it acts, it just disperses and circulates on the way to getting there. Or else they circulate randomly, but only act on cells that are in a certain state and therefore concentrate their effect there. DMacks (talk) 08:48, 7 June 2012 (UTC)[reply]
In my country, Australia, one of the commercial products based on Ibuprofen runs advertising clearly designed to suggest that the drug somehow knows where the pain is and goes directly there, rather than everywhere in one's body. I suspect it's bullshit. HiLo48 (talk) 16:53, 7 June 2012 (UTC)[reply]
We get the same adverts in the UK and I've been told by a doctor that it is nonsense. Perhaps it only acts near the site of the pain, but it goes everywhere your blood goes, just like anything else. --Tango (talk) 21:07, 7 June 2012 (UTC)[reply]
Severe repetitive stress injury (of which carpal tunnel syndrome is one) can certainly force someone to change careers. If you're experiencing pain on a regular basis, you should definitely do something about it. I've been struggling with a relatively minor case for a while, and I believe that I tend to get better when I'm spending large chunks of time without touching a computer keyboard or pointing device. I sometimes use Dragon NaturallySpeaking to browse the web and read my email, but I don't think there's any way I could stand programming with it. I've seen a doctor and physical therapist through workers' compensation, which I highly recommend. I use Workrave to take breaks from computer usage, and I've started getting exercise to try to improve my overall health. I believe that I'll be completely back to normal eventually, but if I'd not taken action, I'd probably be permanently injured and no longer be a programmer. Paul (Stansifer)22:04, 7 June 2012 (UTC)[reply]
Just speaking from experience, I've had CTS in various grades for 15+ years. In the beginning, when it was very painful, I wore wrist splints at the advice of a doctor, and worked to correct many orthopedic problems with my workspace (for a long time I had my mouse on a different level than the keyboard — a real no-no). Anyway, over the years it improved on its own with these adjustments, and today it only rarely manifests as a dull ache. (And the fact that I can't bowl — for whatever reason, bowling triggers every weird CTS symptom in me, and I temporarily lose feeling in multiple fingers. I wasn't any fan of bowling anyway, so no big deal as far as I'm concerned.) I still have a career in which I am constantly using computers and constantly typing. Results will vary given the individual and the severity of the case, but anecdotally, on the face of it I wouldn't conclude that CTS would make computer-based careers inaccessible, but one would really need to be proactive about managing the CTS and correcting the conditions that have created it in the first place. If you have CTS or suspect you do, definitely talk to your doctor about it, there are lots of relatively simple things that can be done to mitigate it, along with non-simple things if those don't work. Separately, I couldn't imagine coding with voice command software, personally, though I wouldn't be surprised to hear that some people can manage it. --Mr.98 (talk) 00:27, 8 June 2012 (UTC)[reply]
density
Is density more/less proportional to the distance between nuclei or the number of protons and nutrons in the nuclei or are they equally proportional. What s the relationship? — Preceding unsigned comment added by 165.212.189.187 (talk) 15:18, 7 June 2012 (UTC)[reply]
Density is mass per volume. The nucleus is essentially all of the mass and that mass is essentially the total of the proton and neutron masses, so the mass (and therefore the density) is proportional to the number of protons and neutrons assuming changing these does not change the distance from one nucleus to the next (i.e., the ionic or covalent radius in the material). Volume is the third power of length, so the density is inversely proportional to the cube of the distance between nuclear centers. DMacks (talk) 15:51, 7 June 2012 (UTC)[reply]
"inversely proportional to the cube of the distance between nuclear centers." is just the simple-case situation of assuming a cubic lattice, etc. DMacks (talk) 18:59, 7 June 2012 (UTC)[reply]
Density implies a finite amount of free space in each atom. Someone on this ref desk said there is no way to measure the amount of free space in an atom. Could someone clarify?165.212.189.187 (talk) 15:24, 7 June 2012 (UTC)[reply]
Density does not require finite free space, merely the ability to say "the vast majority of the mass is somewhere within a certain volume". That is, it doesn't matter whether it's a high-mass point particle (nucleus) surrounded by perfect vacuum, or if the mass is evenly distributed for a certain size, or a fuzzy blob that has no distinct boundary as becomes less dense as it extends outward, or even if we cannot actually describe in "real world" macroscopic ideas what is happening a little ways away from the center. Given a large enough space, we can still confidently say "the total mass is X and the total volume is Y within it" and calculate the average density for that object. When we talk about density of a chemical (unless you're doing x-ray crystallography or something), we're talking macroscopic, not just one or two atoms, so the mass distribution at the atomic level is orders of magnitude too small to make a noticeable difference. DMacks (talk) 15:57, 7 June 2012 (UTC)[reply]
Again, it doesn't matter what is in the space, whether it's extended nucleus or a field generated by nuclear or electronic wavefunctions, or virtual particles or "nothing at all". DMacks (talk) 18:46, 7 June 2012 (UTC)[reply]
(edit conflict) That's true and I'm not trying to answer the question. My response is to dispute your "Density implies a finite amount of free space in each atom" premise on which the question and confusion appears to rely. DMacks (talk) 18:59, 7 June 2012 (UTC)[reply]
OK, I'll try. To define density you need only define the mass and the volume of the total "thing" you are trying to get the density of. The distribution of matter within the volume is irrelevant. When people talk about the density of a nucleus, they are defining a somewhat arbitrary volume within the atom to be "the nucleus". Someguy1221 (talk) 18:56, 7 June 2012 (UTC)[reply]
I am not concerned with the density of the nucleus, just the atom or any amount of a certain element for that matter. once you define the volume haven't you also determined the (free)space that atom/material occupies?165.212.189.187 (talk) 19:20, 7 June 2012 (UTC)[reply]
(linking back to the previous discussion for context) True, once you have defined a volume of interest, you have (rather tautologically) determined the space that things within that volume reside in. Some of it is probably "free space", though it's not clear what you mean by that phrase, and as noted in the prior discussion, "free" will depend not only on the things themselves but also what they interact with -- see the prior discussion's example about neutrinos being able to consider pretty much anything to be "free space". There's also the problem, as noted before, that things at the quantum scale do not have precisely defined boundaries. How big is an atom? We can only speak statistically.
Ultimately, though, I think it will be most helpful if you clarify your intent/meaning regarding "free space" and its specific inclusion in your questions, as it appears to me that the rest of the questions have been ably answered. — Lomn20:09, 7 June 2012 (UTC)[reply]
165, it just gets back to the problem that there is no obvious definition of "free space" in modern physics. At some level, you have to arbitrarily define the volume of a piece of matter. You can do this based on the statistical probability of locating a particle in a certain region of space, or on the distance over which a particle can exhibit a certain type of interaction, or some other equally arbitrary boundary. You can stuck at both ends, actually. Even when you're probing an empty vacuum, something you would consider "free space", there is always a probability of finding an electron that shouldn't be there. And if you probe right at the center of what you think is a proton, there is always a probability of finding nothing. Someguy1221 (talk) 00:20, 8 June 2012 (UTC)[reply]
Thanks, although you question my definition of free space (point taken), I question your definition of "nothing". Really, the electron that you find "shouldn't" be there?165.212.189.187 (talk) 12:53, 8 June 2012 (UTC)[reply]
Someguy is simply illustrating the problem with simplistic models when compared to the real world. In the context of a discussion about "empty vacuum", anything found there (such as the electron) "shouldn't" be there in terms of the model (else it's not empty). This is not a statement imputing motive to the electron, nor is it a statement that electrons shouldn't be found in nature, but rather a recognition that models tend to be imperfect abstractions. — Lomn13:21, 8 June 2012 (UTC)[reply]
Most of the evidence is consistent with a very strong placebo effect. There's a lot of research outlined at Acupuncture. Some of the bigger findings include evidence that the location of needle placement is unimportant (evidence against the importance of meridians or particular points being associated with any particular malady) and that actual needles need not be used at all (poking, but not breaking, the skin with toothpicks performed as well as inserted needles). Add to that the fact that meridians and qi have never been shown to exist and that most of the successful trials involve only subjective outcomes (e.g., pain reports) and/or inadequate control groups, the fingerprint is one of a placebo treatment. — Scientizzle20:09, 7 June 2012 (UTC)[reply]
I'll add that "working" and "a placebo effect" are not necessarily exclusive. If a treatment can reliably and repeatedly achieve the desired effects, then it "works", even if we are pretty sure there is no valid underlying mechanism. SemanticMantis (talk) 20:12, 7 June 2012 (UTC)[reply]
I've seen a summary of published research linking acupuncture to the release of endorphin. It was found that when test subjects were given something that blocks the action of opiates, the pain-relieving effect of acupuncture disappeared. I don't have a reference to the research but the endorphin article has a section that seems to say the same thing. So it seems that acupuncture is not (entirely) placebo. IMO, the reasoning of experiments using supposedly-sham acupuncture to show that acupuncture is based on placebo effect is flawed. The observations don't necessarily support the conclusion that the effects of acupuncture has no physiological basis. An alternate conclusion is that the traditional procedures and emphasis on the meridians are unnecessarily specific. It could be that the same physiological mechanism is triggered by the "sham" procedures. --98.114.146.125 (talk) 12:36, 8 June 2012 (UTC)[reply]
We do have an article on veterinary acupuncture, the subjects of which one might imagine would be less prone (although perhaps not immune) to the placebo effect. The article is dismally referenced, however, so all you can really take from it is that people think it works on animals too. But then people believe all kinds of nonsense. -- Finlay McWalterჷTalk20:39, 7 June 2012 (UTC)[reply]
Pets can be prone to a kind of placebo-by-proxy. The owner things the pet should be getting better, so behaves differently, and that difference in behaviour makes the pet better (or seem better). You need a blind study, with the owner not knowing if the treatment has been done or not, to get useful results. I'm not sure if any of those have been done - our article doesn't say. --Tango (talk) 21:12, 7 June 2012 (UTC)[reply]
Strong OR notice here: I can not speak to whether acupuncture is legitimate science or a placebo effect, but having had numerous treatments, it has been effective for me in relieving pain, stress and other conditions. → Michael JⓉⒸⓂ06:20, 8 June 2012 (UTC)[reply]
I have had treatments and I liken it to the feeling you get after working out. They also say that the placement of the needles does matter because it causes a "micro wound" which triggers the white blood cells and other "healing" chemicals in our body to concentrate there to begin to heal that area.165.212.189.187 (talk) 12:58, 8 June 2012 (UTC)[reply]
I had accupuncture administed by my physiotherapist to treat a tear in my anterior supraspinatustendon. The rationale was that tendons heal poorly due to limited blood flow, and inserting a foreign object triggers an immunoresponse that leads to an increase in blood moving into the area. 203.27.72.5 (talk) 07:10, 9 June 2012 (UTC)[reply]
It's also abusing standard terminology: increasing a quantity "by addition" or "by factors (multiplication)" is not a clear or common way to phrase a geometric operation. Factorization has a very precise mathematical definition, explained in our article. Nimur (talk) 23:22, 7 June 2012 (UTC)[reply]
It looks like it is trying to say the increases are "additive" and "multiplicative", however it's all multiplicative just with different powers. So, it's neither a proof nor correct. I'm not sure what "the geometry file" means, either. --Tango (talk) 23:24, 7 June 2012 (UTC)[reply]
I think the person writing this appreciates that the area of a sphere is two-dimensional, and thus, when the radius is increased by a factor n ("additive"?) its area increases by a factor of n2. The "proof" is that he has sketched two perpendicular axes to represent this area. Of course, this falls short of a clear mathematical proof of the proposition, though someone's intuition is on the right track. I think it would be taken as more of a proof if he considers that the sphere can be divided into many little nearly-square sectors (square in the limit as it is divided up infinitely fine) in which case the proportionality to r and r2 can be proved plainly since they're a defined shape; otherwise the proof needs to incorporate a proof of the area of a sphere of a given radius. Wnt (talk) 11:25, 8 June 2012 (UTC)[reply]
proof of the Pythagorean theorem I would compliment the artist on the nice diagram, and point the OP to the notion of "proofs without words", such as this proof of the Pythagorean theorem. For more, see here [8], and/or google it. Indeed, the originally-posted image could be modified into an essentially rigorous proof, as others have suggested. Basically, it is not necessary to make formal verbal claims to have a "proof", if one is willing to rely on the reader's background a bit (after all, very few proofs are entirely self-contained). SemanticMantis (talk) 15:21, 8 June 2012 (UTC)[reply]
Is Skin Cancer More Prevalent Now Than in the Past?
It seems that every summer we are warned of the harmful and dangerous effects of sunlight on unprotected skin. SPF numbers increase every year. Even the
slightest exposure to sunlight is discouraged, often in a nearly hysterical tone.
What seems odd about this is that, until very recently, constant exposure to the sun was the norm for mankind. Whether it was building the pyramids, growing
crops, sailing, etc., people spent their entire lives working in direct sunlight, without any more protection than clothing and a hat, if that.
I know that scientists, using modern medical technology, are able to determine the ailments of mummified Egyptian pharaohs. Is there any evidence that historic
skin cancer rates were comparable to today’s cancer rates?Phidias007 (talk) 22:42, 7 June 2012 (UTC)[reply]
It might be interesting to look at the average age that skin cancer is likely to occur vs. the average length of life. That is, if the skin cancer rate were lower, it might be simply that something else bit them first. ←Baseball BugsWhat's up, Doc?carrots→ 22:54, 7 June 2012 (UTC)[reply]
My first thought was the same as Bugs', but according to our article on skin cancer people under 19 are the most likely to get skin cancer (at least children from UK). The bigger factor perhaps is ozone depletion. Our ozone depletion article discusses the resultant skin cancer increase at length, but for some reason our skin cancer article only mention ozone depletion once, at literally the very last sentence. Anonymous.translator (talk) 23:26, 7 June 2012 (UTC)[reply]
You've misunderstood the article. It's comparing children in the UK with children elsewhere in Europe. There is no comparison between different ages (I've checked the source). I'm reworded that bit of the article to make it clearer. My first thought was also the same as Bugs' and I suspect we are all correct. Serious infectious disease is so much rarer now that pretty much every other medical condition is more common than it used to be simply because more people are surviving long enough to get it. --Tango (talk) 23:37, 7 June 2012 (UTC)[reply]
Northern Europeans, with their light skins, are far more vulnerable than any other population. It is only recently that large numbers of light-skinned Caucasians have lived at tropical and subtropical latitudes. See our melanoma article for more information. The ancient Egyptians were actually pretty dark-skinned. Looie496 (talk) 23:19, 7 June 2012 (UTC)[reply]
It wouldn't surprise me if skin cancer was quite prevalent in the past and people just ignored it. People get all sorts of harmless things in their skin, and get left with all sorts of scars after catching various poxes, that they wouldn't have noticed melanomas as anything remarkable. Melanomas usually only become a serious problem when they metastasise, and you're not going to know that the symptoms of the new tumour(s) have anything to do with the skin lesions. --Tango (talk) 23:41, 7 June 2012 (UTC)[reply]
We are told here in Australia that we have the highest prevalence of skin cancer in the world, largely because we've plonked a whole bunch of people with northern European ancestry into the sunniest continent on Earth. We've also had a sun, beach and outdoor sports loving culture for most of the life of this nation. And yes, we live longer than our ancestors, so the skin cancer has time to appear before we die of something else. HiLo48 (talk) 23:47, 7 June 2012 (UTC)[reply]
Another issue is that in pre-industrial times most Caucasians had constant exposure to the sun, so they built up tans gradually, and did not burn, which is what causes the most damage. Nowadays people live and work indoors and burn on weekends and holidays. μηδείς (talk) 03:18, 8 June 2012 (UTC)[reply]
Two broad thoughts: 1. Increasing SPFs is a matter of marketing as much as anything else. Beware of confusing advertising with reality. 2. Increased incidence has to be squared away with increased diagnostic ability or changed diagnostic criteria; just because you suddenly measure more or something doesn't mean its base rate has changed. --Mr.98 (talk) 11:47, 8 June 2012 (UTC)[reply]
The amount of UV incident in a given area at the "top of the Earth's atmosphere" is a function of the sun's sunspot activity. This runs in cycles - over the last 100 years, the approx 11-year cycle on top of the gradual rise since the last Maunder Minimum. The variation due to this in the context of skin cancer is just about negligible. More important factors are waht affects attenuation of UV in the atmosphere. This is mainly the rise in particulate polution since WW2, which decreases UV incident at the ground, and currently significant only in certain cities, and the depletion of the ozone layer in the last 20 years of so, which increases UV incident at ground level, more in some locations that at others. See http://en.wikipedia.org/wiki/Ozone_depletion, which does not include the actual UV increase due to ozone depletion, but does give data on the consequent increase in skin cancer. Incident UV affects the output of photovoltaic electricity generatiion. In some areas where ozone depletion is significant, output is in recent years of the order of 2 to 5% higher than expected, but in a quick search I could not find a definitive online reference. Wickwack124.178.139.104 (talk) —Preceding undated comment added 11:33, 9 June 2012 (UTC)[reply]
June 8
who edit Wikipedia in a sensitive subject... who is right ??
Hi:
According to Wikipedia in English, in the page about "omega 3" they said that are really controversial the good effects that brings omega 3 in any mammal... if fact they cite research that say that is not a really big help for the human body.
In Wikipedia in Spanish, they say exactly the opposite, omega 3 is really a big help for the human body. Who is right ?? who edit this articles ?? what about if I am a mediocre doctor who just write something that I learned 100 years ago ?? to whom I have to believe is a kind of articles that involve a live risk ?? many thanks in advance. chau and sorry for my funny English — Preceding unsigned comment added by 186.2.50.237 (talk • contribs)
Because Wikipedia can be edited by anybody, you should always have doubt about what you see in a Wikipedia article. If you want to check, you should look at the sources that the article cites. If there are no sources, you should have a lot of doubt. Anyway, my understanding is that the English article is correct. (For convenience, our article is Omega-3 fatty acid. Looie496 (talk) 03:02, 8 June 2012 (UTC)[reply]
It's difficult to evaluate the claims in the Spanish article as instead of citing sources, they simply provided links, and those links are now broken. Someguy1221 (talk) 04:10, 8 June 2012 (UTC)[reply]
Spanish wikipedia has horrible administrators, some of them, revert all Ip's edits without reading them at all. Many requests are ignored. They take community decisions by votes instead of arguments.. though I have to admit that there is a lot more vandalism there than here. 65.49.68.173 (talk) 16:28, 8 June 2012 (UTC)[reply]
I think the English Wikipedia is not so reliable on these sorts of medicine related issues, because it gives far to much weight on the Institute of Medicine (IoM) reports; these reports are extremely conservative when it comes to accepting claims of benefit, while the burden of proof needed to include possible negative health effecs is extremely low. While this may be a good thing for compiling reports meant as advice to health care workers, what we need on Wikipedia is a balanced approached, one that gives equal weight to equally reliable evidence. Count Iblis (talk) 17:53, 8 June 2012 (UTC)[reply]
It's too unstable. HeH+ is already the strongest known acid, i.e. it is more likely to dump a hydrogen ion than any other compound ever discovered. There are a handful of funky looking sources claiming you can get HeH+ to accept a hydrogen atom, but not a hydride ion. My guess is that even if you could, the second hydride would be dumped almost immediately. Someguy1221 (talk) 05:47, 8 June 2012 (UTC)[reply]
Careful. The relevant unit would be a simple proton (H+), not "hydride" (H–), which would instead give the neutral helium dihydride result. But the massive instability is certainly the key. Our helium hydride ion discusses the species mentioned by Someguy, and also evidence for and stability of HeH2+ and others in this monocationic series. One could certainly so some ab initio calculations on HeH22+ to see what would be happening there. DMacks (talk) 14:22, 8 June 2012 (UTC)[reply]
And just to prove I'm not just making stuff up, HeH22+ (CAS #12519-50-5) has been studied in this manner, and is unstable in normal situations but is stable in the presence of high magnetic fields (like "surface of a neutron star")--see doi:10.1103/PhysRevA.81.042503. Note that all refs I found consider it as a chain--not sure if exactly linear in all cases, but anyway more structurally related to beryllium hydride than the trigonal trihydrogen cation you mention. DMacks (talk) 14:46, 8 June 2012 (UTC)[reply]
Water on the line
I work 50m away from a railway line in SE England. Yesterday, after a day of rain, when a train passed along the line, there was a spectacular noise followed by plumes of steam that stretched for a couple of hundred metres and reached well above the tree line. I've lived and worked next to this stretch of track for 17 years and have never seen anything like this before.
I have not heard on the news that a train-load of people have been fried on the London to Dover line yesterday afternoon, so was the train acting like a Faraday cage and were the passengers in any danger? 83.104.128.107 (talk) 15:51, 8 June 2012 (UTC)[reply]
There's some missing info in your question. Is this an electric train, powered by an electrical rail ? If so, the electricity would want to go towards earth/ground, and standing water by the non-charged rail, in conjunction with electrical connections between the rails through the train's undercarriage, might have allowed that. There would be little "motivation" for the electricity to go into the passenger compartments, so a Faraday cage isn't necessary. StuRat (talk) 17:20, 8 June 2012 (UTC)[reply]
(@StuRat - Most of South East England is a 'third-rail' electrical pick-up system.)
At OP - Can you clarify more closely where you are in SE England and where this incident occured? Your description sounds like some kind of flashover effect, but I hadn't heard anything about such through my railfan contacts, and most modern trains have a specific trip so they don't produce the kind of effect you saw. Sfan00 IMG (talk) 15:49, 10 June 2012 (UTC)[reply]
Hello Sfan00, It was on the stretch of track between Faversham and Selling at around 16.40 on 14th June. The line crosses a farm where I saw it, I checked the train times and no journey in either direction seems to tie in, but there are often freight trains passing. I would love to know what the passengers/driver felt or saw! A friend did suggest that I might have seen a steam train, but there was no 'chugging' and there was no majestic view of it in the distance when I dashed up to the bridge. I am hoping that with more rain to come and the ground still saturated it might happen again this week. 80.176.84.184 (talk) 17:28, 10 June 2012 (UTC)[reply]
@80.176, you said that this happened on June 14, which, unless you are talking about last year, will not occur for another four days. What did you intend? Falconusptc21:03, 10 June 2012 (UTC)[reply]
What cause brownian movement?
What makes gas molecules extert brownian motion ..? is it related to the electron shell?, proton/neutron core? quarks? the energy at least intermediately has to be stored somewhere. Electron9 (talk) 16:15, 8 June 2012 (UTC)[reply]
The energy isn't stored anywhere. It's a manifestation of the thermal kinetic energy of the individual molecules impacting the larger object.
In simpler language, you should know that temperature is really a measure of the average kinetic energy of the individual molecules in a substance. For solids this is just molecules vibrating, but in a fluid like water or air the molecules are free to move around each other. So when you put a small enough particle in a fluid and look at it under a microscope, only a few molecules are going to be hitting the (relatively) larger object at any one time, and the odds are good that they won't be hitting symmetrically. Therefore they impart a small amount of net kinetic energy to the larger object, and it moves slightly. It is a random process, so averaged over infinite time there will be no net motion, but it can still jiggle the object around a great deal over short time scales. -RunningOnBrains(talk)16:39, 8 June 2012 (UTC)[reply]
If brownian motion is plainly a manifestation of thermal vibration. What parts of the atom is vibrating? electron shell? nuclei? quarks? Electron9 (talk) 17:44, 8 June 2012 (UTC)[reply]
Further: Any macroscopic analogy is going to be at least partly incorrect, because we are dealing with individual molecules and atoms here, which are subject to quantum mechanics. However, I will do my best.
In a gas or liquid, molecules are never still. They are constantly moving, at a velocity that can be predicted by the temperature of the gas. As a 2-dimensional analogy, imagine a whole bunch of billiard balls flying around a giant pool table, with no friction to slow them down. They will stay at a constant velocity until they hit either the walls of the table or another billiard ball, and then they will go off at a different velocity in a different direction. It is very nicely illustrated by the image I posted at right: You have to remember that there is no friction at this scale, so unless the material is cooled or warmed, the average thermal velocity of the molecules is going to stay the same. Now if you introduced a larger object to the table, say a bowling ball, it's going to be jostled by the constant collisions, and so if you were standing far enough away that you could only see the bowling ball, it would appear to be vibrating randomly, just like in Brownian motion. -RunningOnBrains(talk)17:59, 8 June 2012 (UTC)[reply]
Thanks! (maybe should be added to the article), does thermal vibration cause the nuclei and electron shell to vary their distance to each other? ie will the atom deform in some way like air does for sound? Electron9 (talk) 19:43, 8 June 2012 (UTC)[reply]
Not really, but here we're really getting to the point where macroscopic analogies break down, because electrons really aren't in one place at any time, and electron shells aren't really a physical object: see electron cloud. I am also probably extremely unqualified to speculate on the exact quantum mechanical processes which take place when two atoms collide; it probably depends strongly on which atoms we're talking about. But it can be safely described as a purely elastic collision for the point of describing Brownian motion.-RunningOnBrains(talk)20:09, 8 June 2012 (UTC)[reply]
Actually, I disagree with Runningonbrains here. Absolutely, thermal vibration causes variations in the position of the different atomic particles relative to one another. As Runningonbrains rightly pointed out the (average) velocity of the particles that make up an object can be predicted from the object's temperature. The velocities of the individual particles vary according to the Maxwell-Boltzmann distribution. The deformation of the atoms or molecules leads to a higher energy state i.e. electrons repel one another altering the shape of their orbitals. These high energy states are relatively unstable and if a more stable configuration can be assumed, then it will be. Many molecules that decompose do so more quickly at higher temperatures. This can be modelled by saying that those molecules that have high energies are deformed by the motion of their constituent particles and assume a more stable state by breaking chemical bonds. The number of molecules with high energies is a function of temperature as predicted by the boltzmann distribution. Electrons can even be heated so much that they leave the atom all together, as in a thermal plasma. 203.27.72.5 (talk) 07:49, 9 June 2012 (UTC)[reply]
That would mean that the atomic nuclei (protons-neutrons) will deform in a plasma just like electrons does at a lower temperature? Electron9 (talk) 08:27, 9 June 2012 (UTC)[reply]
Somewhat more important in practice is that some thermal energy is stored by the rotation of molecules at temperatures above about 600 -700 K, (ie angular momentum, as well is the linear momentum mentioned above) and by the lengthening of inter-atomic bonds with increasing temperature, which further increases the fraction of heat energy stored in rotation. This is evidenced by the fact that noble gasses show specific heat independent of temperature, but large molecules have considerable variation in specific heat throughout the measureable temperature range. Only the fraction of heat energy stored in linear momentum drives brownian motion. Wickwack124.178.139.104 (talk) 11:42, 9 June 2012 (UTC)[reply]
A side question, would a tube between two vessels with a diameter just slightly larger than a single atom (or molecule) and a length significantly less than the average brownian motion distance. And funnel on one side make more atoms to move to one side than the other? especially when the mol/m³ is low. Electron9 (talk) 19:43, 8 June 2012 (UTC)[reply]
Do objects which absorb radiation reemit it? How does this work? — Preceding unsigned comment added by 176.250.228.38 (talk)
It depends a lot on what type of radiation you're talking about. All kinds of matter absorb and emit all kinds of radiation all the time. It is a continuous process. --Jayron3218:54, 8 June 2012 (UTC)[reply]
Most things which absorb radiation with then give off radiation as well, although it may very well be a different form of radiation. For example, if you shine visible light (one form of radiation) onto a black object, it will radiate the energy back out, not as visible light, but as infrared light/heat. StuRat (talk) 19:02, 8 June 2012 (UTC)[reply]
Yes. I know in chemistry, there are lots of experiments that don't do what a hypothesis (based on literature precedent for similar experiments) says they "should"--exceptionally low yield, different geometric form, different part of a complex molecule reacts, nothing happens at all, or a totally different reaction occurs instead. There are probably a near-infinite number of combinations of experiments that would follow some not-yet-detected pattern, or an observed pattern that does not have any known underlying cause, where the data is all "out there" but nobody has even looked yet (i.e., to explain apparently random variations in yield, etc.). DMacks (talk) 20:06, 8 June 2012 (UTC)[reply]
This answer isn't exactly right. The answer to your question really depends on what question you're asking.
If you're asking if there are exceptions to the established laws of physics and chemistry, the answer is no. You're not going to get gravity to be different from one experiment to the next, and you're not going to get sodium and chlorine to react and form anything other than sodium chloride. They are called "Laws" for a reason.
However, if you're asking whether or not experiments can produce unexpected results, the answer is most definitely yes. You can never (probably) have an experiment that is completely controlled, where you know every bit of information about the initial conditions. You can get a different yield than you were expecting from a reaction, but this would be due to some contaminant you didn't know about, or some environmental factor that was different like temperature or moisture in the air. Maybe the yield is highly sensitive to the initial ratio of reactants, and the expiriment Maybe even some unforeseen quantum mechanical effect could change the expected results, if it's an especially complicated chemical reaction. Or, incredibly rarely, maybe your physics experiment has discovered a whole new particle or effect.
However, if you do exactly the same experiment every time, you will get exactly the same result. The reason that biology is such a messy science with many unexpected results is that there are just too many unknown factors to take them all into account; an organism is unimaginably complex, certainly not as easy to describe with simple laws as E=mc2 and F=ma.-RunningOnBrains(talk)20:22, 8 June 2012 (UTC)[reply]
Another important point is that science is neither a set of laws, nor a primarily deductive enterprise. Science is a method for inductive reasoning. --Stephan Schulz (talk) 20:31, 8 June 2012 (UTC)[reply]
Running, your secondthird last paragraph sounds like determinism, which is not a necessary component of science, and indeed is contrary to some widely held interpretations of quantum mechanics. --Trovatore (talk) 20:36, 8 June 2012 (UTC)[reply]
I guess I interpreted the OP's question as a macroscopic sort of thing. You're right in that there is always some QM-related uncertainty (even if miniscule at macroscopic levels); but those would not be unexpected to an experimenter, and certainly follows the "known rule" as the OP put it. I would say that my final sentence above really sums up my point; maybe I shouldve just stuck to that :) -RunningOnBrains(talk)22:46, 8 June 2012 (UTC)[reply]
It depends a lot on what the OP means by "rules". Most scientific "rules" are actually models of some sort, and all models are approximations of reality, so there will always be real examples that lie outside of the predictions of the model. --Jayron3203:04, 9 June 2012 (UTC)[reply]
As a former analogue electronics engineer, now involved in certain aspects of chemistry, I must say there is an immense difference between electronics and chemistry at an engineering/design level. In electronics, everything is ultimatey based on a limitted number of component parts - resistors, capacitors, inductors, conductors, and active devices (transistors etc). The behavior of these devices behave according to well established simple laws - so simple that a 12 year old can, if sufficiently interested, design a stereo (I did when I was 12). Real parts don't exactly follow these laws, but they are close enough. Understand those laws properly, and you can understand anything in analogue ectronics.
Chemistry is very different. The "almost fundamental" component parts of chemistry are the atoms. The behaviour of atoms in any situation can (in theory) be predicted by quantum mechanics. In practice, that's just too hard, so the "laws" of chemical engineering are fortutious theories like the kinetic theory of gasses, and the theories of chemical kinetics with regard to reaction rates. These theories have so many gaps, exceptions, an approximations, that to former electronic engineer, it is very frustrating. To calculate current in a circuit, I can alway do that to at least 3 figures accuracy - 6 figures, if I need it, is not hard. To calculate the rate of a chemical reaction, chemists are doing well if they get within the correct order of magnitude. As a further example, the kinetic theory of gasses pupports to give an understanding of specific heat (thermal capacity) and how it varies with temperature. It accuately gives the specific heat for noble gasses (but who cares), and is roughly right for low valency atoms, but seems to be very inaccurate otherwise.
In electrical enginering, if say, a power company wants a $100M EHV transmission line and distribution system, the engineers do some caculations, order the materials, get it built, and it will work just fine. In chemical engineering, if a company wants a new $100M processing plant, the engineers do some calculations, scour the world for somebody who has done something like it, tweak the calculations, then build a pilot plant and muck about with it untill it works. Then with that experience, do more calcs, scale it up, then order all the materials etc and build the BIG ONE. Then sometimes find out it doesn't work at all well, and $100M has been wasted.
In short, chemistry must conform to valid scientific laws, but those laws are too difficult, so in practice, rough semi-empirical approximations are used. And things don't always go according to plan.
If you understand "basics" like chemical kinetics (and that is not at all easy), you still may not be able to understand real world applications. By "understanding" I mean able to calculate and preduct accurately what will happen.
It's a subtle and important point - every science has different objectives to deal with different topic matter. Physics sets hard-and-fast absolute rules of what is impossible. Chemistry is more about figuring out what is practical. And biology is a science of the possible, where every "rule" has an exception. The continuum continues further in disciplines like psychology, where there is doubt if it is even a science, and perhaps, even to the tropes of fictional writing. Wnt (talk) 15:26, 9 June 2012 (UTC)[reply]
Look at Oxymercuration reaction in which a cyclic mercurinium ion is formed, Hg has lost an electron and has three bonds, two with adjacent carbons. Still no double bond though, and I could find no evidence of it on google either. Graeme Bartlett (talk) 07:28, 9 June 2012 (UTC)[reply]
"Mercuroketone" is probably not a good term for it, since mercury not very electronegative. More likely "methylene mercury" (or other coordination/description of the carbon part, as usual for ligands on metals) or a "mercury carbene" (161 google hits) complex or something like that. Hg+=CH2 (apparent Hg(III) species), CAS#1234574-43-6, has been studied theoretically. But I'm also seeing lots of examples where what your type of connectivity is written as a Hg(II) ylide, for example, Hg+–C–H2, rather than a Hg=C double bond. As for some of the coordination examples, you have to be careful not to exceed an electron-count of 18 (the transition-metal analog to the octet rule used in main-group elements). DMacks (talk) 14:42, 9 June 2012 (UTC)[reply]
Mmm, can't comment on the specific molecules here, but I remember from my chemistry undergraduate that transition metals can form some pretty funky organometallics from time to time. In principle these molecules look plausible. LukeSurltc09:34, 10 June 2012 (UTC)[reply]
Have you yet attempted to model the proposed molecules? Nevermind the mercurylidenemethylidene group, the angle strain of tetragonal carbons alone, is enough to destablise most of these molecules. Plasmic Physics (talk) 10:13, 10 June 2012 (UTC)[reply]
Oooh yes, that triangular arrangement in 4 looks especially painful. I'm assuming the question is mostly about the idea of the C=Hg bond rather than the carbon skeletons I'm guessing the OP has just made up. 10:30, 10 June 2012 (UTC)
No, you are 3 orders of magnitude out. To calculate db[μV] take the Log of the voltage in microvolats and multiply by 20, thus 56 mV corresponds to 95 dB[μV] and 31000 mV (31V) corresponds to 150 dB[μV]. 35 dB[μV] corresponds to 56 μV. 90 dB[μV] corresponds to 31.6 mV. However, 35 dB is rather high for the required input at the teminals of a TV set. A modern analogue TV should get a good picture with 20 dB[μV] or even less. 35 db[μV] would be good at the input to the antenna distribution cable system in a high rise building, where there is significant loss in the cable runs and in the splitters. A digital TV should in theory do rather better but in practice you need to allow a good margin to avoid dropouts and friezes. Keit120.145.6.122 (talk) 06:36, 9 June 2012 (UTC)[reply]
Like I suspected then, at least the multiplier 20 was correct. Btw, do you have any sources regarding the 20 dB[μV] level? maybe there's even a general difference between analog (CVBS) and digital (DVB) in regard to minimal signal strength? Electron9 (talk) 06:49, 9 June 2012 (UTC)[reply]
For some years now, every time the allegedly oldest person in the world dies, the person's age has been reported to be 114. Just today we saw this item, saying this woman was the oldest person in Europe, and she died today at the age of 114. Why always that same age and never 113 or 115? Michael Hardy (talk) 02:38, 9 June 2012 (UTC)[reply]
It's often 115, that number you saw was just for Europe. I would argue it's simply a statistical issue. If you look at the US Social Security Administration's most recent actuarial table, they calculate the probability of making it from 115 to 116 to be only 25%. And making it from 114 to 115 is only a measly four percentage points better. So I would look at that and say that you start with a fixed population maximum of people born in 1897, and have that population experience greater-than-exponential decay from age 10 onward (it is less-than-exponential prior to that). The reason the "oldest person at the moment" is almost always 115 is that the decay function, although a bit noisy at those ages, would predict less than one survivor for all ages past 114. You'll see that creep up in the future as the starting population for each given year is increasing, as is post-adulthood life expectancy. Someguy1221 (talk) 02:53, 9 June 2012 (UTC)[reply]
I have seen data on that the maximum life expectancy is 130-150 years for a human, so 114 years is getting close. And thus the deterioration of the body is likely becoming exponential. Electron9 (talk) 03:02, 9 June 2012 (UTC)[reply]
Now, modifying the human might allow indefinite lifespan, like genetically rebuilding it and adding repair nanobots. So I'm assuming they're talking about without genetic engineering, cyborgization, or reanimation. (perfect preservation already existing, in the form of liquid nitrogen). So how do they propose living 27.5 years over the record? Calorie restriction? Sleeping through most of it?(/coma/hibernation/near death) Hysterectomy? Sagittarian Milky Way (talk) 18:54, 9 June 2012 (UTC)[reply]
I guess the disagreement is probably due to low sample size, but this study suggests that mortality per year (i.e. chance of dying in any given year of life if you live that long) is 50% from age 110-115, and they speculate that that number may even increase beyond this age. -RunningOnBrains(talk)04:41, 9 June 2012 (UTC)[reply]
The small sample sizes are, indeed, a problem. The mortality tables I use professionally are only based on actual data up to age 95. After that, there just isn't enough data to get robust results, so they arbitrarily extrapolate from age 95 up to age 120, which they set as having a mortality rate of 100%. A detailed explanation of the process can be found here (be warned, it is quite technical). --Tango (talk) 16:31, 9 June 2012 (UTC)[reply]
I guess that it is a matter of time. A fresh solution of potassium permanganate should be neutral. As time passed the permanganate decomposes slowly, as it does, the solution becomes more and more alkaline. Plasmic Physics (talk) 10:57, 9 June 2012 (UTC)[reply]
What is the decomposition mechanism for permanganate? O=[Mn-](=O)(=O)=O.O=[Mn-](=O)(=O)=O → O=[Mn-](=O)(=O)OO[Mn-](=O)(=O)=O → O=O.O=[Mn-](=O)=O.O=[Mn-](=O)=O ? Plasmic Physics (talk) 13:17, 9 June 2012 (UTC)[reply]
It isn't really decomposition, it's oxidation: The permanganate will oxidize just about anything, producing manganate and some sort of oxide, or elemental oxygen. The manganate will spontaneously disproportionate to permanganate and managanese dioxide, so given any trace reductant, there should develop an equilibrium between manganate, permanganate, and manganese dioxide which will account for the rising pH. --Jayron3215:41, 9 June 2012 (UTC)[reply]
In that case, I don't think that it is in an equilibrium - it is not a closed system, oxygen escapes from the solution as singlet oxygen, which then decays into triplet oxygen. According to your process, I think that the reaction would be:
Make a simple visual search of places where things can be quickly hidden first. Then, search using a radio scanner. Set the scanner sensitivity fairly low so you don't waste time on legit radio transmissions. See http://en.wikipedia.org/wiki/Scanner_(radio). Scanners are very good at picking simple radio transmitter bugs because when they get to the right frequency while auto scanning, you'll hear your own voice(s), and or the scanner will howl. Sometimes bugs placed by professional outfits or govt angencies are placed that display great inguinuity and cannot be found with scanners. Books have been written about this. I'm not trying to imply anything about your goodself, but be aware that there is a common pschological condition, often occuring in people who are otherwise normal, and sometimes brought on by stress, where people believe that they are being spied on, when they are not. Sometimes businessmen think they are just so darn good that the opposition must surely be spying on them. Very very few actually do so with bugs. Most industrial intelligence is obtained quite legally and simply by employing specialist researchers scanning documents in the public domain. I've learnt what I needed to know about competition by sharing a beer in a pub combined with monitoring employment adverts and press releases. Did you check our wiki article http://en.wikipedia.org/wiki/Covert_listening_device ? Also be aware that it is possible to write computer malware that activates the microphone in a laptop and monitor your voice, as well as your keystrokes, without you knowing, over the internet. Always have reputable computer security installed, and make sure your software firewall is set up with optimised port restrictions. If your PC has Vista, that's good - make sure that installing software requires an administrator password. Wickwack124.178.139.104 (talk) 11:55, 9 June 2012 (UTC)[reply]
You are correct, however a spectrum analyser requires a trained operator or electronic technician to operate it and interpret the display. Anybody with at least average intelligence can operate a radio scanner, and if there is an ordinary FM bug, the scanner operated in the same room as the bug will make the presence of the bug obvious. Wickwack124.178.60.220 (talk) 03:00, 10 June 2012 (UTC)[reply]
Note that radio scanners only work for devices which transmit radio signals. Other approaches are a hardwired bug, with the wires going outside the house to where somebody taps into them, a device that records and is retrieved later, or one that uses existing communication channels, like the phones lines, cable, wireless internet, cell phones, etc., to transmit signals. StuRat (talk) 18:52, 9 June 2012 (UTC)[reply]
Yes, that's another one, where you bounce a laser off the window at a shallow angle, and the refracted laser beam vibrates as the window vibrates, due to sounds in the room, allowing someone at the receiver to listen in. StuRat (talk) 23:45, 9 June 2012 (UTC)[reply]
I think that might be an urban myth, Stu. Being a bit of a nerd, I once made a laser communication system. I tried it out bouncing the beam off a window. I did detect sound from within the room to degree, and noise from inpinging wind to a greater degree, but as for understanding converstions, it was a dead loss. That doesn't mean someone else could have succeeded, but it seems unlikely. The laser I used was a visible wavelength. A covert device would have to be infrared, which window glass doesn't reflect as well. Wickwack124.178.60.220 (talk) 03:10, 10 June 2012 (UTC)[reply]
Indeed, this technique is easy and cheap, as it can be done using a laser pointer, photodiode, and any old device with an audio-in jack. My physics advisor proposed that we make one as our senior project, even had all the materials ready, but I (stupidly) chose a different one. -RunningOnBrains(talk)03:14, 10 June 2012 (UTC)[reply]
Was your physics advisor talking about a voice link (ie from laser pointer to photodiode, nothing in between except air) or a bounce-off-glass eavsdropper system? I just tried my ear against the window (with my other ear plugged) while my lady was inside talking on her phone. All I could hear was the wind. Wickwack124.178.60.220 (talk) 03:25, 10 June 2012 (UTC)[reply]
Correct, you need a window. We were going to have a laser pointer, and a photodiode hooked up to the audio-in jack of a computer set up on one end of the hallway, with the laser pointer pointing at the glass on the other end of the hallway, reflecting back to the photodiode. Theoretically you could use other surfaces too, but windows are best because they are relatively rigid, often single-layer, and not weight-bearing so they are free to vibrate. In retrospect it probably would have been a bad idea for privacy reasons, since that glass was to my advisor's office :D -RunningOnBrains(talk)05:52, 10 June 2012 (UTC)[reply]
You would have been lucky to get that to work. The window vibration will not impart amplitude modulation, it will only impart phase modulation (http://en.wikipedia.org/wiki/Phase_modulation) via doppler effect. A photodiode connected to the audio input of something will not inherently recover the audio, unless you position it just right so that the beam is centred not quite on the diode. That would be extremely critical. In the system I tried, I used circuitry to pulse the laser diode on/off at 40 MHz. That allowed me to use a 40 + a bit MHz oscillator at the recive end to mix with the incomming beam signal and recover the doppler modulation imposed by the vibrating glass. Wickwack120.145.7.109 (talk) 06:32, 10 June 2012 (UTC)[reply]
Crepuscular rays
I saw some interesting atmospheric phenomena the other day (gallery is here; I've included all the shots I took, but the first, fourth, and fifth are the best). So, I'm assuming these are some kind of crepuscular rays, but the shape of them is what I'm curious about. I've seen crepuscular rays on countless occasions, of course, but the shortness of these is something new to me. The sky looked like someone had gotten crazy with a clone brush. What exactly is at work here? Are the rays only showing up in places where there's a certain amount of humidity/water vapour and then disappearing in the drier air below? These pictures were taken around 8am, facing (roughly) east; the sun is off-frame to the left. As you can see, there was a variety of clouds out that morning. I'm afraid all I had on me was my iPhone, so the quality is less than ideal. The images have not be manipulated in any way (other than the standard jpeg compression). Matt Deres (talk) 13:26, 9 June 2012 (UTC)[reply]
The weird thing is, sun rays (crepuscular rays) are lighter than the background, while yours appear darker. I might say they were smoke in the upper atmosphere blown into lines, but that doesn't explain why they would appear to radiate from the Sun. StuRat (talk) 18:47, 9 June 2012 (UTC)[reply]
What you are seeing is the shadow of the cloud. I suspect that you are right that there is a layer of smoke that is causing the short length of the rays. The normal cloud free condition of the sky would be for it to be as bright as the crepuscular rays, and the cloud shadow makes it darker. It would be easier to see a bright ray against a dark background than a slightly darkened sky against the bright sky, but the difference in this situation would be the smoke making higher scattering of light in that part of the sky, bringing up the contrast. Graeme Bartlett (talk) 23:15, 9 June 2012 (UTC)[reply]
I'm not sure what smoke has to do with it. I didn't see any smoke, nor smell any (not that I necessarily would, if it was that high) and we'd just had a few days of gentle rain, so there's no particular reason to think there'd be a forest fire or something. Also, none of the pictures appear smokey. I'm wondering if the cloud on the right is involved; its left edge seems to indicate that it was gradually thinning out towards the phenomenon. Matt Deres (talk) 15:42, 10 June 2012 (UTC)[reply]
Are you sure that it did ? (As opposed to only evolving once and being passed down to all species which evolved from that one.) Our evolution of sexual reproduction article says, in the 2nd sentence, that "All sexually reproducing organisms derive from a common ancestor which was a single celled eukaryotic species.", and provides source(s) to back up that claim. StuRat (talk) 17:40, 9 June 2012 (UTC)[reply]
That sentence by itself wouldn't mean much more than that all sexually reproducing species are eukaryotes. More strongly, it seems very unlikely that meiosis -- the special type of cell division involved in sexual reproduction -- evolved more than once, since it requires a large number of special mechanisms in order to happen, and as far as I know those mechanisms are always implemented in essentially the same way. Looie496 (talk) 19:06, 9 June 2012 (UTC)[reply]
It appears that isogamy was the first stage of sexual reproduction that involves gametes that look alike and so cannot be classified as "male" or "female." In several lineages (plants, animals), this form of reproduction independently evolved to anisogamous species with gametes of male and female types to oogamous species in which the female gamete is very much larger than the male and has no ability to move. There is a good argument that this pattern was driven by the physical constraints on the mechanisms by which two gametes get together as required for sexual reproduction, see Isogamy#Evolution. DriveByWire (talk) 19:21, 9 June 2012 (UTC)[reply]
If you mean meitoic reproduction, just once, per above. But the transfer of genes between bacteria and the recombination of genes in viruses are separate phenomena. μηδείς (talk) 18:04, 10 June 2012 (UTC)[reply]
What is: "colonic sorting" in biology?
thanks.
As I understand it, it's a mechanism by which the colon of certain animals such as rabbits separates small particles and fluids from larger, less digestible particle. Looie496 (talk) 18:56, 9 June 2012 (UTC)[reply]
a simple table which sums up all types of reproduction?
1.monoecious: types, dioecious: male-female, male or female with intersex. thanks.
Doesn't the angle between the camera body and the tracker matter? What are the guidelines for tilting and aligning the camera, once the polar finder has been aligned with the poles? 76.104.28.221 (talk) 19:10, 9 June 2012 (UTC)[reply]
I'm not sure I understand your confusion. Where else would the camera be pointed, if not towards the object you're trying to photograph? See this photo: http://www.astropix.com/BGDA/SAMPLE2/207A.JPG
The barn door tracker acts like a simple equatorial mount with only a right ascension adjustment knob. If you imagine moving the angled board around a full circle, the direction in which the camera points will also move in a full circle. The size of the circle depends on how far the camera lens' direction is from the axis. If it's pointing along the axis, the circle has 0 size, and the camera only rotates. If it's pointing 90 degrees away, the camera traces a great circle around the sky. This is exactly how a star behaves in relation to the celestial pole. --140.180.5.169 (talk) 20:14, 9 June 2012 (UTC)[reply]
Jamming the GPS system on a global scale
Apparently, it's quite easy to jam GPS signals, GPS jammers are even available commercially. But this raises the question of how reliable this system is. Could e.g. China shut the GPS system down globally using satellites in case of war with Taiwan? Count Iblis (talk) 23:31, 9 June 2012 (UTC)[reply]
Yes, along with GLONASS and GALILEO. But most major militaries have frequency hopping positioning satellites ready to be switched on in an emergency. 71.212.248.104 (talk) 04:47, 10 June 2012 (UTC)[reply]
Do you have a source on the frequency hopping GPS satellites? Even if this technology actually exists it's not available in any of the portable GPS receivers that the army uses. Anonymous.translator (talk) 05:30, 10 June 2012 (UTC)[reply]
Or a weak transmitter on board an airplane, or near the up-link which probably is the weakest system point. With a huge (state) budget one could employ satellites for the job. Electron9 (talk) 11:32, 10 June 2012 (UTC)[reply]
June 10
Liquid air energy storage
Thermal energy storage doesn't mention condensing and cooling air to a liquid, storing it, and then boiling it off to recover stored energy. How efficient is that for power storage? I know pumped hydro is about 80%, but it doesn't have a very high capacity on flat terrain. What I really want to know is whether building water towers for pumped hydroelectric is more cost efficient per kilowatt hour than building giant thermos tanks for condensing liquid air. 71.212.248.104 (talk) 04:45, 10 June 2012 (UTC)[reply]
Insulating and maintaining liquid air isn't easy or efficient. In order to keep it cold enough and under enough pressure to be useful will make the system vastly less efficient than pumped hydroelectric. Besides, you don't need to build tanks for pumped hydroelectric. You just need to dig a big hole at high elevation. --Jayron3204:53, 10 June 2012 (UTC)[reply]
Ah-ha! Clearly you're right, because for pumped hydro on flat terrain you only need a big hole in the ground, since the gravity pressure is only necessary at a generator turbine which could be at the bottom of the hole. (Unlike everyone's plumbing which is why water towers need to be in the air, or tanks on hills, etc.) 71.212.248.104 (talk) 05:01, 10 June 2012 (UTC)[reply]
... but if you build it on the coast you can get "more than 100% efficiency" with careful tidal timing (really gleaning energy from the earth-moon orbital system). Dbfirs 08:42, June 10, 2012 (UTC)
Um, what I'm saying is, you just need two (sets of) water tanks, one far underground, and one at the surface, which is fairly inexpensive and would work in Kansas. That is much less expensive than using a water tower, which is the traditional way municipal water pressure is powered in flat areas. You can't predict water usage and you can't predict electricity usage, but you can shape the load with pumped storage tanks in both cases, for only 20% overhead lost to heat at the turbine, pump, and piping in the case of electricity. (Modern engineers would probably want to try to recover some of that lost heat, since it's fairly localized.) 71.212.248.104 (talk) 10:52, 10 June 2012 (UTC)[reply]
If you happen to have a convenient deep underground cavern with easy access, then that idea sounds workable, but you are limited by the size of the underground tank. Dbfirs11:29, 10 June 2012 (UTC)[reply]
Subjective experience and age
How does subjective experience--by which I mean consciousness--change with age? I'll give a few examples to indicate what I'm looking for. Children are easily excited, and their emotions are more extreme. Children are happier, sadder, and angrier than adults in happy, sad, and frustrating situations, respectively. Teenagers and people at mid-life tend to be more depressed. The young perceive time as passing much more slowly older people; one year seems like an eternity to a 6 year old, but passes quickly for a 70 year old.
Does anyone know of a more complete list? My examples above are mostly based on my experience, which is limited because I'm quite young and represent only 1 sample point. --140.180.5.169 (talk) 09:26, 10 June 2012 (UTC)[reply]
I don't know of a more complete list. Except for a couple of things which I'll note, I think you are wrong anyway. I have passed retirement age. I dissagree that children have stronger emotions - what changes is perception about what is sad, happy etc. To a child, not getting your favorite food today can be something to cry over. To an adult, it's unimportant. This is not about strength of emotion - its about intelligence and experience. Children know that by over-acting, they can persuade adults. Acting angry or sad is not being angry or sad, even if you convince yourself. Psychologists have written learned papers about depression in teenagers. Never the less, I think it's nonsense. Having been through it, depression when a teenager is not deep and is quickly overcome. What's different for teenagers and adults in recent years, is that doctors will more readily prescibe for it, and that is a bad thing. On the other hand, depression is recognised as real these days, and that's a good thing. You won't know what sadness is until you loose a loved one after years of being together.
One thing that DOES change is perception of size and distance. When I was about age 3 to 5, my mother and I used to go and spend a week now and then with grandfather at his house, in another city. When there, Grandfather would walk me to a shop and buy me an icecream. It seemed a long walk. We also used to go to the beach - another long walk. Grandfather died when I was 5. When I was 18, the company I worked for sent me to that city. I decided to retrace old steps. I was greatly surprised that the distance to the shop was 1 block! The distance to the beach was 1 block! Recently, my old primary school was opened as a tourist office. I had a look. I was suprised how small the rooms were.
You are correct that time passes more slowly for children.
Another thing that changes is your view of the general competence of adults. When you are a child, you think adults are 100% capable. When you are a late teenager, adults frustrate you whith their slow decision making and inability to learn new things quickly. When you get to retirement age, your perception changes - older people now seem competent, but teenagers seem rash and impatient. But, thinking about it objectively, those pesky teenagers are right!
When people get really old and start to loose their faculties, then the more deeper and more evolutionary early parts of the brain take over. Then, emotions can get rediculously strong. I had a relative who lived in an old-folk's home - the sort where nurses were on duty 24 hours a day. I was there once when the kitchen was 30 minutes late in serving lunch. A number of inmates staged a riot (in their wheelchairs, in slow motion)! You'd never get that at any other age, and you'd never get that in people who are mentally fit - they would just accept that sometimes things go wrong.
What does change as people get older is the rate at which emotions can change. In management training, I was taught that when given bad news, people go thru 5 stages of grief: disbelief, why me/anger, sadness, resignation/acceptance (sometimes bargaining), and finally, seeing an advantage. See http://en.wikipedia.org/wiki/5_stages_of_loss. Some psychologists don't accept this, but you get that. In my experience it is right - it has helped me a lot in counselling employees and family. Now, children can go thru all 5 stages in seconds. Teenagers can take longer. Middle aged folks can take months. Some folks remain stuck at one stage - that can happen with young adults, but not very often. I don't think children ever get stuck.
"One year seems like an eternity to a 6 year old, but passes quickly for a 70 year old" is something that everyone says, but personally, I don't think it's true. What is true as we age is that events we remember can be longer ago, and that any given length of time is a smaller fraction of our total experience. When people say "time goes by so fast these days!", they're always reminiscing about the past, not looking to the future; "1 year ago" doesn't seem like such a big deal now as it did when you were 6, but on the other hand "next week" will take just as long to get here now as it did then. FiggyBee (talk) 14:36, 10 June 2012 (UTC)[reply]
I would disagree with that point, the time until next week does grow shorter with age, the "are we there yet" question shows a slower movement of future time as well. The OP should look for information at Gerontology or in these journals. 65.95.22.197 (talk) 17:02, 10 June 2012 (UTC)[reply]
I have to agree with IP 65. Knowing you had to wait A WHOLE HOUR!!! for something as a kid was torture that took forever. Nowadays that's hardly even enough time to get ready for anything. μηδείς (talk) 18:00, 10 June 2012 (UTC)[reply]
I changed your heading from the non-informative "!!!!!!!!!!!". I guess you refer to something from 2012 phenomenon which says: "Scholars from various disciplines have dismissed the idea of such cataclysmic events occurring in 2012." PrimeHunter (talk) 12:31, 10 June 2012 (UTC)[reply]
From a scientific perspective, nothing unusual, besides an astronomical observation, should happen on 21 December. Don't believe every thing you hear. Plasmic Physics (talk) 13:38, 10 June 2012 (UTC)[reply]
See section 3 page 12 and further of this article. The universe is making a transition to a state of exact supersymmetry, we are now in the false vacuum state, after the transition we'll be in the real vacuum. A long time ago, a bubble containing the exact supersymmetric vacuum nucleated and started to expand at the speed of light. We're now just a quarter of a light year from the edge of that bubble. When we cross the boundary of that bubble, a huge amount of energy will be released killing all of us in a fraction of a second.
However, since we all have an infinite number of identical copies, the closestone located about about 10^{10^29} meters from here see here, and they may be further away from the edge of the expanding bubble that exist there, they may survive into next year. Because these copies are identical to you, you will actually survive as your copy without having any knowledge about being killed here. So, it will look like this was all a doomsday myth. Count Iblis (talk) 16:29, 10 June 2012 (UTC)[reply]
How would Uranus have been pronounced in the 1780s / 1790s
Johann Elert Bode suggested the name Uranus. However, the Wikipedia (and many other places) always refer to this as the 'Latinized name'. I assume this is intended to distinguish it from the Greek pronunciation of the name. (Something like 'oo-RA-nos') My question is, how would an english speaker in the 1780s/1790s have pronounced the latinized version of the name Uranus? --CGPGrey (talk) 15:22, 10 June 2012 (UTC)[reply]
Not a definitive answer, but Greek οὐρανός would have been pronounced, in English, with the stress on the first syllable (OO-rah-nohs), since the vowels in the last two syllables are short. I assume that the Latinized version would have been pronounced something like YOO-ruh-nuhs if the normal rules for such things were being followed. Deor (talk) 17:41, 10 June 2012 (UTC)[reply]
We don't have a speech recording from the 18th century. Those who spoke of the planet would likely have known enough Greek to argue about how the ancient Greeks pronounced Ouranos. Since both the English words "anus" and "your" were current at the time I'm sure someone could have thought of a joke that would have been for once original. Anyone embarassed by that hilarity might persist in calling the planet Georgium Sidus (King George III's star), as did British Victorian nautical almanacs as late as 1850. 84.209.89.214 (talk) 17:55, 10 June 2012 (UTC)[reply]
From the linked report one can gather only that someone (who?) has hailed a schoolboy a genius for "cracking puzzles (what?) that have baffled the world of maths (who?) for 350 years. I see no more to support this claim than an article title translated as "Analytical solution of two unsolved problems in fundamental particle dynamics" on a PC screen, and the utterly non-newsworthy information that "scientists can now calculate the flight path of a thrown ball and then predict how it will hit and bounce off a wall". Poorly educated journalists should keep away from subjects that they are not equipped to understand. 84.209.89.214 (talk) 17:38, 10 June 2012 (UTC)[reply]
Isaac Newton obtained his law of gravitation in which gravity works over unlimited distance, by observation and experiments that are readily demonstrated on laboratory scale. The law also allows planet movements to be explained. We cannot explain fully how gravity works and in extreme cases, such as over great distances and velocities aproaching the speed of light, Newton's classical description must be modified by Einstein's geometric theory of curved spacetime. We can provide these references but cannot give you proof of what appears to be a universal Physical constant. 84.209.89.214 (talk) 17:15, 10 June 2012 (UTC)[reply]
Physics numerical
uniform electric and magnetic fields with strength E and B are directed along the y axis
a particle with specific charge q/m leaves the origin in direction of x- axis with an initial velocity Vo
Find
a) the co ordinate of the particle when it crosses the y- axis for nth time
b) the angle alpha between the particle velocity vector and the y-axis at that moment — Preceding unsigned comment added by 101.214.105.47 (talk) 17:19, 10 June 2012 (UTC)[reply]
Welcome to the Wikipedia Reference Desk. Your question appears to be a homework question. I apologize if this is a misinterpretation, but it is our aim here not to do people's homework for them, but to merely aid them in doing it themselves. Letting someone else do your homework does not help you learn nearly as much as doing it yourself. Please attempt to solve the problem or answer the question yourself first. If you need help with a specific part of your homework, feel free to tell us where you are stuck and ask for help. If you need help grasping the concept of a problem, by all means let us know.
Shelled Pistachios lose their flavor?
On Amazon.com there's two types of pistachios they sell, shelled and unshelled.
But the shelled pistachios have significantly worse reviews, with most people saying it tastes worse, and even rancid. They are both made by the same company, so I'm wondering if shelling a pistachio would have a real impact on its flavor, and if so, why? ScienceApe (talk) 17:53, 10 June 2012 (UTC)[reply]
I vaguely remember from college that taking excessive amounts of some vitamins and minerals isn't a problem as they can be easily broken down and excreted by the body while others build up and cause damage. If I was to take double/triple the RDA of all the main vitamins and minerals what would be the first symptoms (or should I say which would be most toxic)? Thanks 87.115.195.33 (talk) 19:20, 10 June 2012 (UTC)[reply]
Vitamin A toxicity, see Hypervitaminosis A. The dose that can cause toxicity is the lowest above the RDA for all vitamins and minerals. This may also be due to the RDA being too low for other fat soluble vitamins e.g. vitamin D. The minimum dose of vitamin D that is thought to be capable of producing toxicity is about 40,000 IU/day for many weeks (although the lowest dose that has been observed to actually produce toxicity is 77,000 IU/day). The RDA is just 800 IU/day. However, it may be that the natural dose of vitamin D is 10,000 IU/day as that's what we would get from the Sun were we to spend a lot of time in the Sun at Noon time in the tropics. Count Iblis (talk) 19:43, 10 June 2012 (UTC)[reply]
is a twelve minute mile impressive
I know there was a time that the four-minute mile was record-setting, so I'm wondering if a twelve-minute mile is considered impressive IF the person weighs the same as three average-sized people (or if the average weight is obese today, then three averaged-sized people from 50 year ago) for that height and age? This is not a trollish question, if yout think it's bad just answer it matter-of-factly okay. --80.99.254.208 (talk) 20:07, 10 June 2012 (UTC)[reply]
The world record for a mile has dropped a fair bit below 4 minutes now, see Mile run world record progression. For info an average human Walking speed is about 3 miles per hour (a mile every 20 minutes) so a 12 minute mile (5mph) is not quite twice as fast as normal walking pace. Read into that what you will. Over a short distance twice walking pace is nothing, over a mile it does become a question of endurance so I suppose it depends on who you ask...I'm 'relatively' sporty and i'd like to think I can do a faster than 12 minute mile but i've honestly never tried...Maybe I will tomorrow and then reply with my time...might end up embarrassing myself though! ny156uk (talk) 20:43, 10 June 2012 (UTC)[reply]
Yes, in my younger days, I used to walk a mile in 12 minutes (occasionally overtaking slow joggers), but for someone who is three times normal weight, I'd say "well done" and keep up the exercise! Dbfirs21:25, 10 June 2012 (UTC)[reply]
Does the Plastic to Oil Machine really work?
I am curious about Japanese inventor Akinori Ito and his invention that turns
plastic bags into oil.
What can you do with the oil product once produced?
I have closed this discussion, per policy. As Roger said, consult somebody qualified to adress this. I wish you the best of luck in resolving this. Falconusptc20:48, 10 June 2012 (UTC)[reply]
I use a pharmacy solution of salicylic acid and lactic acid. It works wonders for both warts and moles. In Australia it is sold as Duofilm, made in Ireland. Plasmic Physics (talk) 22:33, 10 June 2012 (UTC)[reply]
Photoelectric threshold frequency
I'm trying to design sunglasses that convert UV light into solar energy. Is there a cheap material with a photoelectric threshold frequency in the blue-violet (or near-ultraviolet) range? 68.173.113.106 (talk) 20:57, 10 June 2012 (UTC)[reply]
If each sense has its own cortex (primary visual, primary auditory, etc.) what about the "others" such as equilibrioception, feeling the fullness of one's bladder, etc.? Is there such thing as the "primary equilibrioceptive cortex (EQ1)" ?
Egg Centric 00:34, 7 June 2012 (UTC)
— Preceding unsigned comment added by Mitch the amateur scientist (talk • contribs) 21:18, 7 June 2012 (UTC)
Resolved
Please do your own homework.
