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.
I believe that last time, you were refered to the Bristol Stool Scale and informed that feces is far too variable a substance to come up with a single numerical answer for the density of stool. It will vary wildly from person to person, and from sample to sample for even the same person. --Jayron3200:23, 27 September 2011 (UTC)[reply]
Jayron is correct (as usual). If all faeces stools were the same density we would not have developed the "sinker" and "floater" terms! These two words validate what Jayron is saying. Richard Avery (talk) 06:19, 27 September 2011 (UTC)[reply]
Is there any blender currently on the market that can successfully blend a crowbar without breaking? This is related to the Will It Blend? YouTube vids. People keep requesting that he do a crowbar, but he keeps finding ways to avoid doing a crowbar.
Also, would it actually be possible to blend diamonds? I know that he blended cubic zirconia in one vid when people were requesting diamonds (due to the cost), but those are less hard. --Kurt Shaped Box (talk) 01:05, 27 September 2011 (UTC)[reply]
The blades of a blender cannot slice or cut a diamond, but if it strong enough, it can smash the diamonds into dust. While diamonds are hard to scratch or cut, they are quite fragile, they can be smashed using a hammer. Plasmic Physics (talk) 03:16, 27 September 2011 (UTC)[reply]
"Will it Blend" is an advertisement, I think it's safe to say that if that brand of blender were capable of it, he would do it.
But seriously, a crowbar would be tough to get through quickly with anything short of a plasma torch. (And if a blender's blades can't make it through the thing in one go, it'll just jam.) APL (talk) 07:14, 27 September 2011 (UTC)[reply]
I thought that there might actually be a video on YouTube of someone trying to blend a crowbar using a BlendTec blender (and failing), just to diss Tom. Not so. Yet. --Kurt Shaped Box (talk) 11:59, 27 September 2011 (UTC)[reply]
Hypothetically-speaking, how would one go about designing a blender capable of slicing up a crowbar? I'm just picturing the infomercial now - the pitchman being all like "Look! It will even blend a crowbar and still make guacamole!". Billy Mays would have approved. --Kurt Shaped Box (talk) 23:32, 27 September 2011 (UTC)[reply]
Well, it would need to be huge, with blades far bigger than a crowbar (imagine a battleship's prop), and the blades would need to be coated with something hard enough to cut through the steel in the crowbar (corundum would be fine, no need for diamond), and the blades would need to spin at a hellacious speed, which would require a huge amount of energy. StuRat (talk) 00:37, 28 September 2011 (UTC)[reply]
Machines which grind off pavement have steel teeth with carbide cutting tips welded on. The same trick is used on carbide circular saw blades. It is a very old trick. Archeologists found an axe from (if I recall correctly) the dark ages which had a hard but brittle steel welded (by heating and hammering) onto a softer iron base, so that it could maintain a sharp cutting edge but would not shatter as easily as the harder and more brittle edge material. Industrial metal shredders are close to what is desired, shredding thick metal castings, but no indication they would handle a crowbar: [1]. I have seen shredders at scrap metal recycling companies which could handle an automobile, including the frame and axles, and I expect a crowbar wouldn't even be noticed. The cutting part shown at 2:20 has a rotating cylinder with long teeth sticking out which basically shears the metal into little pieces. That video is a toy sized operation compared to some. Others have a pair if meshing cutter arrays:[2]. Here they shred steel rebar ans structural steel such as angle. Seems like such a company could build a "blender" with a vertical hopper and a spinning vertical shaft with teeth projecting from it. From captions on some of the videos, it would likely need couple hundred kw motor. Edison (talk) 15:26, 28 September 2011 (UTC)[reply]
Those videos are awesome. I'm sooo going to be watching those 'Shred of the Month' vids tonight. It's like a document shredder - but on steroids (anyone else here love feeding the document shredder?). Also, car shredder redirect created... --Kurt Shaped Box (talk) 18:24, 28 September 2011 (UTC)[reply]
Homopolar generator
If a conductor were buried at ether the North or the South pole at the rotational axis of the Earth and another at the equator and an insulated wire laid so that a meter could measure any voltage or current between them what would the voltage and current be? --DeeperQA (talk) 01:21, 27 September 2011 (UTC)[reply]
The question is how much power would be generated if such a scheme did in fact work. Could enough power be generated to solve the energy crisis? --DeeperQA (talk) 03:17, 27 September 2011 (UTC)[reply]
It (voltage and current) depends on how much work (watts) you try to extract from it - which in turn is related to the (reaction) force being exerted on the wire due to the motion of the earth - the general equation is given at the top of Lorentz force. You need the magnetic field at the earths surface, and the angle it makes with the surface of the earth (Magnetic dip ?)Imgaril (talk) 11:00, 27 September 2011 (UTC)[reply]
To a first order of approximation, the Earth's magnetic field is fixed in magnitude and rotates along with the circuit, so there's no time-varying change in magnetic flux through the circuit, which means no current is induced. However, there can be geomagnetically induced current due to geomagnetic storms. However, GIC isn't steady or reliable, and hence would make a poor energy supply. Red Act (talk) 16:41, 27 September 2011 (UTC)[reply]
What if you forget about the equator and build some superconducting hexagons over the north magnetic pole? As it moves you sap the energy from one hexagon, dismantle it, and move it around in the direction the pole is going. How much energy would that produce? Is it possible to trap the pole and keep it from moving with enough superconducting hexagons cutting across the field lines - or even deliberately give it freedom to move only in a direction you want? (Note: this is theory only - real superconductors have limits to the current they can carry, which I think in practice would severely limit the amount of power extracted) Wnt (talk) 18:51, 27 September 2011 (UTC)[reply]
Some responders have been too quick to dismiss the question. When a solar flare reaches the Earth, utilities sometimes experience hundreds of volts and hundreds of amperes through neutral conductors when the transformer neutral is grounded at each end of the line. This is seen in lines under a hundred kilometers, at various orientations. There are also earth currents of electricity of varying magnitude, due to different points of the Earth';'s surface being at varying potentials. They were studied by Sir Humphrey Davy, Becquerel and others from the beginning of the 19th century. W.H. Barlow in the 1840's found them strongest in a N-S orientation. They vary during the day and during the year, and are influenced by electrical storms, and relate to the auroras, and to sunspots.. They were sometimes large enough to interfere with telegraph operations. They generally prevented telegraphy from August 29-September 4, 1859. It produced 800 volts on a 600 km line in France. It produced an effect near Boston equal to 200 Grove cells, or about 380 volts. Telegraph lines could in fact be worked without batteries, powered only by the earth currents. In Europe that particular storm had its strongest effect in a NE-SW direction, but would have likely had a huge effect from pole to equator. In 1881 a similar storm produced 1.1 volt per km in England. You could take 1.1 volts/km times 10,000 km from the pole to the equator as a representative value when conditions are favorable, and expect 11,000 volts between the ends of the line. If the ionospheric currents which induce the earth currents have more localized circulation (on the scale of 1000 km,) then over such a long line there could be some cancellation. As a result, a somewhat longer or shorter line might have a higher end-to end voltage. A telegraph line in the 19th century had appreciable resistance, which held the current to .28 amperes in one report from the last work cited, in a 200 mile line from New York to Providence, corresponding to 644 volts. The ground resistance at the terminals and the resistance per unit length of the conductor would be the limiting factors holding down the current.Today researchers use the term "Telluric current"in place of the older "earth current," to indicate that current in seawater is included in the topic. The latter source, from the 1980's indicated that a 2650 volt potential had been measured along a transatlantic cable in the 1950's. It also notes (p245) that in 1921 a geomagnetic storm produced 1kv over a 100 to 200 km line, with 2.5 amperes of current available (probably one or the other conditions would apply: high voltage at zero current, or high current in a short circuit condition.) The source discusses telluric problems on long power lines, including 100 amps of telluric current in 1972, sufficient to damage power transformers. Edison (talk) 19:12, 27 September 2011 (UTC)[reply]
Red Act's answer above is incorrect. It's such a common misconception that it has a name--the Faraday paradox--that is also discussed in the homopolar generator article. The first error is saying that the Earth's magnetic field 'rotates with the planet'. Since the field is circularly symmetrical, there is no way of knowing whether it's rotating or not; in fact, 'rotates' has no physical meaning in this case. You can't paint a mark on the field so that you can watch it going round. The error may be caused by people imagining that the field is a bunch of field lines rotating, when in fact field lines are a drawing aid and don't actually exist. The field is really a continuum. The second error is the belief that the generator works by the relative motion of the rotor and the field: it does not. What matters is the relative motion of the rotor and the pickup wire, both of which are within a constant (unipolar) magnetic field. I'll stop here because half of you probably don't believe me, but I leave you to read the relevant articles. --Heron (talk) 19:37, 27 September 2011 (UTC)[reply]
Are you saying that the difference in potential between the center of the copper disk and the edges of the copper disk is such that a wire attached to the center of the disk will "drain" that potential at each point on the edge of the disk that it touches and that continuous flow then depends upon the wire touching new points on the edge of the copper disk? --DeeperQA (talk) 19:57, 27 September 2011 (UTC)[reply]
Not quite. The potential (I think it could be called electromotive force or EMF) does not exist until the disk rotates. It is created by the rotation of the disk relative to the collector. Imagine that you cut away almost all of the disk leaving just a narrow wedge between the axle and the collector. Now you rotate that wedge through say 1°. From the start to the end of that rotation, the geometry of the electric circuit has changed slightly. Before the rotation, let's say that the axle, wedge, collector and external wiring were all in a single plane. After the rotation, the circuit is no longer in a single plane, because the wedge is now out of the plane, and the current has to return through a tiny arc of the collector and back into the original plane. This is equivalent to tilting the plane of the original circuit slightly. The tilted plane is now not quite parallel to the magnetic field lines, so it has gone from having none of the flux passing through it to having a tiny bit passing through it, and that is how it picks up an EMF by electromagnetic induction. Now, if the collector circuit were rotating with the disk, none of this plane-tilting would happen and there would be no EMF. This is all hand-waving stuff and probably sounds oversimplified to a proper physicist (I'm not one). The real answer is due to the Lorentz force on electrons, but I can't remember how that works so I've gone for the geometrical explanation instead. I believe you need to use special relativity to explain the Lorentz force properly. --Heron (talk) 20:44, 27 September 2011 (UTC)[reply]
I follow you up to the point of the plane. In my imagination this plane is horizontal and perpendicular to the axis of the disk which is vertical. To follow what you are saying though it seems I am using the wrong orientation for the plane. The plane you are referring to seems to be in line with and parallel to the vertical axis on which the disk is rotating such as is a knife used to cut a piece or wedge of pie. Is the plane you are referring to in line with the axis of the disk or in line with the disk? --DeeperQA (talk) 01:03, 28 September 2011 (UTC)[reply]
The plane is in line with the axis and perpendicular to the disk. If the axis is vertical then so is the plane. Therefore the magnetic field vector is parallel to the plane and does not pass through the circuit. If the plane tilts slightly then a small amount of flux will pass through it. This change in flux through a loop, according to classical electromagnetics, is what induces the current. --Heron (talk) 13:20, 28 September 2011 (UTC)[reply]
So you are saying that because the disk is turning for a moment in time the circuit moves with it and becomes slightly tilted to the plane such that opportunity is presented to the magnetic flux to pass through it? --DeeperQA (talk) 19:54, 28 September 2011 (UTC)[reply]
So I can leave the disk and the magnet stationary and just move the wire around the perimeter of the disk to generate electricity? Homopolar Generator--DeeperQA (talk) 22:30, 28 September 2011 (UTC)[reply]
What's the best way (if any) to store and/or transmit sensitive scientific data in such a manner that said data would be immune to tampering (or nearly so) during storage/transmission? If such a method exists, would it be possible to transmit the data in real time from an aircraft in flight over water? 67.169.177.176 (talk) 03:47, 27 September 2011 (UTC)[reply]
One more thing: If the plane's RNAV system is integrated with the scientific instruments and the recording/transmitting computer, should the navigator/scientist enter his private key before or after aligning the RNAV? (I'm guessing before, but I need to be sure.) 67.169.177.176 (talk) 04:52, 27 September 2011 (UTC)[reply]
An aircraft's flight-critical or flight-essential systems (such as RNAV) should not be integrated with a mission payload. While they might reside on the same computing system, their operation should be independent such that no misconfiguration of mission equipment impacts the safety of the aircraft. — Lomn13:08, 27 September 2011 (UTC)[reply]
Assuming, though, that we're talking appropriate levels of independence, and that the mission payload is pulling some read-only data from RNAV -- I'd have to say, in the very general case, that you always pull data after initialization. It's not at all clear, though, what RNAV data you'd be pulling (a flight plan?) or what "aligning" RNAV would mean, particularly in the context of cryptography. — Lomn14:01, 27 September 2011 (UTC)[reply]
This is correct -- the RNAV feeds position data (including groundspeed/drift angle) to the mission computer, which then uses it to correlate the scientific data (temperature, humidity, solar radiation, levels of CO2 and trace gases, etc.) to the plane's position. So in case of a computer crash, the scientific data is wiped out, but the RNAV system is still operational. As for "aligning" the RNAV system, this involves spinning up the gyros, torquing the accelerometers, and then feeding the plane's current position into the system; this is done before each flight, and is the navigator's responsibility. 67.169.177.176 (talk) 00:07, 28 September 2011 (UTC)[reply]
Inertial nav systems -- gyros, accelerometers, and the like -- are not permissible for primary RNAV inputs, as they cannot maintain appropriate NSE. Note also that you're talking positional data, not navigational data. It appears that you ought to just pull from a GPS box. — Lomn12:33, 28 September 2011 (UTC)[reply]
The advantage of the inertial navigation system (supplemented by a Doppler system), in this context, is that it provides a continuous readout of the plane's acceleration, which in turn allows the computer to calculate the direction and strength of air currents -- a valuable piece of information in the context of studying climate change. GPS cannot give this information in the raw form like inertial and Doppler can. So even if inertial won't be needed for navigation (most of the time, anyway), it will still be used as a scientific instrument. 67.169.177.176 (talk) 05:16, 29 September 2011 (UTC)[reply]
To detect deliberate tampering (i.e., tell legitimate data from bad) you need either a message authentication code or a digital signature. Digital signatures are more expensive to calculate but have some advantages that may or may not matter in your application. To protect against tampering (i.e., ensure that all legitimate data gets through) is impossible in general because the attacker can always just block the signal completely. Error correcting codes might help in certain situations. I can't answer your followup question because I know zilch about aircraft control systems. -- BenRG (talk) 08:00, 27 September 2011 (UTC)[reply]
In this case, I am not concerned about the signal getting completely blocked; my main concern is how to prevent the data from being fraudulently altered during storage or transmission. FYI, the aircraft in question is performing a flight around the equator as part of a mission to study climate change (as well as being a memorial flight for Amelia Earhart) -- so the main concern in this case (as far the the data is concerned) is to prevent either the crew or the scientists from altering the data at will for political purposes, just in case they have a mind to. 67.169.177.176 (talk) 00:07, 28 September 2011 (UTC)[reply]
It is possible to prevent tampering, rather than just detect it, by using narrow beam transmission and directional antennae at both ends, say to and from a series of satellites. The only way to block such a transmission would be to physically place the jamming signal generator between the plane and satellites, or send a massive EMP to overwhelm the signal. StuRat (talk) 00:22, 28 September 2011 (UTC)[reply]
Thanks for the ideas, everyone. Now, how big is a typical narrow-beam directional antenna? Please understand that it would have to fit into an Electra 10-E, which is not only pretty small to start with, but already crammed full of extra fuel tanks (which take up the forward 1/3 of the cabin), navigational equipment, scientific instruments, etc., etc... 67.169.177.176 (talk) 06:13, 28 September 2011 (UTC)[reply]
Question about the Galileo Statement
Galileo theorized that in the absence of air, all things would truly fall with the same acceleration and 300 years later demonstrated this by the crew of Apollo-15 on the Moon (which has gravity but lacks air) by dropping a hammer and a feather.
As moon was seen from both the feather as well as hammer with two different gravitational fields [g] therefore just wondering what was the falling acceleration of moon towards aforementioned feather and hammer from the following possibilities [if i'm not wrong]?
1- Moon had the gravitational acceleration of hammer i.e "g" of hammer
2- Moon had the gravitational acceleration of feather i.e "g" of feather
3- Moon had the net "g" of feather and hammer
So, is Galileo's statement correct [Theoretically] if the senario is considered in the absence of all other gravitational attraction.?68.147.43.159 (talk) 03:52, 27 September 2011 (UTC)Eccentric Khattak#1[reply]
The force on the moon will be the gravitational attraction from the hammer plus the gravitational attraction from the feather. The moon will accelerate towards the centre of gravity of the (hammer+feather) system, a point which will be much closer to the hammer than the feather: so the moon will move very slightly in the direction of the hammer. (The hammer and feather will also accelerate very slightly towards each other, but this tiny movement won't significantly affect the moon's trajectory.)
Since F = (G m1 m2)/r, and a=F/m the total acceleration on the moon will be G (mfeather+mhammer)/r, or with a 1kg hammer and 10g feather about 4x10-17 m s-2 (by my calculations). If you drop them from one metre up, the moon will move something of the order of the diameter of a proton. The moon's surface will be much bumpier than this (plus there will be stray gas and dust molecules on the way down), and its gravitational field is non-uniform (because it's not a perfectly homogenous sphere) so in theory the moon will hit the hammer first, but in practice the difference in gravitational attraction will be unimportant to the outcome. --Colapeninsula (talk) 10:40, 27 September 2011 (UTC)[reply]
Is it correct to say that even if force experienced by hammer is more due to its more mass, acceleration gained by the hammer is same as that of feather because hammer has more inertia due to its more mass? - 61.16.182.2 (talk) 04:03, 28 September 2011 (UTC)[reply]
I just noticed that the ratio for men's and women's world records in the marathon is 1:1.0953 and that of the 100 meter sprint is 1:1.0959. I find it fascinating that they are currently equivalent to the thousandth decimal. Is this a coincidence or is this performance ratio present is other events? The Masked Booby (talk) 03:55, 27 September 2011 (UTC)[reply]
Just crunching numbers for a couple other events, I got a ratio of 1.1215 for the 800m, 1.1319 for the mile, and a ratio of 1.1893 for the high jump, so it looks like a coincidence. Googlemeister (talk) 13:46, 27 September 2011 (UTC)[reply]
Bringing back an old topic. In hyper valent molecules, is it correct to say that the hypervalent atom only contributes to some of the orbitals, meaning that the remaining ligands effectively borrow the otherwise LUMO orbitals to complete their valencies? Take pentachloro-λ5-phosphane or pentachloridophosphorus as an example. According to my theory, the phosphorus only has a total bond order of 4, despite having 5 ligands. The three equatorial bonds have a bond order of one each, the orthogonal bonds have a bond order of a half each. To make this arrangement work, the two half order bonds will have to be spin paired. Essentially, phosphorus does not contribute to the HOMO. This would should result in a triagonal bipyramidal molecule with a 1+ charge on two chlorines each, and a 2- charge on the phosphorus. The two positively charged chlorines, will move to opposite locations to minimise electrostatic repulsion. The remaining chlorines will move to equatorial positions. Because half bond order bonds are lower in energy than 1 bond order bonds, they are longer. As a result, the orthogonal bonds should be longer than the equatorial bonds, which is what is observed. This is my own idea, the three-centre-four-electron article doesn't make much sense to me, so I synthesised this one. I call it orbital borrowing, since the ligands borrow an orbital(s) from the host, without the host contributing. Another thing to note, is that this orbital borrowing should only be a stable system for molecules where the gap energy between the HOMO and LUMO orbitals of the host is small. So orbital borrowing shouldn't be happening for 1st period elements, except under extreme conditions. is orbital borrowing correct, and is it a useful description? Plasmic Physics (talk) 11:56, 27 September 2011 (UTC)[reply]
Orbital "borrowing" is a useful heuristic to describe what happens in either hybridization theory or molecular orbital theory. However, it is important to remember that in MO theory, the idea is to describe all orbitals of the molecule as belonging to the molecule as a whole; without giving actual care as to where they "came" from. When a molecular orbital has a character which resembles what would be expected from the mathematical combination of two atomic orbitals, we can say it "formed" from those orbitals; likewise if a molecular orbital is largely identical to the atomic orbital in an unbonded atom, we can say that the orbital "did not participate in the bonding" of the molecule; but these are heuristic approximations. Strict MO theory actually treats all electrons (and thus all orbitals) as belonging to the molecule as a whole and describes the orbital space around the entire molecule, and does not pretend to assign electrons or orbitals as "originating" at any particular atom. MO diagrams, in this way, present a good heuristic for predicting what the MO structure is going to look like; but to be scrupulously correct, actual molecular orbitals are described purely by the wavefunction of the molecule in exactly the same manner as the atomic orbitals are described purely by the wavefunction of the atom. And you've also gotten way past the level of mathematics I have experience with, so I'm not sure I can get more detailed than that. The concept of "hypervalency" is a way to jibe the mathematical results of the wavefunctions with the heuristic predictions of the simpler models, like the "octet rule" and Lewis/Valence bond theory. In actuality, MO theory doesn't treat "hypervalent" molecules as a special case; they just are what they are. --Jayron3216:55, 27 September 2011 (UTC)[reply]
If I combine this theory with fractional bond theory, I came up with a new Lewis diagram for sulfuric acid. Two single bonds connecting the hydroxyl groups, and two double dashed bonds connecting the oxo groups. A double dashed bond is a description of two fractional bonds, where the electrons are not spin paired. A similiar bond exists in triplet oxygen, a single and double dashed bond combination. In this case, it represents a four electron bond, where only two electrons are spin paired. So, a double dashed bond which indicated a double half-bond in the case of sulfuric acid, should have a higher energy than a single bond, but less than a double bond. It should be, but is it? This revised Lewis model predicts a longer bond length for the oxo-sulfur bond than the older version, is the prediction more accurate? Plasmic Physics (talk) 13:52, 30 September 2011 (UTC)[reply]
Love hurts
Our articles on limerence and being lovestruck mention chest pain as a "symptom". Having experienced this on more than one occasion, I know that it's not made up. My question is, what's the physiological basis/mechanism for this pain? What's making my chest hurt when I'm in love/lovesick? ElMa-sa (talk) 12:53, 27 September 2011 (UTC)[reply]
This is OR, but my impression is that the ache actually arises from the solar plexus, not the heart. In fact, when people have actual damage to the heart, the pain sensation is frequently referred to other body parts, such as a shoulder. Looie496 (talk) 14:27, 27 September 2011 (UTC)[reply]
That article focuses on the worst cases, but in general, stress weakens the heart, allowing for some ballooning to take place, which causes chest pain. There are many kinds of stress and this particular kind tends to follow deeply emotional events - which is why it is referred to as broken heart syndrome. -- kainaw™14:45, 27 September 2011 (UTC)[reply]
I scanned over the news-website you linked; noticed strong claims; and so as I am typically inclined to do, I searched for the original paper. The article didn't mention where it was published, so I turned to Google Scholar search. All I turned up is ... nothing. It appears User:Edison also searched for an original publication back in April and independently reached the same conclusion: who is "Peter Braun"? Where does he work? Where did he publish his research? Absent such clarifications, we can't really evaluate the claims. This is a case of serious shortcoming of internet "journalism." The original "newspaper report" does not actually answer any of these questions. I think we should probably call "citation needed" on that Telegraph article. Nimur (talk) 16:55, 27 September 2011 (UTC)[reply]
Have you looked at the amount of mercury involved? Yes, it is technically correct to say that a CFL contains mercury, but the amount involved is minuscule. For comparison, the amount of mercury released into the environment by smashing a CFL is far less than that released by burning coal to power an incandescent bulb for the 10000 hours or so that a CFL will last. --Carnildo (talk) 01:11, 28 September 2011 (UTC)[reply]
Yes, because of the soldering. That mercury and lead and tin and zinc are all considered hazardous materials these days is just a ploy to make you get rid of those for free so they can recycle them. The official reasoning is usually just some chemical compounds containing those metals that are actually harmful on your health, but you can find harmful compounds of any element; or allergic reactions, but you can find allergy and skin irritation for any material as well. I wonder which classical metal will be turned dangerous next: iron, copper, silver, or gold? – b_jonas19:40, 28 September 2011 (UTC)[reply]
Solder is less of an issue than it used to be. Industry is moving toward lead-free solders, because products with lead solder can't be sold in the E.U.--Srleffler (talk) 18:53, 30 September 2011 (UTC)[reply]
They're considered toxic waste because when you put thousands and thousands of them into a landfill, that little amount of mercury becomes a significant environmental problem. It's not because any individual bulb is that dangerous. It's the aggregate that's the issue, and it's an environmental issue more than a "you are going to die" sort of issue. --Mr.98 (talk) 11:58, 29 September 2011 (UTC)[reply]
I never understood why people accept that argument. It's like saying that an open fire without a chimney in your house would be better than electric heating because it produces less CO2. Outside air has 6 to 30 nanogram mercury per cubic meter, while tests with broken bulbs give: Mercury concentration in the study room air often exceeds the Maine Ambient Air Guideline (MAAG) of 300 nanograms per cubic meter (ng/m3) for some period of time, with short excursions over 25,000 ng/m3, sometimes over 50,000 ng/m3, and possibly over 100,000 ng/m3 from the breakage of a single compact fluorescent lamp.http://www.maine.gov/dep/rwm/homeowner/cflreport.htmDS Belgium (talk) 02:18, 1 October 2011 (UTC)[reply]
Emission of carcinogens at switch on
When electronics heat up it will give of carcinogens that are normally in solid state. Small amount, but the process exist.
Skin is affected by ultraviolet radiation
Affected yes, in any meaningful way.. NO
Eye irritation, though new bulbs are widely claimed to have no perceptible flicker
If 50-60 Hz then maybe, but most CFLs are in the > kHz range asfair. A more substantial risk is from energy rich UV light because the "white" light is produced by florescent material being irradiated by UV internally and some leaks through! see High-energy visible light. There might be one answer to the mechanism thoe, energy pulsed at proper time intervalls might interfere with the cellular ion-pumps.
Pollution to environment by mercury
The idea here is that electric power is generated by burning coal which contains mercury. And that the CFL saves enough to make up for their own mercury content. This assumption is invalid where nuclear or hydropower is significant.
If it did, it would have reduced the nitrogen atom from the +5 oxidation state to the +4 oxidation state, which would mean that something else would have had to have been likewise oxidized. Look at the oxidation states of all of the elements in your mixture, and see if any is likely to be oxidized. --Jayron3218:23, 27 September 2011 (UTC)[reply]
From personal experience, no. Potassium nitrate and excess sulfuric acid is reacted to produce nitric acid, which can be disilled off as an azeotrope. No brown gas is evolved, the nitrate ion is preserved in the substitution reaction. Plasmic Physics (talk) 01:28, 28 September 2011 (UTC)[reply]
Generally no - as described above - however the nitric acid produced will decompose (slowly) in the presence of light - liberating small amounts of NO2 Imgaril (talk) 00:35, 30 September 2011 (UTC)[reply]
Depends on whether you heat it or not. If heating it to make HNO3 at atmospheric pressure, you'll get RED fuming nitric acid (10 to 30% N2O4, and the fumes are spectacular when using improvised methods. You can avoid this by using vacuum distillation. DS Belgium (talk) 02:38, 1 October 2011 (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. --Sean17:55, 27 September 2011 (UTC)[reply]
It does not sound like it is necessarily a homework problem. A sensible homework problem would be more likely to ask how the color changes when the compound reacts with something. It is not a violation of the rules to just answer a question here with a referenced answer. I see lots of sites with discussion of similar sounding compounds; maybe there are various nomenclature conventions. Is "iron(II) thiocyanate", also discussed here as an indicator, different from "iron (II) isothiocyanate?" The latter site discusses Fe(SCN)x compounds; are these different from the OP's "CNS" compounds? Edison (talk) 18:40, 27 September 2011 (UTC)[reply]
Thiocyanate is an ambidentate ligand (I took freshman chemistry with the guy who discovered this general property in organometallic chemistry). Thus, SCN is different from NCS in that SCN bonds to the iron at the sulfur, while NCS bonds at the nitrogen. I just noticed that the OP is asking about CNS, which I am unfamiliar with, but doing a quick search indicates that CNS is an alternate way of notating the thiocyanate ion (I always use SCN), apparently used in applications where the actual bonding is irrelevent or unknown. --Jayron3218:48, 27 September 2011 (UTC)[reply]
I remember a ton of similar problems when I took inorganic chemistry, based on ligand field theory. Different geometries and numbers of ligands, and different orbital energies of the ligands and different oxidation states of the metal all play a role in predicting the energies of the transitions among electronic states. That's the same as saying "what are the UV/vis spectral characteristics?", which is really just "what color is it?". One could even analyze this pair based on HSAB (different iron charge and effect of added "S" have on the nitrile ligand)? That latter is sort of like Edison's idea, except the "reaction" is the removal/addition/electronic-tuning of the ligands on one to get to the other:) DMacks (talk) 02:18, 28 September 2011 (UTC)[reply]
Anecdotally, I know several people who have adopted small dogs from shelters, and many such small "rescued" dogs have a pathological fear and hate of strange men, maybe due to their life experiences, or maybe because they're nuts. I've known several dogs whose owners got tranquilizers prescribed for the animals to calm them down so they don't bounce off the walls. I have known several nutty cats which go berserk and start scratching or biting for reasons known only to themselves. Maybe behavior which seems nutty to us makes perfect sense to the animal. Edison (talk) 19:51, 27 September 2011 (UTC)[reply]
Most of the cases of 'animal mental illness' I've come across inevitably have the animal kept under the close auspices of, you guessed it, humans. Cf. dogs, zoo animals. Nietzsche wrote a little about how he thought animals would regard humans as the 'sick animal, the laughing animal'. Vranak (talk) 00:56, 28 September 2011 (UTC)[reply]
Here's the full quote:
I fear animals regard man as a creature of their own kind which has in a highly dangerous fashion lost its healthy animal reason – as the mad animal, as the laughing animal, as the weeping animal, as the unhappy animal. The Gay Science. Vranak (talk)
Are you kidding? The only dogs I have known that did not go into kill mode when hearing or seeing a person riding a bicycle are those who have been trained and conditioned by a loving owner to accept such persons as friends. Dogs come into this world as completely neurotic killing machines until guided by their owner with love through the most normal of human activities and situations. --DeeperQA (talk) 03:19, 28 September 2011 (UTC)[reply]
What you call neurotic Cesare Millan might called 'prey drive' -- which would be very useful if a dog had to go feral and catch its own dinner. Vranak (talk) 03:55, 28 September 2011 (UTC)[reply]
In terms of prey most domesticated dogs (family pets) are too well fed for food to be the driving force behind their neurosis. Rather their neurosis is to establish dominance over the territory they inhabit in the mind of anything and everything that enters it, right up to the point of who shares their food bowl. What love and conditioning and training and food do is merely raise the trigger point to a higher level. For a feral dog or cat that level may be only slightly above that of a wild animal. Check news articles for the past month and you will find one in which a family pet mauled and killed a new born because the parents were expecting the dog to automatically accept it as a family member without introducing the dog to the child. For the dog the infant was something that had entered its territory without introduction and anyone's demonstrated approval and in need of being killed and eaten like a person cuddling a rabbit they are about to set down in front of a pet snake for its next meal. --DeeperQA (talk) 05:51, 28 September 2011 (UTC)[reply]
How does that mean neurosis? Human beings usually automatically kill any snake or spider they find in hysteria that it might be venomous. Does that make humans neurotic as well? You are making the classic anthropocentric mistake of classifying animal behavior by how they relate to humans. -- Obsidi♠nSoul06:37, 28 September 2011 (UTC)[reply]
So you are saying that puppies are completely neurotic killing machines. Seems harsh. :) For interest, in my area there are several packs of feral dogs and my experience of their behavior doesn't really match your descriptions. They are effectively wild. They live, move, hunt/scavenge as packs and they are rather good at finding snakes it seems. They invest a lot of time establishing their dominance hierarchy or playing (...hard to tell the difference) and sleeping of course (..it's hot) to the extent that they are almost nocturnal. They don't really interact with people or seem very interested in people at all. They seem to reserve barking for ritualized, quite rare interactions with neighboring packs and encounters with potentially dangerous snakes. They are oddly quiet. Their territories seem to be rather fluid and change quite rapidly over time. The area is rural so there isn't really any kind of population pressure. I cross various pack's territories on foot quite often and if anything, I think the packs tend to react with indifference or moderate fear when they see a large primate crossing their territory. They certainly don't go into fight or flight mode or act aggressively. Sean.hoyland - talk06:50, 28 September 2011 (UTC)[reply]
Trigger Point Ally, Trigger point... I am saying is that all that changes is the threshold. For puppies it is not until the first really painful bite on the nose by a litter mate that reveals puppy love may not be the only thing a puppy is made of. Dogs that have not been conditioned, trained, and reinforced with food and shelter to tolerate others and even strangers in their environment may have a very low Trigger Point of becoming neurotic over the presence of undesired guests. Even those which have been conditioned may "loose it" if their mind is in control and not their owners when faced with the need for action. Dogs which are left on their own forced to rely upon their own decisions may have no threshold of tolerance at all. Hence dead baby brought to you by an otherwise loving pet. --DeeperQA (talk) 07:52, 28 September 2011 (UTC)[reply]
Read Animals in Translation the works of Monty Roberts if you want to read about animal neurosis and do not include me or my family in any so called humanity which automatically kills spiders or snakes. My reaction is to photograph them, or capture them and release them a day later. μηδείς (talk) 00:31, 29 September 2011 (UTC)[reply]
Every particle is made up of fields and the relationship is not one to one. Most of those fields have components that may or may not be counted as separate fields depending on how you choose to count them so counting fields is not a trivial matter. For instance, quarks come in three different colors and each color possibility should be counted separately for some purposes such as statistical mechanics where the number of independent states of an ensemble must be accurately calculated. Often times though identical quarks with different colors are not counted separately since they are so similar. Dauto (talk) 21:31, 27 September 2011 (UTC)[reply]
The second question has an even more subtle answer. Most of those fields are believed to be massless at high enough energies, but at low energies, after electroweak symmetry breaks, many of them acquire mass through the Higgs mechanism. Only massless particles propagate at the speed of light. Dauto (talk) 21:35, 27 September 2011 (UTC)[reply]
The real answer is that fields are a concept which act as a model that allow us to probe the details of how the universe works in a mathematically consistent way, and which best approximates the actual behavior of "reality". (It should be noted that, in this way, fields are not any different than "particles" or any other physical concept). Fields can be said to exist for any force or property which propagates through space (or space-time) and decays. We use fields because they work. --Jayron3221:41, 27 September 2011 (UTC)[reply]
In modern physics "field" has a pretty clear definition and it is reasonable to ask how many there are (though, as Dauto said, there are different ways of counting them). -- BenRG (talk) 07:16, 28 September 2011 (UTC)[reply]
The number may increase dramatically if supersymmetry is discovered. And since supersymmetry is broken, there then likely exists hidden sector fields corresponding to particles that unlike the supersymmetric particles don't interact with Standard Model particles at all (except via gravity). Count Iblis (talk) 22:18, 27 September 2011 (UTC)[reply]
Yeah, LHC has been pretty hard on some of the more creative ideas, but supersymmetry is still alive and well despite the rumors of its demise. Dauto (talk) 13:18, 28 September 2011 (UTC)[reply]
Although, if the Higgs is not found soon, that is then evidence for new physics. E.g. there could exist many different types of dark matter particles to which the Higgs can decay. That then hugely broadens the resonance signal for the Higgs, making it much more difficult to extract from the background. Count Iblis (talk) 15:52, 28 September 2011 (UTC)[reply]
The LHC has closed many windows in the mass spectrum for the Higgs, but the range from 115 to 130 GeVs is still open because of the large amount of background. There seems to be a peek forming in that region but it is still too early to tell. That peek seems broader then expected which goes along with your remark about peek broadening. That broadening might be spurious though. - Heck, the whole peek might be spurious. We may have to wait at least another year before things become more clear. Dauto (talk) 18:31, 28 September 2011 (UTC)[reply]
Suppose I was to wind a coil of wire around the equator lots and lots of times. In this configuration with the Earths magnetic field the coil takes on the same (albeit linear rather than "U" shaped) voice coil or solenoid. Could I do anything useful with this coil such as listening to the Earth's magnetic field or sending dance tunes to aliens in the far reaches of space? --DeeperQA (talk) 20:57, 27 September 2011 (UTC)[reply]
Not that I'm aware, although you could use up the world's copper output for a while. Electricity is generated when a wire loop moves with respect to a magnetic field. Yours would be pretty static. Regards, RJH (talk) 22:50, 27 September 2011 (UTC)[reply]
You could probably pick up fluctuations in the field, such as from geomagnetic storms, though. Of course, you could probably do almost as well with a much smaller coil... --Carnildo (talk) 01:14, 28 September 2011 (UTC)[reply]
Silver mining
My crew of dwarves that I'm running a dnd quest for is about to liberate a silver mine from a band of marauding orcs who took it over from the human kingdom recently. They've mentioned desire to actually take it over and manage it during down time from adventuring so I thought I'd investigate what would be involved. I've taken a look at the silver mining page and have decided that the mine is actually a lead mine and the silver is extracted by melting the lead with zinc. My question is this: About how much lead could one expect to need in order to get a measure of silver? Are we talking about a tonne of lead for a pound of silver type ratios or would it be much greater than that? Thanks for any insight :) 142.244.35.91 (talk) 21:21, 27 September 2011 (UTC)[reply]
If it helps any, an example on pg. 41 in Principles of Mining by a certain former president gives 20 ounces of silver per ton, of which 15 ounces is recovered. You could just use 2d20 ounces/ton. Regards, RJH (talk) 22:08, 27 September 2011 (UTC)[reply]
Make sure you ask your DM for the maximum rainfall in the local climate so you can take precautions against flooding (e.g. dig out lower levels under your main shafts so your miners have time to escape.) Also you should be able to hire some low level henchmen magic users or clerics to help locate the mineral seams and maybe set up some kind of a magic smelting furnace to cut down on operating costs. 69.171.160.139 (talk) 22:59, 27 September 2011 (UTC)[reply]
I'm able to track down information on the past and current crystallization of the Earth's outer core along the boundary with the inner core, but I can't find anything about future projections for inner core expansion and the freezing out of the magnetic dynamo process. If memory serves, there was a story about that in the press a while back (perhaps in a science magazine), but I can't find it anywhere. Does anybody have a reference I could use, or recall where that story appeared? Thank you. Regards, RJH (talk) 22:01, 27 September 2011 (UTC)[reply]
Yes I think that might be it. Thank you. Unfortunately, in skimming through it, that doesn't appear to answer the question I'm trying to resolve. I.e. how long it will take for the core to freeze solid. I guess then that may be just too difficult to model at the moment. Regards, RJH (talk) 22:43, 27 September 2011 (UTC)[reply]
Given the current rate of 0.5 mm/yr and assuming that rate stays constant (which is probably not a good assumption as the proportion of iron drops and the radius increases), I get (3,480 − 1,220 km) × (1,000,000 mm/km) / 0.5 mm/yr. = 4.5 gyr. But either way it's not something I can reliably cite, alas. Regards, RJH (talk) 21:47, 28 September 2011 (UTC)[reply]
It's an energy loss problem, so the scaling factor ought to be change in volume rather than change in linear scale, but yeah I don't see any thing one can cite either. Dragons flight (talk) 22:05, 28 September 2011 (UTC)[reply]
It's an energy flux problem. The exact composition of fissile material in the core is not known, and ignoring such heating factors will thorw off any estimate. μηδείς (talk) 00:27, 29 September 2011 (UTC)[reply]
Actually, the composition doesn't matter much (not to order of magnitude anyway) since whatever exists now has already had 4.5 billion years to decay. So any isotopes still present in significant quantities are going to have a long half-life. You might be wrong by a factor of several, but not orders of magnitude, simply by assuming the fluxes stay constant. Dragons flight (talk) 20:06, 29 September 2011 (UTC)[reply]
A reference was suggested to me that gives an estimate of 3-4 billion years, so I'm going to go with that for now. Thank you. Interestingly, I found another source that says the end of the tectonic plate system (with the advent of the "moist greenhouse") would change the net heat transfer rate, which may end inner core growth and thus shut off the dynamo. Regards, RJH (talk) 21:17, 29 September 2011 (UTC)[reply]
I took the wrapping off several sets of batteries and now I can't tell which ones are rechargeable (and some of them are) and which ones are not. Not one of them says "rechargeable". I think the Energizer Lithium AA 03 2023 batteries are the rechargeable ones, but I can't seem to find anything on Google that says they are or they aren't, just lots of specs about their uses in cameras. Anybody out there know? Thanks Bielle (talk) 03:04, 28 September 2011 (UTC)[reply]
Having kept looking at batteries in general, I have concluded that if they don't say "rechargeable" writ large, then they are not. Thus, the Energizer Lithiums are not rechargeable. Comments appreciated. Bielle (talk) 03:15, 28 September 2011 (UTC)[reply]
The chemistry in rechargable batteries is very different than the chemistry of other batteries. Attempting to recharge a non-rechargable battery involves unfun events like "leakage" and "overheating" and "explosion". Don't do it. Every rechargable battery is always clearly and unambiguously labeled as such, so if it doesn't say "rechargable" in a giant freindly font on the side of the battery, don't stick it in a recharger. --Jayron3203:37, 28 September 2011 (UTC)[reply]
Lithium is a violently excitable substance if it is abused. A lithium fire would probably ruin your day. It cannot be put out by common extinguising substances such as water or carbon dioxide. Roger (talk) 09:26, 28 September 2011 (UTC)[reply]
In other words, say hello to the new battery-sized hole in your table. And to the huge blackened areas in your floor and ceiling. And I bet that any smoke entering your lungs will be completely healthy and won't leave you coughing for weeks. --Enric Naval (talk) 10:31, 28 September 2011 (UTC)[reply]
As a further note, the only common rechargable AA batteries are NiMH, NiCd and NiZn. Of these, NiMH are what you're most likely to encounter, NiCd are rare nowadays for various reasons including them containing cadmium, their general low capacity, the memory effect etc. NiZn are still relatively new and should be used with care since their nominal voltage is fairly high, may be too high for some devices. NiMH and NiCd batteries can usually be charged in the same chargers but NiZn need their own chargers. You can get rechargable Lithium-ion batteries (there are actually several kinds of rechargable lithium ion batteries) in AA size equivalent, more commonly called 14500 but these are not intended to be used with most devices accepting AA batteries because of the much higher voltage. These need special chargers and aren't really intended for the consumer market (although are popular in some circles) and instead are generally used for packs and in devices where they aren't intended to be removed, and you will not find Duracell ones. Because of the risks, I wouldn't recommend them if you don't know what you're doing. As has been noted, lithium batteries should be treated with care, even primary ones. Nil Einne (talk) 12:41, 28 September 2011 (UTC)[reply]
Normally if batteries are not rechargeable they will say "do not recharge" on them in small print somewhere. I have plenty of batteries which are rechargeable but don't explicitly say "you can recharge this battery" on them 82.43.90.142 (talk) 09:54, 28 September 2011 (UTC)[reply]
Thanks for all your help. I did some checking in with electronics stores today and they all, without exception, agree with Jayron: if it is rechargeable, it will say so; if it does not say so, do not try recharging for all the reasons shown above. I am concerned that 82.43.90.142's response is backwards, and may be dangerous, though perhaps this varies from country to country. Bielle (talk) 02:37, 29 September 2011 (UTC)[reply]
solar wind movement and direction
solar wind is the current of ejected particles from sun corona , those particles have 250~750 km/s velocity , which is further than sun gravity field escape velocity (180km/s).In addition the particles obey electromagnetic field rules which make them to have spiral rotation , suppose any particle (for example proton)how will it move at its way in interstellar space?akbarmohammadzadeIRAN--78.38.28.3 (talk) 03:36, 28 September 2011 (UTC)[reply]
I don't think so. They are only related in the sense that plasma can sometimes behave like a type of liquid. The current, widely-accepted stellar model holds that the cores of hydrogen fusing stars are entirely composed of plasma matter. At the millions of Kelvin needed for nuclear fusion of hydrogen, how can it be otherwise? Ergo, the properties of liquid matter don't appear directly relevant. Regards, RJH (talk) 21:55, 28 September 2011 (UTC)[reply]
Remember, there is pressure also to contend with. Plasma is not a state of matter, at least it is not distinct from solids, liquids, or gases. After all, you do get solid plasmas, liquid plasmas, and the more familiar gaseous plasmas. Plasmic Physics (talk) 22:06, 28 September 2011 (UTC)[reply]
Yes, that is quite clear indeed. The misconception of plasma as a fourth state of matter is quite widespread, but has no basis in reality. Even if one were to dismiss ionic liquids as not being liquid plasmas because they derive their plasma-like quality from anions rather than electrons, there's still e.g. liquid metallic hydrogen which forms at tremendous pressures, and which derives its "plasma-like" quality from electrons, and as such is not actually "plasma-like" at all, but a clear example of liquid plasma. And as if that's not enough, metals in the process of conducting electricity fit the exact description of a solid plasma, to the point where the electron oscillations due to the current are literally called "plasmons".
When the fighter aircraft pass over our heads they make such terrible sound, but why does the big Jumbojet(or Airbus) don't make so much noise though it has more and far bigger engines, when it is landing or taking off near us at airport etc, though it is quite close. 124.253.129.113 (talk) —Preceding undated comment added 04:15, 28 September 2011 (UTC).[reply]
The planes don't always break sound barrier. Breaking sound barrier is a momentary phenomenon that last a fraction of a second...most of the time they are traveling below speed of sound 124.253.129.113 (talk) —Preceding undated comment added 05:02, 28 September 2011 (UTC).[reply]
Jumbo jets have high-bypass turbofan engines, which derive most of their thrust from pumping cold air through the bypass ducts and out the exhaust nozzle; this is for better efficiency at medium-high speeds (500-600 knots or so), but it also muffles the roar of the hot gases from the combustion section of the engine. Jet fighters, on the other hand, have straight turbojet engines that derive most if not all of their thrust from the hot gases; these produce more power at very high speeds (> Mach 1), but obviously don't have the same muffling effect from the cold air. 67.169.177.176 (talk) 05:32, 28 September 2011 (UTC)[reply]
The above is totally correct. If it's worth mentioning, this is not an accident, noise consideration is a considerable factor in commercial jet engine design. If anything, apart from absolute performance, fighter jets are, in complete contrast, designed to shock and awe. Vespine (talk) 06:15, 28 September 2011 (UTC)[reply]
Not as a general statement; it depends on what kind of stealthing the aircraft is concerned with. For instance, a helicopter operates a low altitudes where being noticed by foot soldiers can be a threat. Thus, stealthing a helicopter reasonably includes sound-deadening measures. Stealth aircraft operating at high altitudes, on the other hand, won't care about sound output to nearly that extent. If they're quiet, it's more likely as the result of a happy design accident (perhaps aforementioned high-bypass turbofans are good for stealth purposes due to heat of exhaust) than intentional design methodology. — Lomn18:50, 28 September 2011 (UTC)[reply]
I was going to say a simiar thing that noise is not generally a factor on "stealth aircraft" but our article on the state of the art F-22_Raptor#Stealth seems to disagree. It says The aircraft was designed to be less visible to the naked eye; radio, heat and noise emissions are equally controlled. Vespine (talk) 23:31, 28 September 2011 (UTC)[reply]
I'll provide a "yes, but" to that. Reading the source document, there is but a token generic statement ("To make a stealthy aircraft, designers had to consider... muffling noise...."). That's it. The rest of that article discusses relevant stealth features in detail -- radar signature, visual signature, emcon, and heat signature. There is zero mention of noise reduction as an actual meaningful objective of F-22 design. I note also this image, from this F-35 acoustics test report, notes that the F-22 is the noisiest fighter in the modern US arsenal at both minimum and military power. The F-35, itself a stealth design, is also relatively loud. Finally, note that the acoustics study is for purposes of noise exposure to maintenance personnel and the general public around military airfields. It's not a combat consideration at all. — Lomn03:18, 29 September 2011 (UTC)[reply]
Planes in supersonic flight produce a sonic boom continuously. It is not a momentary effect when they "cross the barrier". The sonic boom is highly directional, however, so an observer at a fixed location on the ground only hears it briefly, but observers everywhere along the plane's path will hear it. It's true, though, that military jets do not always flight faster than the speed of sound, particularly when traveling over populated land.--Srleffler (talk) 19:02, 30 September 2011 (UTC)[reply]
Although the noisiest aircraft in the world seems to be a propeller plane, the Tupolev Tu-95:
Its blades, which rotate faster than the speed of sound, according to one media source, make it arguably the noisiest military aircraft on earth,[2] with only the experimental 1950s era Republic Thunderscreech turboprop powered American fighter design as a likely rival.
See Republic XF-84H: "Unlike standard propellers that turn at subsonic speeds, the outer 24–30 inches of the blades on the XF-84H's propeller traveled faster than the speed of sound even at idle thrust, producing a continuous visible sonic boom that radiated laterally from the propellers for hundreds of yards. The shock wave was actually powerful enough to knock a man down."Alansplodge (talk) 00:58, 29 September 2011 (UTC)[reply]
The thrust of an engine is equal in magnitude to the change in momentum it imposes on the air. The change in momentum is the mass flow of air times the change in air's velocity. Since the minimum energy required for that grows quadratically with the change in speed and only linearly with mass flow (kinetic energy is m * v^2 / 2), the most fuel-efficient way to generate a given thrust is with a high mass flow (= high bypass ratio) and a smaller change in air velocity. Fuel efficiency is of high concern to airliners, so this is exactly what they have done. It's a nice side effect that a flow with a smaller velocity tends to make less noise. --145.94.77.43 (talk) 11:26, 2 October 2011 (UTC)[reply]
Question about GRAVITY?
Galileo was first to demonstrate that all objects fall at the same rate in the absence of an atmosphere. As it is said that the earth and the apple fall toward each other but apple looks a lot to falls to the earth as compared to the falling of earth toward the apple which is so tiny to be detected.
Let's imagine earth is a homogeneous sphere therefore in the following cases, what would be the direction of direction [falling] of earth in the absence of all other gravitational attraction including atmosphere?
1- If two apples start falling simultaneously from ANTIPODEAN trees from same altitude.
2- If two different masses [say one apple and other big asteroid] start falling simultaneously from same ANTIPODEAN altitude.
I think there is some very fundamental goof here. He says apple and earth fall towards each other. The apple is OK, it falls i.e. it moves in a straight line towards earth, but does earth move towards apple ? I think no, as it is already "falling" - it is moving around the sun - that is technically falling, no ?. Now, how can something move towards two different directions at same time ? 124.253.137.182 (talk) —Preceding undated comment added 05:49, 28 September 2011 (UTC).[reply]
Different vectors. Any object in space is moving along a path that's the "sum" or "net effect" of its vectors (there may be a different word for that, but I can't think of it just now). Earth and moon orbit around a common point. That point orbits around the sun. The sun presumably orbits around the center of the Milky Way Galaxy. And the Milky Way is presumably heading along an approximately straight line away from wherever the Big Bang occurred. So the earth, viewed in isolation and relative to all these different forces, would be taking a very interesting path. As noted in the earlier section, the falling apple has a theoretical effect, but so small as to be negligible. ←Baseball BugsWhat's up, Doc?carrots→ 05:52, 28 September 2011 (UTC)[reply]
Fair answer except for the moment where you talk about a place where Big Bang happened since Big Bang did not happen in a place. Big Bang is an expansion of the whole universe, not an event within the universe with specific time and space coordinates. Dauto (talk) 14:03, 28 September 2011 (UTC)[reply]
Again you are essentially setting up a Three body problem. What don't you understand? It's fairly simple. In scenario 1, given "perfect conditions, the earth will stand perfectly still, pulled equally by the two apples. For 2, assuming the asteroid is more massive then the apple, then it will have more of a gravitational effect on the earth then the apple, however unless the mass of the asteroid is a meaningful fraction of the mass of the earth, the effect will still be insignificant. I don't actually understand what you mean by Also, would gravity "g" of two equal planets cancel each other if placed on each other? No, if your adding their masses, their gravity would also add up, not cancel out. Note, the gravity would NOT double with a doubling of the mass, it's one of those inverse square law situations. Vespine (talk) 06:09, 28 September 2011 (UTC)[reply]
The force on an object due to gravity does double when the mass is doubled. This means that the acceleration due to gravity remains the same. Depends exactly what the OP meant, though, which isn't clear. Grandiose(me, talk, contribs) 09:42, 28 September 2011 (UTC)[reply]
With both planet's particles being present, the various forces these generate will be present too, but they can create a static equilibrium with forces opposed to each other. If you lived in the place where the two equal planets are joined together, the net force due to gravity is zero. See for instance Lagrangian point #L1. In a similar situation, gravitational acceleration decreases in a tunnel or well with increasing depth, see Gravity of Earth. Living between the two planets should be interesting, being nearly weightless and being able to easily move around but with a significantly greater rate of fall as you climbed out. The building materials on the surface would need to be progressively stronger too. The L1 position is unstable, so apples will fall towards the planet that is nearest, but one might become adapt at calculating the inertia needed to miss the surface in a fall (see Douglas Adams' novels, :-), for more detail). Once you reached either "end of the world" the gravity due to both masses will not be doubled on account of the one planet being further away. So locally, there is no doubling, but as you leave the system, the difference in the distances to each planet can become so insignificant that the gravitational acceleration due to both can be practically twice that of only one planet. --Modocc (talk) 17:54, 28 September 2011 (UTC)[reply]
faster than light question
Just to clarify the predominant understanding of traveling faster than the speed of light actually means I take it then that if I am traveling on the back of a light beam and a faster than light is coming up behind me that from my perspective it is in my past and when it passes me it is in the future while I remain in the present? --DeeperQA (talk) 07:37, 28 September 2011 (UTC)[reply]
In many ways, FTL travel introduces many real paradoxes which cannot be resolved by a simple explanation in the English language (or indeed any language). --Jayron3213:20, 28 September 2011 (UTC)[reply]
You keep asking the same question. Do you expect to get different answers? Speed is relative - that is: different observers can see objects moving differently. For someone in a train, their luggage is just sitting their on the floor, not moving at all, but that same piece of luggage is observed to be moving by someone at the train station. It turns out that the rules for transforming speed between different observers is more complex than believed by 19th century physicists and an object seen moving faster than light by one observer will be observed moving backwards in time by another observer leading to severe paradoxes. The simplest and most logical way to avoid those paradoxes is to assume that faster than light travel is not possible. Dauto (talk) 13:32, 28 September 2011 (UTC)[reply]
Also, if you are traveling on the back of a light beam, you have no past and no future since your time stops due to an infinite time dilation factor. From the point of view of a photon, the universe is flattened into a 2-dimensional pancake and the photon is created and absorbed in the same spot simultaneously. Dauto (talk) 13:57, 28 September 2011 (UTC)[reply]
I can't see anything there that justifies answering this question with a yes. Nutrition does matter, but to my knowledge there is no evidence that muscular exercise affects height. Looie496 (talk) 16:27, 28 September 2011 (UTC)[reply]
I did explicitly say bone mass. While it does not (or at least only very slightly) affect bone length, surely you can agree that being a bit more healthy can make the difference? -- Obsidi♠nSoul16:53, 28 September 2011 (UTC)[reply]
From my knowledge of developmental biology, increase in bone length is not caused by increase in muscle mass, and the diversion of resources to the muscles and premature bone closure by increased hormone levels might instead lead to stunting. If this is a serious question ask a specialized pediatrician. There are far to many factors and risks to take comments here as above the level of nonsense. μηδείς (talk) 00:24, 29 September 2011 (UTC)[reply]
Ok now you're both putting words into my mouth. Read my original post again. First I never said anything about bone length. I said bone mass - i.e. bone strength, bone health. And I, again, explicitly said with proper nutrition and that it all depends on genetics in the end. And no, muscular mass is tied intimately (literally) to the bones they're attached on. You can not exercise any major muscle without affecting bone as well. And to the contrary, just as exercise has no effect on increasing bone length, neither does it stunt it. Unless we're talking about one of those crazy parents pushing ten-year olds into bodybuilding with steroids or malnourished kids working high-intensity jobs in sweatshops. I mean, jeez, why do you think we had PE classes? To stunt our growths? Nevertheless, I agree, talk to a pediatrician. Overenthusiastic and incorrect exercise can result in injuries that can stunt growth.-- Obsidi♠nSoul03:04, 29 September 2011 (UTC)[reply]
What manner of substance is this?
So, I was watching videos of shredders shredding things and found myself on this one, which shows large chunks of metal igniting and burning and popping and sparking whilst going through the machine. The video just calls it 'speciality metals' and there are a few guesses in the comments section - but does anyone here know for sure what the hell that stuff is? --Kurt Shaped Box (talk) 19:19, 28 September 2011 (UTC)[reply]
Looks a lot like magnesium to me; for liability purposes they may not want to name the metal, which is why they use the term "specialty metals". Magnesium is a fairly reactive metal, and it is flammable, even in pure CO2. (google "magnesium and dry ice" for some fun videos). The bright flashes look a LOT like magnesium when it burns, and grinding it up like that could easily produce enough heat via friction to ignite it. --Jayron3219:28, 28 September 2011 (UTC)[reply]
I'd just settle for shredding propane tanks. Man, if I owned a big shredder, I'd be tempted to throw random objects in there, just to see what they look like whilst being ground up. I'd be going around looking at things and thinking 'will that shred?'. Hmmmm, I wonder if those guys could shred a (battle) tank whole? --Kurt Shaped Box (talk) 14:08, 29 September 2011 (UTC)[reply]
Mass M is required to be levitated (stationary) in a gravitational field of strength g by pumping out a fluid of density rho over a cross-sectional area A (e.g. a ducted fan blowing out air). Assuming no losses, how much power is required?
Clarification: In case unclear, I mean that the fan is attached to (part of) the mass M and is also hovering (as opposed to blowing upwards from the ground). 86.179.118.99 (talk) 22:16, 28 September 2011 (UTC)[reply]
If it helps, this is called a gravity burn. "Assume no losses"? The entire thing is a loss! No work is done on Mass M - it isn't moving - which means that 100% of the work is done on the exhaust-gas! Nimur (talk) 22:38, 28 September 2011 (UTC)[reply]
I mean no losses to heat/noise/etc (other than after the fluid has escaped and done its business, obviously). For example, a real fan would not convert all of the energy it consumes into moving the air in the required direction, I assume. Thanks for the link to the article, but it doesn't really help to me verify that exact formula, as far as I can tell. 86.179.118.99 (talk) 23:21, 28 September 2011 (UTC)[reply]
I concur with your formula. It is a measure of the minimum rate at which downward kinetic energy must be imparted to the air in order to stay stationary in the limit that air is incompressible. In practice, there will be corrections (potentially large ones) from the fact that air is a compressible fluid and the fact that the amount of energy used to operate any real fan will always be significantly larger than the amount of downward kinetic energy acquired by the air. Dragons flight (talk) 01:38, 29 September 2011 (UTC)[reply]
Unless I'm making some gross error of intuition, wouldn't it be more? Think of what a propeller does in incompressible water vs what it does in air. Vespine (talk) 04:19, 29 September 2011 (UTC)[reply]
I dunno. There are viscosity and density to take into account too... Plugging the density of water versus air into my formula above makes a huge difference. 109.156.50.129 (talk) 13:36, 29 September 2011 (UTC)[reply]
Right, I see what you mean. Ok, so I still think the answer is more. Planes need more power to fly high in thin atmosphere which is more compressible, right? I wasn't sure if thin atmosphere counts as more "compressible" but I think by the definition in our Compressibility article, it does. Vespine (talk) 01:23, 30 September 2011 (UTC)[reply]
Normally, cantennas DIY projects require that you have a USB-WiFi stick + cable or some other way of connecting the cantenna to your WiFi circuity. However, would it be possible just to do a cantenna without that? Just imagine a normal catenna + some other improvised devise forwarding the signal to your laptop (without messing with the laptop). 22:55, 28 September 2011 (UTC)
I wanted actually an inexpensive solution for connecting a laptop too far away from the router. A directional antenna (cantenna wa my first thought). I'll try to find a DIY repeater. Quest09 (talk) 23:46, 28 September 2011 (UTC)[reply]
I outright reject the usurpation of the term "Cantenna" for some upstart waveguide doo-hickey used with Wi-Fi. The word has a long and honorable history as a descriptor of a RF dummy load embodied by a 50 Ohm resistor in a gallon can of mineral oil, used to adjust an amateur radio transmitter, and able to dissipate a kilowatt. Can the upstart handle a kilowatt? Hunh? Edison (talk) 04:23, 29 September 2011 (UTC)[reply]
Times change. In the past a computer was as big as a shipping container, but, alas, we still use the same word for a completely different machine. Quest09 (talk) 16:30, 29 September 2011 (UTC)[reply]
September 29
Criteria for mountain pass
Is there some sort of defined criterion or criteria for what does and doesn't constitute a mountain pass? Our article on the topic says that there are precisely 42 passes in the 5300-km-long border between Argentina and Chile; it would seem that some sort is necessary in order to say that most low spots between mountains aren't truly passes. Is there perhaps a minimum amount of topographic prominence for the mountains on each side of the pass? Nyttend (talk) 01:11, 29 September 2011 (UTC)[reply]
There could be within some limited context, but not in general. For a very non-pass-looking-pass, take a look at Deadhorse Pass, File:CupLakeDeadhorsePass.jpg—it's the dip on the right side of the photo. I wonder about the claim of "precisely 42 passes" on the Argentina-Chile border. Seems fishy. Perhaps there is some context, like "named passes" (although I would think there are many more than 42), or some arbitrary cutoff prominence. Or perhaps Chile and Argentina have some official standard they abide by for whatever the term would be in Spanish... Pfly (talk) 01:33, 29 September 2011 (UTC)[reply]
I would think that one criterion -- maybe the only one -- is that at least a few people have actually used it to get from one side to the other. Looie496 (talk) 01:41, 29 September 2011 (UTC)[reply]
"For passes with roads, it is also customary to have a small roadside sign giving the name of the pass and its elevation above mean sea level. An example of this is Argentina and Chile that share the world's third longest international border, 5,300 kilometres (3,300 mi) long, running from north to the south through the Andes mountains, having a total of 42 mountain passes between them"
--emphasis mine. I think the article is saying that there are 42 passes with roads, each of which has a sign with a name and elevation. I have no idea if that's true, but it does seem to imply how the count was arrived at. SemanticMantis (talk) 01:55, 29 September 2011 (UTC)[reply]
mental excercise
What mental exercises ramp up the burning of calories to the max such that a person who is wheel chair bound can burn off as many calories as possible? --DeeperQA (talk) 02:12, 29 September 2011 (UTC)[reply]
CT scans are said to be used to find places in the brain where activity has increased as the result of seeing an appealing human form which have been said to cause excitement. The method I am told relies on the places in the brain where the greater activity is related to increased cerebral metabolism, i.e., the burning of calories. --DeeperQA (talk) 03:41, 29 September 2011 (UTC)[reply]
Perhaps playing a musical instrument ? Dealing with music seems to be a global process in the brain and presumably burns slightly more glucose. The movements involved in playing it probably burns more calories though I suppose. Sean.hoyland - talk04:05, 29 September 2011 (UTC)[reply]
I'm pretty sure that the brain is like a TV, at least in that what's on the screen doesn't really influence how much power it uses.. Overall, your brain consumes about 20% of the oxygen you inhale, I've never heard of anyone "puffing them selves out" on a particularly strenuous mental activity. Vespine (talk) 04:13, 29 September 2011 (UTC)[reply]
I'm not so sure. Try doing something which is likely to use as many different parts of your brain as possible: translate this article Polynesian navigation into medieval French (or whatever obscure language you prefer) while balancing a spinning plate on a stick held in your mouth - none of this will involve a great deal of physical effort, but your brain will probably tire rather quickly. If Vespine has never 'puffed himself out' from mental activity, he has either been exceptionally lucky, or has more mental grunt than the rest of us. AndyTheGrump (talk) 04:27, 29 September 2011 (UTC)[reply]
The following lists the amount of energy used during various activities in kJ/kg/h. (Sorry it's not in calories. We're metric here.)
Sitting quietly 1.7
Writing 1.7
Standing relaxed 2.1
Driving a car 3.8
Vacuuming 11.3
Walking rapidly 14.2
Running 29.3
Swimming (4km/hour) 33
Rowing in a race 67
It's the fourth line that's important - Driving a car. We do that while sitting down. Many wheelchair users drive. We all know we get tired after driving for a long time. Obviously it uses more energy than just sitting there. HiLo48 (talk) 04:35, 29 September 2011 (UTC)[reply]
I'd say you can just as easily claim that the second line is critical; presumably writing takes more creative thought than sitting quietly. Driving, on the other hand, requires near-constant use of major muscle groups. "While sitting" and "just sitting" are very different things. — Lomn13:05, 29 September 2011 (UTC)[reply]
Paraplegic or Quadriplegic? In either case, singing along out loud to and dancing (to whatever extent is possible) with some fast paced music. The soundtrack to Saturday Night Fever is excellent. It gets the whole body in motion, including the brainwaves. μηδείς (talk) 19:11, 29 September 2011 (UTC)[reply]
The metabolic rate and the blood flow to different brain regions varies when they are active or inactive, as a function of the task, and the harder a part of the brain is working, the more glucose it metabolizes, just as a given muscle does: [6], [7]. I would be expect that that the overall metabolic activity of the brain, and the energy it uses, and the waste heat it produces, less when one is sitting with the eyes closed and the mind blank. or in a dreamless sleep, than when doing difficult mental activity such as memorizing, learning a complex task, designing something, doing mental arithmetic, or taking a difficult math test. I would even expect that brain metabolism varies during the various stages of sleep, since some dreams represent incredibly complex creativity. One reason for this is that the faster a neuron fires, the more energy is required to repolarize it. How hard the brain is working, and at what type of task, causes less variation in the eneregy consumption than for muscles, since the brain is never really inactive. Even when "idling," it consumes a huge amount of energy. It is about 2% of the body's mass and consumes about 20% of the energy:[8]. Edison (talk) 20:09, 29 September 2011 (UTC)[reply]
I disagree with one of the above comments about driving, there's a bigdifference between getting "tired" and actually doing meaningful exercise; sitting on a couch doing nothing for 3 hours can be very tiring. Similarly, in reply to Andy's comment: If Vespine has never 'puffed himself out' from mental activity, I wasn't talking figuratively, I meant literally puff yourself out, like you would after running around the block. Yes a chess grandmaster might need a rest after playing a challenging game because they are mentally exhausted, but that is a very different kind of exhausted to someone who has just skipped a rope for 20 minutes. I doubt very much that you burn (EDIT: significantly) more calories playing a challenging game of chess compared to relaxing and reading a book, and if you do, it's probably more due to things like heightened stress rather then actual mental activity. Vespine (talk) 22:53, 29 September 2011 (UTC)[reply]
Vespine, that's actually an important difference here. If you found that the game of chess burned more calories than relaxing due to heightened stress, then that would be an answer to OP's question. He or she doesn't ask anything about the brain, he asks specifically about burning calories. – b_jonas17:29, 30 September 2011 (UTC)[reply]
A double planet that shares an atmosphere?
Let's say there were two planets that were both identical to our earth. The only difference is that they were close enough that their exospheres touched (The shortest distance between the two planets' surfaces would be 1 megameter). Let's assume everything else is the same as far as our earth is, except that this double planet shares a single moon which orbits the two earths' center of mass. Now, obviously they would be tidally locked to each other. But how stable would this configuration be? How fast would they be rotating around their center of gravity? Would humans or any life forms be able to survive on this double planet? Assuming that they could, what would be the physics behind traveling from one planet to the other? In other words, given our current technology, how feasible would it be to design aircraft (or would it be spacecraft?) that would be able to fly from one planet to another?
If each planet were Earth size and Earth mass, I think tidal forces would rip both planets apart, and when the dust settled, there would be one large spherical planet. Gandalf61 (talk) 08:17, 29 September 2011 (UTC)[reply]
I'm not really sure how best to calculate it. I'm not sure the usual formulae in Roche limit are going to hold in this case. If I've done the calculations right (somebody please check!), then the tidal acceleration on a loose object sitting on the surface of one of the planets (assuming the planets are solid, rigid objects) is 5.2m/s2. That is less than the 9.8m/s2 gravitational acceleration at the Earth's surface, so it would not be lifted away from the planet it is sitting on. Even if the two planets were touching each other, the tidal acceleration would only be 7.3m/s2. My assumption that the planets are completely rigid isn't true, of course, and it is possible that, if deformation were taken into account, I would get a different conclusion. --Tango (talk) 12:41, 29 September 2011 (UTC)[reply]
Think of it this way. Imagine a spherical boulder sitting in the middle of a flat desrt. How big do you think the boulder could be before it collapsed under its own weight ? The size of a large skyscraper ? The size of a mountain, maybe ? Certainly a boulder the size of the Moon would just crumble. Now imagine a boulder the size of the Earth ... Gandalf61 (talk) 14:54, 29 September 2011 (UTC)[reply]
It's in orbit, so it's weightless. That means it can't collapse under its own weight. In order to be ripped apart, you need tidal forces to be greater than its self-gravitation (and its structural strength). The bigger (strictly speaking, denser, but big tends to mean dense) the object is, the less likely it is to be ripped apart, not more. --Tango (talk) 17:06, 29 September 2011 (UTC)[reply]
Gandalf61's thought experiment is close enough to be perfectly illustrative. An object in orbit is not so much weightless as in free fall--but it is entirely subject to gravity and the tidal forces it generates. Two earths approaching 1000km apart (no such orbit would be stable, it's well below the Roche limit) both would crumble into each other just as Gandalf61's thought experiment implies as well as be spinning around each other at terrific speed.μηδείς (talk) 17:35, 29 September 2011 (UTC)[reply]
There is no one "Roche limit", it depends on the object. For an object with as much self-gravity as the Earth, it's much closer than for, say, a comet. Unless my calculations above are in error (which is possible), then the planets could touch before getting past their Roche limit. --Tango (talk) 20:06, 29 September 2011 (UTC)[reply]
You did, and you've just done it again. We're not talking about Earth-Moon. We're talking about Earth-Earth. The Earth is a lot denser than the Moon, so the Earth-Earth Roche limit will be much closer than the Earth-Moon one. Suggesting that the Earth-Moon Roche limit is relevant to a discussion about the Earth-Earth Roche limit suggests you don't understand that there are different Roche limits for different objects. My calculations are just based on Newton's law of universal gravitation, so you can easily check them yourself. --Tango (talk) 23:05, 29 September 2011 (UTC)[reply]
Whether an airplane (something requiring air) could fly between them would depend on the shape of their atmospheres - it's certain that the atmosphere of each would bulge towards the other, and it's possible that sufficient air would be pulled in between them to form a column that a plane could fly through - I'm not sure how likely this would be, since it would probably result in thinner air at the ground. Rockets would be fine - the space shuttle could get 600km above the earth, and you only need to go 500km to reach the midpoint and begin your descent. Assuming the planets are tidally locked, it may also be possible to build a 1000km tether to bridge the gap - see space tether. --Colapeninsula (talk) 11:33, 29 September 2011 (UTC)[reply]
Such a situation was portrayed fictionally by Bob Shaw in his Land and Overland trilogy, whose second volume was called The Wooden Spaceships. In conversation, he used to say that this scenario would be possible given a different value of pi, but if pressed on what value, he would mutter "Whatever it takes to make my planets work." {The poster formerly known as 87.81.230.195} 90.197.66.236 (talk) 14:11, 29 September 2011 (UTC)[reply]
Can intermediate or stellar mass black holes be detectable gravitational lenses?
Could an ordinary black hole or an intermediate mass black hole (say less than 50,000 stellar masses) produce detectable gravitational lensing? I ask because [9] says "Micro-lensing shows no change in the lensed object's shape, just in brightness." 69.171.160.131 (talk) 05:54, 29 September 2011 (UTC)[reply]
Gravitational lensing cannot tell what the focussing object is, but it can give an idea of the mass and how fast it is moving. Stellar mass blackholes should be possible candidates to observe this way. The main issue will be the probability that one crosses in front of a distant star. There should be quite a number of large mass stars that have collapsed into blackholes, but still they would be much less common than white dwarfs or red dwarfs. A 50000 solar mass back hole is probably very very infrequent, even if it does have a bigger sphere of influence, there would be a slimmer chance of detection. Graeme Bartlett (talk) 10:52, 29 September 2011 (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. Roger (talk) 07:30, 29 September 2011 (UTC)[reply]
2. I carried out TLC a couple of days ago. I don't really know how to verbalize poor spotting techinques in TLC. I held my TLC paper under UV light and cirecled what seemed like the center of the mobile phase.--98.88.80.117 (talk) 07:46, 29 September 2011 (UTC)[reply]
TLC is not well suited for quantitative analysis, or even very good qualitative analysis, because it has really bad resolution. Your spotting technique will affect your resolution; if your spot is too small, you won't see the components as they get carried up the TLC matrix. If your spot is too large, then it tends to spread out too much and you will get bad resolution, as the spots won't seperate well and won't travel well, and will instead tend to merely "smear out". --Jayron3219:20, 29 September 2011 (UTC)[reply]
How to include more ordinary people in Wikipedia
Aside from the lack of attention to the REAL WORLD KNOWLEDGE of your contributors, the great problem with Wikipedia is that it doesn't provide a REDUCTIONIST option. Nearly everything you reflexively present is "expansionist," branching out into endless NEXT LINKS, with no focus on a universal (fixed) reference point: mine is SCIENTIFICALLY VALIDATED INFORMATION, by people who know a lot about the BIG VIEW in the particular subject being discussed.
I'd prefer to always be given a link to a REDUCTIONIST version of what the article is about. For example, I'd like to have a reductionist version of the Diagnostic Statistics Manual, in which the states of the human mind are reduced in a way that ordinary people can easily grasp what the text is saying. Instead of listing endless speculative categories, I'd prefer your presentation of information to be FOCUSED ON THE REDUCED ESSENCE of the subject being discussed. I'm a PhD psychologist (and an Aspergian), frustrated with the atomization of everything, and wanting to reduce discourse to its core elements. For instance, I'd like to hear from the psychologists who regard the human animal (all animals) in terms of POSITIVE AND NEGATIVE (equivalent to WANTING AND NOT-WANTING) neural systems (limbic structures) that deterministically cause-force-allow humans to approach or avoid cognitive events in the external world and within their own selves.
This is a big subject. Ring me on [redacted], or send a reaction to [redacted].
OP's personal information removed per instructions at top of page: "Do not provide your contact information. E-mail or home addresses, or telephone numbers, will be removed. You must return to this page to get your answer." AndrewWTaylor (talk) 10:12, 29 September 2011 (UTC)[reply]
And please do not shout at us, we haven't done you any wrong that is worth being shouted at. Surely a guy with a PhD has had enough exposure to the internet that he can understand that words in capital letters is the equivalent of shouting?--Lgriot (talk) 14:20, 29 September 2011 (UTC)[reply]
Reductionist? I do not think that word means what you think it means. I get the gist, though, and recognize the common frustration of being led into clicking through a series (or worse, a branching tree) of links to different articles, in the course of trying to understand some unfamiliar subject area. It's difficult to fix, because: although it's desirable, from the point of view of any individual reader, to reduce an article to only its core elements, each reader will have a different idea of what those core elements are. (I use italics instead capitals, they don't seem to upset anyone.) Articles on important subjects have a lot of links because they are important in a lot of different contexts. The core information is the information germane to a particular context. So not much can be done to make such an article better, beyond a certain point, all though all kinds of spurious information could be added to make it worse. Card Zero (talk)14:59, 29 September 2011 (UTC)[reply]
Whether you like Wikipedia or not, you are under no obligation to use it. Would you like help finding other resources that are not as expansive? Wikipedia is the largest encyclopedia in the world. If you would like an encyclopedia with a little less content, you might find one by at the List of encyclopedias, which is ironically an expansive list that links to several other expansive lists. I highly recommend the World Book Encyclopedia, but I recognize at least a few shortcomings (it's geared toward a primary- or secondary-school level of knowledge for most subjects, with a clear bias of coverage toward American topics). Nonetheless, the quality of content and the editorial control are very good. Nimur (talk) 16:58, 29 September 2011 (UTC)[reply]
SMILES
According to simplified molecular input line entry specification procedures, a double bond is "=", a triple bond is "#", a quadruple bond is "$", and a bond and a halffraction is ":". What is the character for half a fraction bond? Plasmic Physics (talk) 10:17, 29 September 2011 (UTC)[reply]
Where do those occur? Can they be implied with % ring closure notation? SMILES has no provision for explicit intermolecular bonds like hydrogenation, for example, which limits its usefulness for many minerals. 69.171.160.237 (talk) 19:34, 29 September 2011 (UTC)[reply]
I think H+ 2 is [H][H+] and while AuCl 3 is actually Au 2Cl 6, its article says the empirical/ion formula is [Cl-].[Cl-].[Cl-].[Au+3] which can't be right; I'd say [Cl-]1.[Cl-]2.[Cl-]([Au+3]1)2. 69.171.160.237 (talk) 21:25, 29 September 2011 (UTC)[reply]
H+ 2 is [H]-[H+] only in Lewis structure (valence bond theory) which requires that elements have whole number formal charge and that bonds must always exist as pairs of electrons. Real molecules are better described by molecular orbital theory which more accurately represents things like fractional bond ordder. Take H+ 2 for example: In reality, why would one H atom have a greater affinity for the electron than the other would? The electron density should be uniform across the molecule. An even better example of where lewis structures really have problems is with diborane. Now, SMILES is based on Lewis theory, so it may have the same sort of problems with properly dealing with fractional bond order as drawing Lewis diagrams on paper does. --Jayron3221:45, 29 September 2011 (UTC)[reply]
You say: "It may have the same sort of problems with properly dealing with fractional bond order as drawing Lewis diagrams on paper does." Although there is a SMILES representation for fractional bonds - for bond orders between 1 and 2, there is ":". Bonds that lie between 1 and 2, are represented in Lewis theory as a double line, one solid, the other dashed. Therefore a bond that lies between 0 and 1 should be represented by a single dashed line in Lewis theory. Plasmic Physics (talk) 01:59, 30 September 2011 (UTC)[reply]
Yeah, but that sort of representation is a sort-of "workaround" for Lewis structures, just like "bond resonance" is. Its a somewhat inelegent solution (for example, the "line w/ dash": What fraction does it represent? How does it distinguish between a 3/2 bond order and a 4/3 bond order?). Lewis structures are useful, to a point, once you understand their limitation. --Jayron3202:20, 30 September 2011 (UTC)[reply]
Well, of course, but then again if you are using the valencies of the surrounding atoms, a single line would do equally as well. In general, the standard practice when working in Lewis structures is to write all possible resonance forms, like this: [10]. The dashed-line thing is a later innovation to save time from having to write out all of the contributing resonance forms. That practice is common, but not part of the "standard" rules for working with lewis structures. --Jayron3203:10, 30 September 2011 (UTC)[reply]
A single line doesn't make sense, because it always depicts a unity bond order. That kind of resonance doesn't exist, you should know that. At all times are there three 1 ⅓ bonds. The bonds, don't alternate between 1 and 2 as is depicted in Lewis structures. As far as I know resonance structures was the norm before anyone actually knew what really happens, like the deferent and epicycle theory in Ptolemaic astronomy, before the heliocentric nature of the solar system was known. The single line approach for fractional bonds doesn't make sense in diborane, it would imply that hydrogen is somehow sharing two electrons. Plasmic Physics (talk) 03:43, 30 September 2011 (UTC)[reply]
Well of course, resonance is not a real thing. That's the shortcoming of Lewis Structures. I believe I already said that. There are workarounds to deal with the problems, but they are still workarounds and not part of the actual rules for working with lewis structurs. That's the whole problem Restating my words back at me doesn't mean I didn't already say them in the first place. There is not an adequate, elegant method for handling fractional bond order in Lewis structures. You can invent concepts (resonance, the dashed line thingy, etc) which attempt to deal with the problem. Those workarounds don't get at the fundemental problem, which is that Lewis structures aren't well designed to deal with fractional bond order, and sometimes (as with dioxygen) they just get it completely wrong in ways that cannot be fixed. "All models are wrong. Some models are useful..." --Jayron3203:52, 30 September 2011 (UTC)[reply]
Not much to go on, but any suggestions would be welcome. Seen today in ancient woodland, just north of London in south-east England. Apparently a beetle with wing-cases that appeared to fully cover the wings, but similar size and shape to a shield bug, or maybe a little smaller. Its back was silver-grey and rough, so that it resembled a piece of lichen. Sorry, no photograph. Any ideas anyone? Alansplodge (talk) 12:55, 29 September 2011 (UTC)[reply]
Without a picture it's impossible. One thing to remember though: true bugs (Hemiptera) have four membranous wings. Their "wing cases" (the scutellum) also do not split down the middle. If this is the case and the scutellum more or less covers the abdomen entirely, then it is a jewel bug/shield-backed bug (Scutelleridae, the hemipterans which resemble beetles most closely).
True beetles (Coleoptera), on the other hand only have one pair of membranous wings. Their "wing cases" (the elytra), are actually hardened forewings. So unlike scutellerids, it does split down the middle. Another thing to look at is mouthparts. Hemipterans will invariably have sucking needle-like mouthparts (a rostrum) while beetles will have chewing or biting mandibles (although in some cases, like weevils, they may be so greatly modified or small, it's hard to recognize them).-- Obsidi♠nSoul13:09, 29 September 2011 (UTC)[reply]
It's a really diverse woodland that was enclosed into a park about 150 years ago. Dominant species are oak, hornbeam, sycamore and alder with a mid-layer of hazel, hawthorn, elder, sallow; plus old specemins of spruce, beech and sweet chestnut. Easily 20 species in about 5 acres. Not much help huh? Alansplodge (talk) 16:59, 29 September 2011 (UTC)[reply]
No, it sits on London clay and Reading Beds which is a sandwich of sand, clay and pebbles. I scoured the internet and some general guidebooks before posting, but couldn't see anything like it. It was found by a 9 year old from a primary school class and I said that I'd find out what it was. I was hoping that the lichen-like appearance would ring a bell with someone. Alansplodge (talk) 09:44, 30 September 2011 (UTC)[reply]
Me again; after another web search, I think I might have got it. The Fungus weevil seems to fit the bill. "...a very strange one. It looks like it is covered in fungi and it even has small brushes of hairs growing on top... They are closely related to the ordinary weevils, only their snout is large and flat."[11]. The lack of a weevil-type nose stopped me from looking in that department before. The picture doesn't look exactly like mine, but there are many species. Many thanks for all your efforts, and apologies that I led you up the "shield bug" path, but it was the nearest thing I could think of. Alansplodge (talk) 10:28, 30 September 2011 (UTC)[reply]
The previous question brought up another one for me: why is Hemiptera the order of "true bugs"? I mean, what's special about about them that they get to claim to be the only "true" bugs, and everyone else is just a misnomer?
The OED says that the word "bug" has been used generically to mean any miscellaneous insect as early as the 17th century. It's not clear from it (to me) when the specific usage of "bug" for only Hemiptera became important. The use of "bug" in the context of "bed bugs" seems to go back very far, but it's not clear to me why that term would then be generalized to that specific order (as opposed to their family).
Anyone have any real insight on this, or is this just some silly scientific custom? I've always found it very strange to insist, "oh, that's not a true bug..." --Mr.98 (talk) 14:51, 29 September 2011 (UTC)[reply]
First of all, it's not a scientific custom at all, it's a silly laypeople custom. Biological sciences themselves do not rely on common names. The reason for the ambiguity is because in everyday language, names for organisms are very very arbitrary (also why I always prefer to create taxonomical articles with their scientific names unless they have an unambiguous common name widely understood).
Oh, and bed bugs are also hemipterans. And I'm not sure what you mean by "generalized to that specific order (as opposed to their family)"... bedbugs were not the first insects called 'bugs'. Obvious in the way it requires a modifier - 'bed'.
Though "bug" certainly had more widely applicable etymons, there was a certain time when they begin to be applied specifically to certain kinds of insects (though I do not know when), which turned out to be hemipterans. People back then weren't exactly discriminating when it comes to which is which in such things. Another example is 'hornet', which was once applied to any large noisy flying insect in Old English. If someone calls a beetle a 'hornet' now, would you also think it silly if someone says "oh, that's not a hornet..."? What about if someone calls a marmoset an "ape", a dolphin as a "fish", a harvestman a "spider", a hyena a "dog" etc.? Adapting common names to point to an exact taxon minimizes confusion.
In taxonomy, they are simply hemipterans and coleopterans. That's it. No ambiguity whatsoever despite their superficial similarity to a layman.-- Obsidi♠nSoul15:31, 29 September 2011 (UTC)[reply]
My understanding from the OED is that bed bugs were called "bugges" well before the "bed" got added. Apparently that goes back to the 17th century as well. Anyway, the above doesn't really explain anything other than saying, "it's arbitrary and I don't know when it started." --Mr.98 (talk) 16:16, 29 September 2011 (UTC)[reply]
Go back further than that and it was used to refer to beetles, grubs, and lice. Its application to bedbugs in M.E. may have influenced later usage specifically for rounded large insects which have sucking mouthparts. It's really common sense isn't it? Biological taxa have very specific delineations where one organism can not be a member of another, and hemipterans and coleopterans are quite distantly related, despite both being insects. Unless you're quite adamant with the "science is silly" thing, then discussion is irrelevant.
Like mentioned, dolphins and whales were also "fish" just a few centuries ago. In some languages today, they still are. Should we go back to that and also call them fish, and call people who insist that they are not "true fish" silly? There's a reason why mere morphology is not used for classifying organisms. Otherwise, we wouldn't have bats or birds, we'd have mouseflies and featherflies.-- Obsidi♠nSoul16:40, 29 September 2011 (UTC)[reply]
I'm asking a historical question — when "true bug" became a meaningful distinction. I'm not saying "science is silly". I'm asking when it started. You don't seem to know and don't seem to have the means to find out. I am not looking for a lecture in generic taxonomy. I know what taxonomy is and how it works. --Mr.98 (talk) 18:55, 29 September 2011 (UTC)[reply]
Then this should be in language shouldn't it? If you're wondering why I feel obliged to answer when I don't think there even is an answer is that I was the poster who replied with the "oh, that's not a true bug". And your OP makes it feel like I had no basis in doing so, which is more than a little insulting. Also 'generic taxonomy' refers to genus-level taxonomy. I'm sure that wasn't what you meant, since we aren't exactly talking about genera here.
Anyway, when Linnaeus separated Insecta into different groups, he based his classification on wing form. He established Coleoptera and Hemiptera ('sheath wing' and 'half wing' respectively) in the 10th edition of his Systema Naturae (1758 - 1759). When you look at the insects he separated (see 1806 English translation), you realize that members of the first group tend to have common names in English that end in beetle (particularly those known to bite - rove beetles for example). Also see William Fordyce Mavor's The elements of natural history (1806) and Oliver Goldsmith in A history of the earth, and animated nature (1804).
It's not a stretch to imagine that Hemiptera became 'true bugs' to distinguish them from 'false bugs'. Including the beetles, which during that time were still lumped together with the term 'bugs' (cf. ladybugs, june bugs, horn bugs). When the 'false bugs' have been stripped out, you really are left with a group that is mostly composed of hemipterans - hence true bugs. And like I said earlier, it's M.E. application for bedbugs may have influenced subsequent adoption as a group name for Hemiptera since bedbugs were the best known of the insects already called bugs as well. And they exhibit a number of distinctive features shared by all Hemipterans - piercing mouthpart, shape, smell - which are so evident even from cursory inspection by a layman that they probably became the diagnostic criteria for the group. Other group common names have originated in similar ways.
What else is there anyway? All I know is that calling beetles 'bugs' as if they belong to the same insect group is confusing and wrong. Like calling a panda a 'bear'. -- Obsidi♠nSoul21:54, 29 September 2011 (UTC)[reply]
Some know-it-all declaring that only this type of bug is really a "bug" is reminiscent of the half-educated person declaring that "classical music" was only composed from 1750 to 1830, and that Baroque or romantic music is not "classical." Broadly speaking, classical music is classical music, and bugs are bugs. I listen to one on the classical music FM station, and I squash the other. (Figuring out which is left as an exercise for the reader). Edison (talk) 06:00, 30 September 2011 (UTC)[reply]
Go ahead and call anything that crawls a 'bug'. You sound very educated and doesn't-know-it-all-ish when you do so, broadly speaking. Next time someone asks for an ID of an insect here, just say "that's a bug, squash it". In fact, let's just reinvent taxonomy altogether. I propose we start calling Arthropoda "creepy crawlies"; Insecta as "bugs"; Coleoptera "fat bugs"; Lepidoptera "pretty bugs" (adult), "fugly bug" (larva); Hymenoptera "run away from this bug"; Orthoptera "noisy bugs"; Diptera "annoying bugs"; Odonata "bigass bugs"; etc. After that, we all dump our biology classes (if you haven't already) and just listen to FM to get our full dose of edumacationz. -- Obsidi♠nSoul10:53, 30 September 2011 (UTC)[reply]
In England, people say sometimes say "bug", but they generally know that it's American slang. In general parlance, most people say "insect", or "creepy-crawley" for anything else that obviously isn't an insect but is in that general class of creature. It often comes as a surprise to laymen to find that there are things that can properly be called "bugs". Alansplodge (talk) 11:56, 30 September 2011 (UTC)[reply]
Can it be said that ethanol is an energy carrier rather than an energy source?
It can certainly be said. It could then be debated, but it seems reasonable on the surface. Note, however, that "dead organic material" will then reduce to "sunlight". — Lomn14:59, 29 September 2011 (UTC)[reply]
It probably depends on what you mean by "efficient" (particularly, how far back in the manufacturing process are you looking, how far forward are you willing to look for future gains, and what are you optimizing?) Note also that, for portable fuels, energy density is a particularly relevant value. Per our chart, batteries are particularly bad, ethanol is middle-of-the-road, and gasoline/diesel quite useful. Hydrogen is an interesting case in that it's exceptional on a per-mass basis but handicapped per-volume since you can't conveniently store it as a liquid. — Lomn15:17, 29 September 2011 (UTC)[reply]
The article, Relativity of simultaneity, is also a good introduction to this conceptual idea. The trouble is, "the moment of departure" is not a universal "moment." It is a coordinate - an (x,y,z,t) coordinate for a specific position and time. One of the first things you need to know about relativity is that coordinates are relative to an observer's frame of reference. The Lorentz transform provides the most simple method to calculate a coordinate transform between two observers. However you want to look at this: if an object moves faster than light, either causality is violated, or the equations we currently model spacetime with are violated. The most succinct way I can think to state this: for v > c, proper time is poorly-defined (it is the square root of a negative number). Nimur (talk) 17:12, 29 September 2011 (UTC)[reply]
I can imagine there are perhaps methods to restrict or restore causality using strings?... and there appears to be possible extensions to the Standard model, but I am not familiar with them. What I am familiar with though is a model that I've been working on which directly and succinctly addresses this whole topic, and you guys just might like, :-), what I have when you get to see it because I plan to blog on it, although I am not sure yet what date I'll start posting. But when I do I'll put a link, to my blog, on my Wikipedia user page. I almost never update that page thus if you watch it and it eventually turns up in your watchlist, its because I either added the link or, alternatively, I'm giving some lame excuse for not following though with this. --Modocc (talk) 20:54, 29 September 2011 (UTC)[reply]
I don't think this question can be sensibly answered. There's a standard answer that assumes Lorentz invariance, but the very existence of faster-than-light propagation would seem to be strong evidence that Lorentz invariance is wrong. The traditional idea of tachyons from the early days of special relativity (1910s, 1920s) just isn't relevant any more. In modern field theory, everything propagates at light speed, even "tachyons". (See this page, particularly the part about unit propagation velocity.) In the wake of the OPERA paper I've heard people suggest that neutrinos could take a "shortcut through extra dimensions", but as far as I know this is simply a way of breaking Lorentz invariance—the shape of the extra dimensions violates the symmetry. Likewise I've heard people say that loop quantum gravity allows for different maximum propagation speeds for different particles, but again LQG breaks Lorentz invariance. The principle of causality, that "things can't arrive before they leave", seems much more likely to survive a shakeup of physics than the principle of Lorentz invariance. Assuming, of course, that the neutrinos really are superluminal, which they almost certainly aren't. Even the OPERA team thinks it's a measurement error. -- BenRG (talk) 20:22, 29 September 2011 (UTC)[reply]
Personally if true tachyons were found, I think it is much easier to accept that causality and time as we experience it are merely an illusion, then to suggest that some version of Lorentz invariance (or at least a close approximation thereof) isn't true. There are ways to cheat by adding extra dimensions and making small modifications to Lorentz invariance, but you can't change it very much without diverging from lots of experimental results. On the other hand, most of physics can be understood just fine without invoking any notion of causation. In that sense, I personally think that causation is more peripheral and superfluous to physics than Lorentz invariance is. Dragons flight (talk) 21:24, 29 September 2011 (UTC)[reply]
Causality is extremely important to physics theories. Without it physics as a science loses the power of predictability since things could happen for no reason at all. That would be a serious departure from centuries of physics dogma. Lorentz invariance violation could possibly be accommodated, but it wouldn't be easy. Dauto (talk) 21:56, 29 September 2011 (UTC)[reply]
No, most physics is deterministic not causal. To give simple examples, Newton's laws "predict" the time evolution of a system as time evolves. However, those laws are completely time invariant. One can say that the future state must be X because the past state was Y, but one can just as easily say that the past state must be Y because the future state is X. The direction of time is irrelevant and no notion of X causes Y or Y causes X is actually needed. Most (though not all) of physics is deterministic in principle, though in practice chaos and the impracticality of measurement make many time evolutions impossible to predict either forward or backward very far. There are of course exceptions, especially in quantum mechanics, that appear to be non-deterministic or not time-invariant, but that could be because we don't really understand the way they are determined. And of course some things that appear non-deterministic are still not causal (e.g. when a quantum entangled state decays it isn't actually meaningful to say that measuring one end caused the other end to decay). Dragons flight (talk) 23:43, 29 September 2011 (UTC)[reply]
If causality is violated then determinism goes with it (and that's the problem) because at any time non-causal unpredictable information might come come from no-where and influence physical states. Dauto (talk) 00:45, 30 September 2011 (UTC)[reply]
No. You can have a deterministic universe that is non-causal, it depends on the nature of the causality violation. If tachyons exist, for example, then the universe is non-causal because events can by influenced by other events separated from them by space-like intervals (and asking which event precedes or follows the other is an arbitrary matter of perspective). However the universe can still be deterministic if all observers would agree on the set of events that must occur (up to changes of reference frame and disagreements about event ordering) assuming those observers are also granted global rather than local knowledge of the system. Dragons flight (talk) 01:10, 30 September 2011 (UTC)[reply]
As BenRG explains below, a non-causal theory has causal loops which introduce unpredictability in the theory because there is no way to tell the state of those loops from previous or future events since those loops are caused by themselves, nothing else. Those unpredictable loops can affect other events of the theory making it non-deterministic. Dauto (talk) 02:41, 30 September 2011 (UTC)[reply]
To me causality just means the absence of closed causal loops, and I have trouble understanding what it would mean for a theory to have closed causal loops. It seems akin to ill-founded induction or a set theory that suffers from Russell's paradox. It's not a property a theory can have so much as it's a proof that you don't actually have a theory. On the other hand I have no trouble imagining consistent theories that are approximately but not exactly Lorentz invariant (such as a luminiferous aether!). That's not to say that I "want" Lorentz symmetry to be violated, and of course right now there's no evidence that it is. -- BenRG (talk) 01:01, 30 September 2011 (UTC)[reply]
the abillity to differentiate
in RGB how many points do I need to move for a human viewer to differentiate between color A to color B?
example:
I differentiate between the color blue and the color pale blue.
Exx8 (talk) —Preceding undated comment added 18:06, 29 September 2011 (UTC).[reply]
According to the International Commission on Illumination, an average human can identify around 2.4 million shades of red/green/blue combinations. By "points", I think you mean points on a 0-255 scale. Then, it would be about 7 points for an average human across the entire scale. A single human tested could be higher or lower. Further, sensitivity to red, green, or blue shades should be very different within the individual. -- kainaw™18:13, 29 September 2011 (UTC)[reply]
It would also depend on what colour and how you're changing it, since different cones are responsibly for different colours. So, for example, one individual might perceive changes in the red range of colours more easily than changes in the green range. --Goodbye Galaxy (talk) 18:30, 29 September 2011 (UTC)[reply]
It is not quite true that "different cones are responsible for different colors." That implies a different cone type for each discernable hue. Instead, each type of cone has a certain spectral sensitivity function, the curves overlapping, and hue discrimination require comparison by the nervous system of the differential responses by the different types of cones. Edison (talk) 19:48, 29 September 2011 (UTC)[reply]
one individual might perceive changes in the red range of colours more easily than changes in the green range. I know that was just an example and would be true for at least "one individual" ;) however just out of interest, in general, people can differentiate more shades of green then any other color. Vespine (talk) 22:33, 29 September 2011 (UTC)[reply]
Hmm. It looks like scale insects only range up to 1/2" long (Magnolia scale), and this thing was easily double that. --Sean20:05, 29 September 2011 (UTC)[reply]
I'm afraid I didn't get a good look at them. I was thinking caterpillar, too, based on the horns. I looked all over the "what's that bug" and "bug guide" sites but didn't see anything similar. --Sean20:28, 29 September 2011 (UTC)[reply]
Its an Isa textula or "Crowned slug moth caterpiller". For the record, I typed "fuzzy green caterpillar" into google images and got the answer that way. Google is still awesome. See [12]. --Jayron3221:49, 29 September 2011 (UTC)[reply]
(EC with below) Not exactly. Sea salt generally lacks any fortifications like iodine that may be present in more ordinary table salt where you live. Such fortifications are usually considered beneficial for a significant proportion of people hence why health authorities and governments mandate them and sea salt (and other unfortified salts) don't tend to naturally have them in significant amounts. I'm not aware any of the additional minerals/impurities present in sea salt are considered particularly beneficial in the quantities present. So it's probably fair to say sea salt may be less healthy for a significant number of people in some places. Nil Einne (talk) 21:04, 29 September 2011 (UTC)[reply]
Meaning, the salt in question is only as healthy as the additives contributed to it which, brings us back to the point: any differences in health, is due to what has been left, or added in. Plasmic Physics (talk) 22:59, 29 September 2011 (UTC)[reply]
I have a tangential question. Which parts of the world obtain their salt mainly from mining, and which parts by evaporation of sea water. I get the distinct impression from both this thread and the Sea salt article that there is some distorted view that "normal" salt does not come from the sea. That may be the case where a lot of our editors live (USA? UK?), but is not the case everywhere. I also note that "Sea salt" is a trendy fashion item in modern western cooking. When something is associated with fashion, logic (and often, knowledge) regarding its best use goes out the window. HiLo48 (talk) 23:14, 29 September 2011 (UTC)[reply]
Virtually all salt in California, at least, comes from evaporation of ocean water. Nevertheless "sea salt" is sold in stores. It's a marketing thing. Notionally I think "sea salt" is supposed to be less processed than "regular salt", but I don't know whether it actually is less processed. --Trovatore (talk) 23:34, 29 September 2011 (UTC)[reply]
Hmm, it's possible I had a misimpression on this. There are large salt ponds near where I live, but they seem to be operated by Cargill, which is not a brand that I see much. The most recognized table salt brand around here is Morton Salt. I don't know where they get their salt. --Trovatore (talk) 00:17, 30 September 2011 (UTC)[reply]
I can't speak for the USA, but given the size of the place and the distance of most of it from an ocean coastline I'd hazard that a large amount of salt in the USA is mined, not gently evaporated from limpid salt pools beside the dreamy ocean. Same in the UK - we have great big salt mines in Britain. See for instance Salt in Cheshire and Droitwich Spa, from where salt has been extracted since prehistoric times. Droitwich is pretty much as far from the sea as you can get in the UK. TonywaltonTalk23:47, 29 September 2011 (UTC)[reply]
Yes, I didn't want to sound rude, but I couldn't imagine that the climate of the UK would be all that conducive to evaporating sea water. ;-) HiLo48 (talk) 00:31, 30 September 2011 (UTC)[reply]
Actually, most of our salt (especially that intended for food consumption) is indeed evaporated from the ocean, mostly in California and along the Gulf Coast, as well as on the southern Atlantic seaboard. (Not all of it, however, is evaporated "from limpid salt pools beside the dreamy ocean" -- some of it (most notably in Tampa, FL) is produced by flash distillation in desalination plants.) There are numerous salt mines in the Midwest, but these produce salt mainly for industrial use (especially for the production of chlorine and sodium hydroxide). As for climate, please note that in Russia, some of the salt is (used to be, at any rate) produced from the water of the White Sea (near Arkangel!), not by solar evaporation, but by fractional freezing followed by hot distillation of the resulting concentrated brine. 67.169.177.176 (talk) 00:58, 30 September 2011 (UTC)[reply]
Salt is a bulk good that is not all that difficult to haul around, so it wouldn't be very surprising for primarily sea salt to be used thousands of miles from the coast. Googlemeister (talk) 13:10, 30 September 2011 (UTC)[reply]
The Salt Institute seems to have the opposite opinion that because it is a low cost high weight bulk item it is most economic when consumed near where it is produced. However their FAQs [15] are heavily weighted to industry, pointing out that the large majority of U.S. industrially-used salt is produced by solution mining wells (not sea salt or rock salt). But they don't answer the question of how much of edible salt is sea salt versus rock salt. 75.41.110.200 (talk) 16:16, 30 September 2011 (UTC)[reply]
Do consider that table salt is probably not the main consumer of salt in the US. I expect more is used for road maintenance in winter, for chemical production or for the raising of livestock, then is eaten by humans. Googlemeister (talk) 18:56, 30 September 2011 (UTC)[reply]
Both Mayo Clinic [16] and Massachusetts Department of Public Health [17] say that table salt is rock salt. ("Table salt is generally sold in very small grains, and is mined from underground salt deposits. ... It is commonly used in salt shakers and during cooking and baking.") do we have any sources for 67.169.177.176's claim that most U.S. table salt is sea salt? 75.41.110.200 (talk) 16:28, 30 September 2011 (UTC)[reply]
It may also be worth noting that nearly all rock salt also ultimately formed by the evaporation of sea water, except that the evaporation happened naturally when a pool of ocean water got cutoff from the ocean, possibly many millions of years ago. Creating salt from sea water evaporation today is a more controlled and arguably cleaner process than what happened to form natural evaporite deposits, but the starting material is the same. Dragons flight (talk) 19:09, 30 September 2011 (UTC)[reply]
For many years (5-10) an oak tree in my house's front yard would be infested with wasps, which caused a fair amount of terror for me in the years I had to mow the yard while they were there. My family never found a wasp nest (although that's not saying that one wasn't there), they just seemed to really like that tree (and that tree only, given that no other trees in the neighborhood had wasps)...enough so that they would remain there after we would spray the tree with wasp killer. The wasps just randomly didn't come back after winter one year, which kinda surprised me, and since then the tree has gradually (for the first time I can remember) started producing acorns. My two questions are:
Why did these wasps seem to have a liking for this oak tree?
I've frequently heard from people an oft repeated saying about wasps and bees that "if you don't bother them, they won't bother you" whenever I get freaked out when there are any around, usually said as a rationale for me to just ignore them and not leave the area as quickly as possible. I find this very dubious based on the fact that, to me, they seem to be "bothered" by my very presence anywhere near them (they won't leave me alone) despite the fact I'm not doing anything that these people would consider "bothering them". Is there truth in the phrase "if you don't bother them, they won't bother you" when it comes to wasps and bees, and if so, what constitutes "not bothering them" given that they don't leave me alone whenever I'm near them?
For reference about species, my location that these questions are based off of is Salina, Kansas for the former and both Salina and Norman, Oklahoma for the latter. I don't know the species of the oak tree, but it appears to be about 25–35 feet (7.6–10.7 m) tall, at most 20 feet (6.1 m) wide and I'm pretty sure it's been there since at least 1995 (the earliest I can remember). Thanks in advance, Ks0stm(T•C•G•E)22:40, 29 September 2011 (UTC)[reply]
Wasps are more aggressively territorial than bees, because stinging doesn't kill them. Generally neither will attack without provocation though, but a noisy lawnmower could have easily resulted in accidental provocation. I suspect that the tree was diseased, allowing the wasps to build a nest inside it, and then the nest became abandoned and the tree was able to recover. 69.171.160.237 (talk) 23:09, 29 September 2011 (UTC)[reply]
What kind of wasps? Oak trees are actually one of the most favorite trees for gall-making wasps, and they have hundreds of species, most preferring oak. You will not find a wasp nest because the wasps themselves are solitary and nest in the oak tissue. It's actually fascinating stuff, a kind of natural genetic engineering.
Wasps inject something into developing acorns, mutating it into something quite unlike an acorn (probably the reason why you've never come across a 'normal' acorn until that year they weren't around). Hard and perfect for developing wasp larva. Other species develop galls in other parts of the plant. Leaf buds, for example, result instead in what is known as oak apples. Still others produce galls on the roots, which are then called oak potatoes. See gall wasp. -- Obsidi♠nSoul23:31, 29 September 2011 (UTC)[reply]
Oh, and like most solitary wasps, they are not aggressive so don't worry. As long as you don't swat one or cuddle one or something, heh. Social wasps, like yellowjackets, which build colonies are the ones to watch out for.-- Obsidi♠nSoul23:36, 29 September 2011 (UTC)[reply]
What a great vespine question :) . Most of this was news to me, I chose my name because I like the word and I like wasps, not because i know particularly much about them. Vespine (talk) 01:08, 30 September 2011 (UTC)[reply]
My experience with yellowjackets is that they buzz around a lot and go for intimidation, but rarely sting. I don't guarantee that that will be anyone else's experience. --Trovatore (talk) 01:11, 30 September 2011 (UTC)[reply]
Gall wasps, which, where I live, mostly infest twigs and leaves, are solitary, and extremely small, like midges. (As for yellow jackets, I have been stung by them more often than by honeybees.) μηδείς (talk) 03:57, 30 September 2011 (UTC)[reply]
Growth of the Universe
Human growth is done by the mitosis of cells. The universe is growing at an infinite rate, but it isn't made of cells. The only real thing that can grow the universe is atoms, but how do atoms grow the universe? These atoms can't just appear, right? 64.229.181.189 (talk) 23:50, 29 September 2011 (UTC)[reply]
The existing atoms are just moving further apart from one another, taking up more space. Oh, and the rate of growth is quite measurable. It's not infinite. HiLo48 (talk) 00:28, 30 September 2011 (UTC)[reply]
Think of the universe as a rubber band. Stretching a rubber band doesn't mean that all of a sudden you have more rubber than before. Everyone would agree that is a silly conclusion, otherwise the band would get heavier. (It does somewhat, but that is due to a different reason.) Plasmic Physics (talk) 00:40, 30 September 2011 (UTC)[reply]
Yeah, but that is not the reason that the universe is growing. Infact, the word "growing" isn't even applicable here. The universe is stretching. It's not gaining net energy/mass (mass-energy equivalence) as what growing involves, it's simply redistributing itself. Plasmic Physics (talk) 01:50, 30 September 2011 (UTC)[reply]
HiLo48 is correct that the rate is measurable and not infinite, but atoms per se are not moving apart. For example, the atoms in your body are not moving apart. They are electrostatically bound, just as the stars within galaxies are gravitically bound. It is the universe as a whole which is expanding (stretching and growing have implications which make those words less than ideal) and the space between gravitically non-bound galaxies is increasing. But there are plenty of galaxies which are actually moving towards each other, such as ours and the Andromeda galaxy. μηδείς (talk) 03:52, 30 September 2011 (UTC)[reply]
Historically, the supposed spontaneous creation of new atoms indeed formed part of the Steady State theory (aka Continuous Creation), which was formulated (by several leading astrophysicists of the day) to explain the then-recently observed expansion of the Universe together with the inferred maintenance of its density over deep time. The atoms' creation could be viewed as a further consequence of the driving force behind the expansion, or - in some interpretations - as the actual cause of it. Subsequently accumulated observational evidence has of course favoured the rival Big Bang theory, and Steady State, despite corrective modifications by its proponents, has largely fallen out of favour. {The poster formerly known as 87.81.230.195} 90.197.66.221 (talk) 14:18, 30 September 2011 (UTC)[reply]
I don't know. When you added glucose to a mixture of hydrochloric acid and iodine in your chemistry lab, what happened for you? --Jayron3219:09, 30 September 2011 (UTC)[reply]
Warning: the topic below will require some basic knowledge of music theory, physics of sound harmonics and radio astronomy.
Hi. Last night, I was experimenting with my radio. It was on the AM radio dial, and at approximately 600 AM (probably a bit lower), an ambient humming sound filled the room. I found that the beat frequency was almost exactly 0.5 Hertz, and that the pitch was at an E natural, more than two octaves higher than middle C. At frequencies of both 640 and 680 AM, I detected a near-monotome, static buzzing hovering at B flat, just below middle C. Experimenting further on my electric keyboard, I found that the distance between these two notes was two octaves plus six semitones. Thus, I found the note between these two, as well as another note of a lower pitch at the same frequency. The chord formed by these four notes is a G-diminished 7th.
My question is: what causes these specific channels to produce these ambient buzzing and humming sounds, even in the lack of clear channel reception? What is the Hertz interval between the two main notes mentioned (B flat below middle C and E natural 2 octaves above)? Moreover, what actually caused the 0.5 Hertz beat? Thanks. ~AH1(discuss!)14:15, 30 September 2011 (UTC)[reply]
You are probably hearing intermodulation distortion (tone injection) due to complicated interference between the transmitter's carrier frequency and some other frequency (possibly your local oscillator, or some other frequency in your radio). Do you know what type of radio you have? Do you know what type of signal you're tuning? (If it's commercial AM radio in the United States, it's probably DSB-SC; but you can still get intermodulation distortion even without a tonal carrier, it's just harder to describe in simple terms).
Also, to convert musical-note to numeric frequency, you only need to know this equation:
...where n is the number of half-tones above f0. For example, Middle C is three half-tones above A440 - so f = 440 Hz × 20.25 or about 523 Hz.
Finally: to diagnose the root-cause of a 0.5 Hz beat, we'll... need some more technical information. How much do you know about your radio? Does your receiver have a built-in frequency auto-tuner? AGC? ... Can you hook a spectrum analyzer up and see what else is going on? If you can't answer those more precisely, we probably can't say any more than "it's probably interference from something." These kinds of questions are better probed with the sort of equipment you find in a HAM base station, rather than an off-the-shelf AM radio. Nimur (talk) 18:25, 30 September 2011 (UTC)[reply]
What is the average size of human tonsils?
I've read the WP article on tonsils including the passage "Tonsils tend to reach their largest size near puberty, and they gradually undergo atrophy thereafter. However, they are largest relative to the diameter of the throat in young children." So what is that largest size, on average? I don't have a medical encyclopedia handy but an initial G-search proved fruitless. §everal⇒|Times19:03, 30 September 2011 (UTC)[reply]
What measure are you looking for? This paper PMID 20304270 reports tonsil size (in terms of height, width, thickness, and weight) in overweight and normal-weight children who have problems with breathing during sleep. If that does not answer your question, perhaps you could provide a more specific question. -- Scray (talk) 12:16, 1 October 2011 (UTC)[reply]
Why exactly does the enthalpy of vaporization fall to zero at the critical point?
Is it because the heat capacity of the gas ends up increasing much faster than the liquid? If so, this qualitative explanation doesn't appear to be included in our critical point (thermodynamics) article, though it delves into derivatives. And if so, why does this overtaking trend accelerate ? I note that 10 degrees Celsius below the critical point, the heat of vaporization is still considerable, but at the critical point, this disappears completely. elle vécut heureuseà jamais (be free) 19:43, 30 September 2011 (UTC)[reply]
The reason is actually tautological. At the critical point, the substance becomes a Supercritical fluid, for which there is no distinction between a gas and a liquid. The fact that the enthalpy of vaporization at this point drops to zero is because vaporization has no meaning above the critical point. The two statements (enthalpy of vaporization = zero, distinction between liquid phase and vapor phase doesn't exist) are simply two statements of the same fact; which is the critical point of the substance. That is what the critical point is. --Jayron3219:48, 30 September 2011 (UTC)[reply]
Well yes, but I was looking for a more quantitative explanation. This doesn't really solve problems of the heap: what's the difference in enthalpy of vaporization for a substance 0.000001 K below the critical point versus at the critical point? Surely we cannot resolve problems of analysing a quantitative trend by merely using hand-waving definitions. AFAIK, the two distinct phases don't exist because Hvap is zero, not the other way round. The two phases disappearing is a result of this fact. So I want to know -- what causes this fact? elle vécut heureuseà jamais (be free) 20:00, 30 September 2011 (UTC)[reply]
What causes the critical point? The article Critical point (thermodynamics) covers that rather well, I thought. The reason that the critical point exists is that the kinetic energy of the molecules overcomes the energy holding the molecules together in the condensed phase. After you exceed that temperature, no amount of compression will cause the molecules to condense. In some ways, you can think of the enthalphy of vaporization as related to this energy difference. In one perspective, you can think of the enthalpy of vaporization as the extra energy needed for that fraction of molecules not yet in the vapor phase to get into the vapor phase. The closer you get to the critical point, the greater proportion of molecules have that necessary energy, so the less extra energy you'd have to supply to get there. The article Enthalpy of vaporization has a very nice graphy which shows that the enthalpy of vaporization decreases until it reaches zero. The difference between the critical temperature and an arbitrary temperature below that is that, at the lower temperature enough molecules have a low enough kinetic energy that they cannot overcome the intermolecular forces holding them together, if there is enough pressure. At the critical temperature, this number of molecules functionally drops to zero. --Jayron3220:56, 30 September 2011 (UTC)[reply]
Yes, but that's all nice and qualitative. I guess I want a semi-quantitative explanation (e.g. a relation like dG/dT = -S) for why heat of vaporization should accelerate to zero near the critical point. What exactly is going on? Why does it accelerate? elle vécut heureuseà jamais (be free) 21:51, 30 September 2011 (UTC)[reply]
Oh. In that case I think you want to look at the Clausius–Clapeyron relation for a general mathematical treatment of phase changes. This page discusses some of the mathematics involved at the critical point. This page here represents some early research on the topic, and discusses the derivation of the relationship between heat of vaporization and critical temperature. --Jayron3223:34, 30 September 2011 (UTC)[reply]
Unidentified bird in France
What is this?
Does anyone know what kind of bird this is? I thought it was some sort of finch, and then I realized, hey, I don't know anything about birds (and even if it is a finch, I have no idea what kind). If it helps, this was taken in Nantes, France. Apologies for the rather inadequate photo. Adam Bishop (talk) 20:44, 30 September 2011 (UTC)[reply]
How would you calculate the change in freezing point (with increasing pressure) without the Clausius-Clapeyron equation?
At first if am only given density difference, difference in pressures, and the freezing point at one temperature. If I assume the change in entropy of vaporization is constant -- +22.014 J/(K*mol) and I know the change in volume per mol is +217.05 cc/mol (yup this is water). Then the difference in atmospheric work between 10 MPa and 1 atm is 2.0785 kJ/mol.
ΔS = ΔH/T
T = ΔH / ΔS
then our new temperature point = ΔH/ΔS = (6.01 kJ/mol - 2.0785 kJ/mol) / +22.014 J/K = 178.6 K = -94.6 C
This is a very large change in the melting point of the ice.
But if I use Clausius-Clapeyron engineering equation (which I don't think I'm supposed to do):
Which brings me below absolute zero (if I flip this -- I get 98.35 C)
If I use the slope equation mentioned in the Clausius-Clapeyron article I get
ΔT = ΔP * T * ΔV / L = (10 MPa - 1 atm) * 273 K * 0.875 cc/mol / 8.68 kJ/mol = -0.272 Kelvin
Which is wayyy less than the ~100K drop predicted before.
Just trying to comprehend something the teacher said in class. So if I have 8 identical point charges (protons let's say) positioned at the corners of a cube (like a physical octopole except all charges have the same polarity), then if I place a positive test charge at the center of the cube, the test charge will be in equilibrium, right? From what I understand of Earnshaw's theorem, the test charge will be in equilibrium but it won't be in a stable equilibrium so if I perturb the test charge slightly (just push it in any one direction), the test charge will fly away as fast it possibly can. My teacher sounded like as if there is no equilibrium and no matter what (even theoretically) the charge just won't stay at the center. I was like how can this be. If I put a charge dead center it will stay there. All of the components of the force vectors should cancel. In the real world of course, I understand its impossible to confine a single charge like this let alone plasma or something. Am I right?128.138.138.122 (talk) 22:38, 30 September 2011 (UTC)[reply]
That isn't quite an accurate description. Instability means that if you push the charge away from the center, it will start to move farther away. It does not necessarily mean that it will move quickly -- and in fact it won't. If you only move it a small distance, it will move very slowly. The point is which direction it moves: away from the center. Looie496 (talk) 22:52, 30 September 2011 (UTC)[reply]
A stable equilibrium will return to the equilibrium point when perturbed (in chemistry, this is called Le Chatelier's principle, but the concept applies to any stable equilibrium system). An unstable equilibrium (usually termed metastable) will drift away from the equilibrium point when perturbed. --Jayron3223:09, 30 September 2011 (UTC)[reply]
Note that Earnshaw's theorem implies that there are no completely unstable equilibria either, because any such point would be a stable equilibrium for a test particle of the opposite charge. So the center point in your example must be stable against perturbations in some directions and unstable against perturbations in other directions. -- BenRG (talk) 23:39, 30 September 2011 (UTC)[reply]
I guess you could argue that since the antigenic material in the vaccine is influencing the prevalence of certain configurations of genes in the adaptive immune system it is "contaminating" but that would be a major stretch. Viruses like HPV are much messier when it comes to "contaminating" DNA, and there has been some speculation that a substantial percentage of human DNA (i.e. Introns) is nothing more than "contamination" by inactive viruses, though that's largely speculation. In short, if you want to preserve your genetic cleanliness (whatever the heck that means), getting vaccinated against viral infections is a good thing because the viruses will cause a lot more genetic damage than a little bit of the adaptive immune system doing what it's supposed to do. SDY (talk) 07:43, 1 October 2011 (UTC)[reply]
I didn't want to mention this per our policy on medical advice but I think it's likely the OP is thinking of the bullshit certain people and sites are spreading that is debunked here [19] not anything to do with effects on the adaptive immune system. Nil Einne (talk) 16:57, 1 October 2011 (UTC)[reply]
You have accomplished nothing with your statement, merely supporting the impression of capitalism I have - the rude, selfish, greedy, sarcastic creature that cares for nothing but its capital and itself. May I also suggest that you take a look at WP:NPA and refrain from losing your temper with floods of barbaric slang?
I think our IP editor may need a little assistance. "Commie scumbag" is a bit clichéd. Here's a bit more choice for you...
Tree sniffing whale hugging pinko lesbian long-haired vegetarian namby pamby do-gooder new age sandal wearing hippie black armband stoned un-Australian pot smoking limpwristed baby boomer subversionary bearded pacifist leftie bleeding heart. Hope this helps. HiLo48 (talk) 07:08, 1 October 2011 (UTC)[reply]
You have just confirmed that communism is not a viable long-term economic system with your link, as well as shown everyone on the ref desk your complete ignorance of world affairs by claiming that N Korea is not communist (it most decidedly is). I don't understand why I should even try to discuss anything with someone as clueless and insanely brainwashed as you? 67.169.177.176 (talk) 06:50, 1 October 2011 (UTC)[reply]
(edit conflict) Uh, no, if the USA hadn't stuffed their nose into the irrelevant affairs of other nations, I assure you that communism would've survived and thrived.
North Korea only thinks they're communist. Clearly, your view of "communism" is the stereotyped one, the one dripping with pro-USA bias. Well, I must agree with the general point of your blatant personal attack - it's pointless discussing anything with anyone who has neither a WP:NPOV or a WP:CIVIL tongue. My apologies, you can go back to counting the money you never earned and your employeesslaves while I work on researching for the Soviet Union article. →Σ⚑ ☭ 07:03, 1 October 2011 (UTC)[reply]
In many parts of the world large efforts are going into the development of alternative technologies for supply of energy. As oil prices rise, these will become viable options. HiLo48 (talk) 06:31, 1 October 2011 (UTC)[reply]
Boy, there's some POV on show here today, isn't there? ;-) I may have compiled a different list (nuclear is experiencing a bit of a hiccup in some areas right now), but the point is still valid. HiLo48 (talk) 06:41, 1 October 2011 (UTC)[reply]
If the amount of oil decreases significantly, the price of energy will probably go up, at least in the short term. The demand for energy will probably also go up. The ways around this are that if the price rises significantly, certain technologies which do produce energy today, but are not competitive with oil, will suddenly become competitive economically. Additionally, if the price of using energy goes up, then there will be additional incentive for using less of it. So in the end it is actually not that awful a thing for the oil to run out. Oil is addictive — it's cheap and bad for you (as a planet). There are lots of alternatives. They are not as cheap as oil (or coal), on the whole, at least in their current iterations. But many/most of them are a lot better for the environment than fossil fuels. (The really stupid but tempting thing to do would be to switch from oil to coal, of course.) Now this kind of transition, if it is very rapid, could cause significant economic disruption, which could cause significant political or social disruption if handled poorly by the powers that be. On the other hand, if the transition is anticipated, and the energy portfolio is diversified, then it's probably only a minor hiccup. I don't see total collapse of industrial civilization as a result in any case, though. This is obviously me gazing into a crystal ball, but it's based on fairly straightforward economic principles (and based on some article I read in Scientific American a long while back). --Mr.98 (talk) 13:30, 1 October 2011 (UTC)[reply]
According to our article on ultra-high-energy cosmic rays, occasionally particles hit the atmosphere at 1.0×1020 eV+, with the vast majority (~99.9995%) of energy in these collisions going to kinetic energy, and the rest going to "interaction with a proton or neutron." Where exactly would the kinetic energy go? As in, which particles, if not protons or neutrons?
If this particle were to crash into a sunbather, would it be cancer inducing? My understanding is that most cancer comes from photons in electromagnetic radiation. If the answer to my previous question was "photons", I imagine the answer would be a resounding "yes" because the collision would produce some very high energy gamma rays.
(1) AFAIK the kinetic energy goes into inducing high-speed movement of air molecules in the ionosphere, as well as their ionization; the other 0.0005% actually goes into transmuting a proton or neutron into any of a whole bunch of exotic particles, along with gamma ray emission.
(2) Yes, any kind of ionizing radiation is potentially cancer-inducing; however, a single cosmic-ray particle would definitely not suffice to cause even one cell to become cancerous. 67.169.177.176 (talk) 06:43, 1 October 2011 (UTC)[reply]
Clarification to part 1: the 0.0005% of the energy goes into transmuting the proton or neutron into such exotic particles as mesons and hyperons, while the other 99.9995% goes into propelling said exotic particle at very high speeds through the upper atmosphere while it collides with other protons/neutrons, transmuting them as well. Eventually all the exotic particles formed in this cascade reaction decay back to protons and neutrons, with emission of gamma rays and neutrinos.
Clarification to part 2: None of the exotic particles formed in the aforementioned cascade reaction ever actually reach the surface -- they all decay in the upper atmosphere, while the gamma rays emitted during the decay process are absorbed by air molecules in the ionosphere, which become ionized in the process and form ozone among other things, as well as re-emitting the absorbed energy as visible light, particularly in the form of Northern lights. 67.169.177.176 (talk) 07:06, 1 October 2011 (UTC)[reply]
What is the word to describe the effect whereby the degree to which something like longitudinal or transverse structures in a 3D ripple pattern are visually enhanced (by light and shadow contrasts) depends on the position of the light source ? For example, if you are looking down on a 3D ripple pattern that has superimposed fairly regular londitudinal N-S structures and fairly irregular transverse E-W structures, a light shining from the east will tend to emphasize the regularity of the N-S longitudinal structures and the pattern will look pretty regular whereas a light shining from the north will tend to emphasize the irregularity of the transverse structures and the pattern will look pretty irregular. I get the feeling that this is some obvious word that I should already know but I can't think of it. Here's an example, the same ripple pattern with 1. left image - light shining from the east, and 2. right image - light shining from the north (although there is a very slight horizontal offset between the images). Sean.hoyland - talk08:33, 1 October 2011 (UTC)[reply]
Those are real ripples created with sand on a plate that was moved in a water tank by a motor repeatedly to reproduce the effects of a current. Sean.hoyland - talk09:04, 1 October 2011 (UTC)[reply]
It is easier to provide a precise answer when the question is specific. What step in blood drawing are you asking about? A pheresis machine draws blood out of people once connected to a catheter that has been inserted into a vein. My guess is that you're asking about venipuncture, and if so I don't think there's a machine that will find and puncture a vein. -- Scray (talk) 12:09, 1 October 2011 (UTC)[reply]
If there are machines that can perform surgery then there must be machines that can find and puncture a vein and avoid the many possible mishaps technicians have when taking sample after sample per hour and per day. I have seen many mishaps with needles sliding out for instance and the technician using the wipe to clean blood from the arm rest before wiping the arm to reinsert the needle - a process an automated system would likely not perform no matter how many samples it took a day. --DeeperQA (talk) 19:22, 1 October 2011 (UTC)[reply]
There aren't machines that can perform surgery. There are machines that surgeons can use to perform surgery. The surgeon is still in control. --Tango (talk) 19:25, 1 October 2011 (UTC)[reply]
Water-added chicken breast
In this week's circular at a local supermarket is chicken breast with 20% added water. Is there a problem of insufficient disclosure in the United States of this added water, or is it well regulated? (If this question belongs in law, please move it.) Thanks. Imagine Reason (talk) 13:16, 1 October 2011 (UTC)[reply]
It's because of how water boils. If you just have a pot of water at sea level and boil it, no matter how much heat you apply, the water will never get hotter than 100 degrees centigrade. Instead, the water will just boil faster. But if you increase the pressure, the boiling point of water is increased, so you can get the water hotter than 100 degrees, which cooks things a lot faster. Rabuve (talk) 17:23, 1 October 2011 (UTC)[reply]
Yes. Essentially, if it were to be at sea level, at 100 degrees, all the heat would go in making steam, rather than cooking the rice. At a higher pressure, the BP is raised, so the heat goes to the rice instead of trying to make steam. Lynch717:35, 1 October 2011 (UTC)[reply]
Miller indices and reduction to lowest integers?
In the 8th edition his book "Materials Science and Engineering: An Introduction", Callister states in a footnote on page 64 that "On occasion, index reduction is not carried out (eg., for x-ray diffraction studies that are described in section 3.16); for example, (002) is not reduced to (001). In addition for ceramic materials, the ionic arrangement for a reduced-index plane may be different from that for a nonreduced one."
I am not able to fully understand why this is the case, and hence, would really appreciate it if anyone could provide me with more details about this kind of "special" cases. If anyone has a document that discusses this kind of issues with respect to either x-ray diffraction or ceramics that would be great.
Sometimes you can deduce the structure without index reduction, and sometimes index reduction will give you an incorrect structure. You have to check your results with what you know about the composition to make sure it makes sense. 75.71.64.74 (talk) 19:52, 1 October 2011 (UTC)[reply]
Please do your own homework.
![[10]](http://chemwiki.ucdavis.edu/@api/deki/files/5228/=CO3-resonance.jpg){kind=link}
