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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]
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]
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]
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 Wikipedia. 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]
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]
Please do your own homework.