Wikipedia:Reference desk/Archives/Science/2012 March 2

= March 2 =

Question about experiment
Firstly, consider a situation in which a client insists that their home is haunted. You have been hired by the clinical team to conduct research that might demonstrate if this was a reasonable possibility before they make a judgement about their client's mental health.

Secondly, describe how you might conduct such an investigation. What are the variables? Which is the dependent and which is the independent variable? How would you measure these variables? How would you describe the experimental and control groups?please answer this question — Preceding unsigned comment added by 206.54.215.76 (talk) 01:32, 2 March 2012 (UTC)
 * Welcome to . Your question appears to be a homework question. I apologize if this is a misinterpretation, but it is our policy 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. -- Jayron  32  01:40, 2 March 2012 (UTC)
 * First you want to come to grips with Experiment and maybe something like design of experiment. Then move onto hypothesis and Dependent and independent variables. There is no simple answer to a question like this. Vespine (talk) 02:59, 2 March 2012 (UTC)
 * Please note that the same question was posted to the math refdesk with the statement "turn in your extra credit answer with your exam", indicating that this is perhaps actually more than just homework. -- Kinu  t/c 04:53, 2 March 2012 (UTC)
 * I would have hoped questions for extra credit would be better than that... whether a scientific experiment suggests a possibility of the house being haunted isn't relevant to whether the client is mentally ill. We don't usually consider people mentally ill just for believing in a god or gods, which is the same situation. Definitions of mental illness normally compare the patients responses to those of a typical, healthy person. The truth is irrelevant, since typical, healthy people can, and often are, wrong about things. --Tango (talk) 12:09, 2 March 2012 (UTC)
 * I would have hoped questions for extra credit would be better than that... whether a scientific experiment suggests a possibility of the house being haunted isn't relevant to whether the client is mentally ill. We don't usually consider people mentally ill just for believing in a god or gods, which is the same situation. Definitions of mental illness normally compare the patients responses to those of a typical, healthy person. The truth is irrelevant, since typical, healthy people can, and often are, wrong about things. --Tango (talk) 12:09, 2 March 2012 (UTC)


 * No one has asked yet: if repeatedly posting such a question (cheating) is a sign of a personality disorder  or weather it falls into the realm of sociopathic disorder or a combination of the two. Note: This is rhetorical and so I ask no for no answer. --Aspro (talk) 20:07, 2 March 2012 (UTC)


 * Who says it's a "disorder"? There was a time when academia was based on high-minded principles of universal access to knowledge, but nowadays a person "earns a degree to make a bigger salary"; it is an investment, and an expensive one.  "Cheating" is merely proof that they have the mettle and initiative to advance their status; conceptually, a cheater demonstrates his willingness to do whatever he is asked, without naysaying, to get a job done.  The question is whether someone who doesn't cheat should be employable.  (After all, nobody in any industry will ever hire a whistleblower, even if he was right and could have averted something like the Challenger disaster; a felon would have a better chance.  So why accept students with a record of unprofessional honesty?)  Of course, getting caught at cheating is another matter, as it marks a lack of skill.  Not saying it should be that way. Wnt (talk) 03:52, 3 March 2012 (UTC)

GRAVITY
What will happen in the absence of gravity. Will the humanity exist or no ?? Will there be — Preceding unsigned comment added by Kotharinandan8 (talk • contribs) 08:44, 2 March 2012 (UTC)


 * No. → Σ  τ  c . 08:49, 2 March 2012 (UTC)


 * Agreed - definitely no. Without gravity there would be no planets, no stars, no galaxies - the Universe would be filled with a very cold, very low density gas of hydrogen, helium and small amounts of other light elements. No possibility of any form of life as we know it. Actually, gravity is such a fundamental part of general relativity and the structure of space and time that I am not even sure the Big Bang would have happened without it. Gandalf61 (talk) 09:07, 2 March 2012 (UTC)
 * In General Relativity, gravity is an inevitable result of the way spacetime, energy and matter are formulated. It isn't something that is specifically added in, so it isn't something you can just remove. Physics, in its entirity, would have to be completely different to have no gravity, so it is meaningless to ask what the universe would be like in that situation since it would depend on what new physics you invent. --Tango (talk) 12:12, 2 March 2012 (UTC)
 * But scientific folks have speculated on what the case would be for the universe if gravity were too weak to allow planets to form, as part of the fine-tuning argument. I don't think it's beyond the pale to ask the question, just because things are different. --Mr.98 (talk) 12:20, 2 March 2012 (UTC)

what if gravity just disappeared suddenly today (the force became epsilon weak). would all our atmosphere drift away, or would momentum keep it with us? would people who didn't get pushed up stay roughly where they were? could they blow upwards to return downwards if they drifted away, or would everyone need jet packs? naturally absent (strong) gravity we would no longer orbit the sun, but wouldn't momentum keep us roughly in the same relative place (relative to each other, the earth, atmosphere, etc). the moon would likewise careen away, but no big deal in my humble opinion, how many werewolves are there anyway. What other serious effects would an instant conversion to epsilon gravity have? (Other than darkness as we careened away from our sun). Also, assuming we did careen away from our sun due to lack of gravity toward it, we would still continue to spin, wouldn't we (as earth) -- could scientists somehow through nuclear detonations put us at a relative stable distance from the sun, so we were just lazily hovering at one AU from it (earth-sun distance) while continuing to spin (rotational momentum) as we always have?

I just don't see how gravity is ALL that important, unless otherwise our atmosphere drifts away. In that case we had better do something quick, like seran-wrap all of earth or somehow blow it back or create negative pressure (suck up) enough atmosphere for our foreseeable future needs...how else could we deal with this effect? Is my premise that hte atmosphere would drift away correct? How fast would this happen? Anything esle I'm missing? 78.92.82.6 (talk) 13:25, 2 March 2012 (UTC)


 * If there was no gravity suddenly, the atmosphere would disappear. The planet would disintegrate because of the rotational forces. The planets would no longer orbit the Sun. All life would be extinguished in a very short space of time. There is no way that scientists could "somehow through nuclear detonations put us at a relative stable distance from the sun". I think you need to really think about this, no offense. Zzubnik (talk) 13:35, 2 March 2012 (UTC)


 * Don't forget that gravity operates at the micro level (atoms etc) as well as the macro level (planets, galaxies), and so if there was literally "no gravity" there would be no aggregations of matter - maybe only quarks, although I'm no physicist so this might just be pure speculation. --TammyMoet (talk) 13:38, 2 March 2012 (UTC)
 * That's just not correct. Gravity has very little to do with holding fundamental particles together, and is almost negligible in the behavior of lumps of matter smaller than asteroids.  The stability of protons, neutrons, and atomic nuclei is governed by the strong force.  Chemical bonds and physical interactions involving macroscopic objects of smaller-than-astronomical scale are dominated by the electromagnetic force.  Gravity only becomes relevant when you're dealing with large masses; it is by far the weakest of the fundamental forces.  Gandalf61's responses (above and below) are exactly right in terms of what would happen in the two scenarios: either no gravity ever (in which case no agglomerations of matter form in the universe), or gravity gets switched off tomorrow (in which case large objects tend to split apart until they reach a size where they are held together by electromagnetic forces).  TenOfAllTrades(talk) 14:57, 2 March 2012 (UTC)


 * What do you think holds the Earth together ? Without gravity, there would be no force to balance the pressure in the Earth's core and the Earth would immediately disintegrate like a burst balloon. So would the Moon, all the other planets, the Sun, and every other star in the universe. All that would be left would be clouds of gas and small rocks. Gandalf61 (talk) 13:39, 2 March 2012 (UTC)
 * So I was almost right then - the main thrust was right but the detail wasn't. --TammyMoet (talk) 16:14, 2 March 2012 (UTC)
 * Tammy, no, not a single thing you said is correct only the part about your statements being pure speculation were correct. Sorry, but it's true. Atomic structure would remain intact. Gandalf, what would even hold clouds of gas together? Unless connected by a chemical bond or attracted through magnetism I don't see how matter would stay together. To the OP, momentum or more specifically inertia is exactly what would cause earth (and all other rotating bodies held together by gravity) to be flung apart. At the surface of the earth you and everything else is moving at 7.9km/s. With no gravity, there's nothing to bend your path around the earth's center of gravity. You would proceed out tangentially from the surface at 7.9km/s. 203.27.72.5 (talk) 20:56, 2 March 2012 (UTC)
 * The main thrust being "so if there was literally "no gravity" there would be no aggregations of matter ". You've just agreed with that. --TammyMoet (talk) 21:31, 2 March 2012 (UTC)
 * Strong force, Van der Waals, London forces, Covalent bonding, ionic bonding, magnetism, electrostatics, hydrogen bonding. Plenty of aggregations of matter. 203.27.72.5 (talk) 22:04, 2 March 2012 (UTC)
 * Everything with interact with directly (rocks, people, pools of water, etc.) is held together by electromagnetism, not gravity. The Earth itself is held together by gravity, but pretty much the whole crust isn't (the crust wouldn't survive if gravity turned off because the core would explode, but it wouldn't be broken up into particularly small pieces - there could easily be mountain sized pieces left). --Tango (talk) 23:10, 2 March 2012 (UTC)
 * Where do you get 7.9km/s from? The Earth's equatorial rotation velocity is 465.1m/s (see the infobox on Earth), and the rotational velocity reduces to zero as you approach the polls. Of course, 465.1m/s is plenty to make you fly off into space - the bit of ground you are standing would (after being separated from the rest of the crust by the core exploding) fly off with you, though. --Tango (talk) 23:10, 2 March 2012 (UTC)
 * A bit like those British Gas advertisements then? I always knew there was something wrong with the physics those guys were working to.  Spinning  Spark  01:39, 3 March 2012 (UTC)
 * As others have said, gravity is important for far more than just keeping the atmosphere together. However, since you are interested in the effect on the atmosphere, let's work that through. The molecules of air are all moving around, bouncing off each other and the ground. Their average speed is determined by temperature (in fact, it's one definition of temperature). For air at 27C, the average speed is about 476 m/s (see Maxwell–Boltzmann distribution). That is extremely fast. Without gravity to keep the molecules down, they would fly off into space in an instant (about half would be moving downwards at the time, but they would bounce off the ground and fly off into space). You wouldn't have time to take even one more breath before there was no longer a significant amount of air. --Tango (talk) 23:10, 2 March 2012 (UTC)
 * I calculated it by taking the equatorial circumfrence in kilometers and dividing by the number of second in a day...but I made a mistake in the artithmetic. Trying again I get 0.464km/s which is ~465.1m/s. Actually I got two numbers confused; the equatorial rotation velocity and the orbital velocity. I had them both written down in front of me. 203.27.72.5 (talk) 23:41, 2 March 2012 (UTC)

Related question
If gravity did suddenly cease, what would happen to black holes and their associated singularities? Would all of the matter that they had previously consumed be spat out? Wouldn't it involve an instantaneous flattening of spacetime? Wouldn't motion then be impossible? 203.27.72.5 (talk) 00:20, 3 March 2012 (UTC)
 * Gravity suddenly ceasing would involve a completely new physics, so anything could happen. The closest thing to what you describe that has actually be theorised is the collapse of a false vacuum. If that were to happen, all the physical constants would change, which I guess could result in gravity be many times weaker than it is now. I guess that would be equivalent to reducing the black hole's mass, which would accelerate the rate of Hawking radiation and the black hole would rapidly evaporate. All the other physical constants would have changed as well, though, so it is difficult to really say what would happen. --Tango (talk) 00:40, 3 March 2012 (UTC)

P-N junctions forward biasing problem.
In a simple P-N junction diode, if the a positive +6 volts potential is applied to the P terminal and a +3 volts applied to the n-terminal, is the diode forward biased or is it reverse biased? Is it essential for the n terminal to have negative voltage only to be forward biased or it could be like more negative compared to the P-terminal in order to be forward biased. Desperately in need of help. — Preceding unsigned comment added by 210.4.65.52 (talk) 14:31, 2 March 2012 (UTC)


 * When a voltage at a certain point is quoted, it means that the point has a voltage difference with respect to some other point. If the other point is not specified, the convention is that it means with respect to "earth" (English term) or "ground" (American term), or to a common reference point.  So, if you quote a voltage as simply "6V" or "6 volts", it means 6 volts different to some common point in the circuit.  Since you have +6V on the P terminal and +3V on the N terminal, it means that the N termal is at +3V with respect to some other (unspecified if you like) point.  As far as the PN diode is concerned, it sees the difference, that is it's P terminal at (6 - 3) = 3V positive compared to its N terminal, and is forward biased.  Note that, in general, 3 volts P to N forward bias would be sufficient to burn out most diodes.  Keit60.228.244.61 (talk) 15:12, 2 March 2012 (UTC)


 * So a proper biasing circuit would place the P how much more positive than the N? Perhaps 0.7 volts for silicon and 0.3 for germanium, if memory from semiconductor courses long ago serves? I agree that 3 volts of forward bias might fry the device. Edison (talk) 16:06, 2 March 2012 (UTC)

How can IR temperature sensors get readings no matter what angle they're pointing at?
If the angle of reflection equals the angle of incidence, how come my infrared temperature sensor, the gun shaped kind bought at Home Depot that shoots out an infrared beam, can get a reflection back into its sensor no matter what angle at which I point to the surface I want to measure? I've pointed it at flat surfaces from a shallow 30-degree angle up to a straight-on 90-degree angle (at which it's easy to visualize the reflected beam returning to the sensor). But for the non-straight-on beams, how's it getting back? Incidentally, after that question gets answered, if it does, what does the temperature of a substance (no matter what that substance is, since these sensors aren't sold with any disclaimers such as "only works on some specific kind of metal | ceramic | etc...") invariantly (to a practical enough degree) do to an IR beam? What physical change in aspect of the beam does the thing compare to the beam that went out to compute what temperature the surface (again, of whatever material) was? 20.137.18.53 (talk) 15:13, 2 March 2012 (UTC)


 * I've never been to a Home Depot, so I don't know what they sell. However, the common "point and measure" gun type temperature sensors sold world wide do NOT put out an infrared beam.  They only sense the natural infrared radiation coming from the object(s) you point them at.  This radiation is emitted from an object in all directions.  The sensors work on the basis of a theory called "black body radiation".  Essentially, they compare the radiation strength at two different wavelengths.  The hotter the object, the stronger the radiation, and, more importantly, the closer in strengths radiation at two different wavelengths will be.  What the gun does is measure the ratio of the strengths of the two wavelengths.  Just the strength of radiation at one wavelength, or the sum of all wavelengths is no good, because that will vary with the distance to the sensor.  Note that the object to be measured does not have to be visibly black, but the assumption is that it is "black", ie has a surface that is a near perfect raditor at the two infrared wavelengths.  A rough surface helps.  Some surfaces will measure more accurately than others.  Ratbone121.215.156.27 (talk) 15:31, 2 March 2012 (UTC)


 * I've used temperature sensors which put out a laser beam (not an IR beam) so it is easy to see the spot whose temperature you are measuring. The laser is a few milliwatts, so it does not appreciably affect the temperature of the object. The instructions for one said to mark the spot with a black marker if it was shiny, like polished aluminum or stainless, to get an accurate measurement. The thermometer has an infrared sensor with limited viewing angle, so it is measuring only the area near where the indicator laser hits the object. Edison (talk) 16:01, 2 March 2012 (UTC)


 * As for "the angle of reflection equals the angle of incidence", while true, this doesn't mean that every surface produces a coherent reflection, at every wavelength, just like a mirror. The reason is that most surfaces are quite rough at a fine scale, and light striking them is really hitting what looks like gravel, under a microscopic.  Thus, the light hits at a variety of angles and reflects at a variety, too. StuRat (talk) 16:55, 2 March 2012 (UTC)
 * I'll be sure to include the phrase for all practical purposes in future inquiries as to things scientific ;) 20.137.18.53 (talk) 17:27, 2 March 2012 (UTC)
 * Right. See the articles specular reflection, diffuse reflection and Gloss (material appearance).  Red Act (talk) 17:24, 2 March 2012 (UTC)
 * It's more intuitive than that. An infrared lamp is just like a flashlight.  Can you point a flashlight at those surfaces and get a reflection?  Then infrared works the same way.  Of course, for a temperature sensor what you're interested in is not the reflected light, but the light that is emitted in infrared because the object is actually glowing something-less-than-red hot. Wnt (talk) 17:34, 2 March 2012 (UTC)
 * It seems a reach to claim that EM radiation of any wavelength is absorbed or reflected identically, if that's what you intend when you compare IR to a visible light flashlight. It is a questionable claim that visible light and IR have the same reflection and absorption by mirrors and filters, windows or lenses. Edison (talk) 23:14, 4 March 2012 (UTC)

Piranha bite
I'm interested whether exposing an elbow and surrounding area to piranha bite (by 1 or 2 fish) would be tolerable (and whether it would be difficult to detach the fish(es)). 178.181.132.121 (talk) 16:35, 2 March 2012 (UTC)


 * What do you mean by tolerable ? It would be quite painful to have a chunk bitten off, but you probably wouldn't die, unless it became badly infected.  I don't think the piranha remain attached after they bite.  StuRat (talk) 16:49, 2 March 2012 (UTC)
 * I believe that whether the piranha nor the bite remain attached, since piranhas tend to bite off a chunk of their prey, whatever that is, although normally they do not attack humans, unless in fiction. XPPaul (talk) 22:08, 2 March 2012 (UTC)
 * A relevant question. How much scientifically accurate was Piranha 3D where peoples' bodies were shown being torn into pieces by piranhas? -- Supernova Explosion   Talk  16:53, 2 March 2012 (UTC)
 * Without having seen the movie, I feel I can safely say "negligibly." - Goodbye Galaxy (talk) 18:41, 2 March 2012 (UTC)
 * I've read that one such fish can bit off a hunk of meat the size of a cigarette lighter, which would be, to most people, quite intolerable. Edison (talk) 22:29, 2 March 2012 (UTC)
 * What does exactly intolerable mean? Just one piranha bite probably won't kill you, it's also not the case that they'll try to bite your jugular. XPPaul (talk) 22:47, 2 March 2012 (UTC)
 * Here you'll find some relevant information. Quoting from wetwebmedia.com:
 * It is actually very rare for wild Piranhas to attack humans, and they only do so in certain places at certain times, and the locals know to avoid those places when there's a risk. When it comes to being dangerous, Piranhas don't come anywhere near such terrifying animals as Honey Bees or Domestic Dogs in the number of deaths they cause per year! ...


 * From the perspective of the aquarist though the Piranha most likely to bite is probably the Black Piranha, a rather grumpy loner that will view the aquarist's hands as a threat to its territory. Such fish need to be confined to one end of the tank using a net or screen before you can work in the tank safely. The Red Belly Piranha rarely bites but it does happen, and because it is by far the most commonly kept species, it is probably responsible for the most injuries. The ones most likely to cause serious harm though might be the vegetation Pacus; these can and do bite, and their incredibly strong jaws mean they can crush bone. In all cases, aquarists who've been bitten need to get medical attention afterwards because of the high risk of secondary infection.
 * --NorwegianBluetalk 10:27, 3 March 2012 (UTC)

The Coriolis effect in relation to dogs (possibly stupid question)...
Even I think this is a stupid thing for someone to believe. :)

Is there any truth to the claim that dogs are compelled to chase their tails anticlockwise in the northern hemisphere, but clockwise in the southern hemisphere due to the Coriolis effect doing something to their inner ears?

Someone I work with swears down that this is true. I thought that they were joking at first when they said it, but apparently not. --95.148.106.57 (talk) 21:45, 2 March 2012 (UTC)


 * No, it's far to weak to influence them. Your coworker is wrong. Whoop whoop pull up Bitching Betty 21:57, 2 March 2012 (UTC)


 * Off the top of my head, I would think its more dependent on weather they wrote with their left or right paw – or rather  the canine equivalent. some sheep dogs are apparently left handed --Aspro (talk) 22:00, 2 March 2012 (UTC)


 * Thanks. I thought that it sounded about as likely as [www.alexchiu.com/cell/7.htm Alex Chiu's theory] that prey animals run from predators due to magnetic repulsion. --95.148.106.57 (talk) 22:11, 2 March 2012 (UTC)


 * (WP:EC) Indeed. (WP:OR) Many dogs who chase their tails have a strong preference for one rotation (or for that matter, a side to sleep on, etc.). To me, this is completely consistent with biological chirality in general and mammalian handedness in particular. SemanticMantis (talk) 22:15, 2 March 2012 (UTC)
 * From the article on the Coriolis effect; "Because the Earth completes only one rotation per day, the Coriolis force is quite small, and its effects generally become noticeable only for motions occurring over large distances and long periods of time, such as large-scale movement of air in the atmosphere or water in the ocean." The motions of fluid in a dog's inner ear are small and occur of short periods of time, so they will feel and extreamly weak coriolis effect (orders of magnitude too low to detect, even for a dog's keen senses). 203.27.72.5 (talk) 22:21, 2 March 2012 (UTC)

Could Wheatley have obtained lunar escape velocity?
At the conclusion of the video game Portal 2, Wheatley and another personality core are sucked directly down into a portal on a flat surface and ejected from another portal on the surface of the Moon. He remains connected to the mainframe on Earth by a cable which is severed when he is protruding a small distance (let's say 1 meter) and he is sucked into the void. Later, both personality cores are shown floating freely in space, implying that they; a) attained lunar escape velocity or, b) attained orbital velocity or, c) were launched fast enough that they will remain in the "air" (not in physical contact with the Moon's surface) for an extended period (let's say at least 1 hour).

For those not familiar with the physics of Portal and it's sequel, I've included a diagram showing how travelling through portals affects kinematics (or more precisely, how it does not).

As I see it, the forces acting on Wheatley before he enters the portal in the laboratory are Earth's gravity and the pressure differental between the Atmosphere and vacuum. Once he exits the other portal Lunar gravity is pulling him down and expanding air (and neurotoxin) exiting the portal is pushing him up.

Given that; Does Wheatley attain enough speed for any of the options listed above to result? Or should he have been shown lying in the dust on the surface of the Moon shortly after the survivors disconnected him? 203.27.72.5 (talk) 23:26, 2 March 2012 (UTC)
 * Wheatley is 1m above the lunar portal when he is released,
 * the portal is an ellipse with a transverse diameter of 2m and a conjugate diameter of 1m,
 * the pressure is 1atm on Earth and negligible on the Moon,
 * acceleration due to gravity is 9.8m/s/s on at the Aperture Science Enrichment Center and 1.622m/s/s on the Moon,
 * the Moon's escape velocity is 2400m/s,
 * I can't see any way he could achieve significant velocity. The air would spread out as soon as got to the other side of the portal, so it wouldn't continue to apply significant pressure on him after a couple of meters. That means he would need to achieve escape velocity (or whatever) within that few metres and there just isn't enough force involved to do that. --Tango (talk) 00:33, 3 March 2012 (UTC)
 * I haven't looked at the presumably fearsome calculations to figure out the rate of air passing through a nozzle, but from Moon landing escape velocity is 2.38 kilometres per second (1.48 mi/s), i.e. 8568 kilometers per hour. That is indeed a difficult thing to picture... Oh - I see the article on Rocket engine nozzles gives a range of velocities nearly that high for the exhaust when there is a pressure differential of 70 atmospheres.  That's a different sort of nozzle, though ... and getting you a precise answer well, is rocket science.  But I'm guessing this fails by a factor of 100 or more. Wnt (talk) 17:30, 3 March 2012 (UTC)
 * what if the cable that attach Whitley is highly elastic? Is it feasible that the force that accumulates in the cable will suffice? 109.64.24.206 (talk) 20:19, 5 March 2012 (UTC)
 * What force do you think is accumulating in the cable? The cable is pulling him down (from the lunar perspective - up from the Earth perspective), so it will reduce his altitude not increase it. Once the cable snaps, it will shrink back to its unstretched length and that will pull him backwards (the cable shrinks towards its middle by conservation of momentum). --Tango (talk) 13:17, 6 March 2012 (UTC)
 * Utilizing an extreme simplification of the problem, the Bernoulli Equation, I calculate that the air rushing out on the moon would be $${p_{atm}\over\rho_{atm}}={v_{escaping}^2 \over 2}$$ which implies that $${{101,000 Pa}\over{1.001 {kg\over{m^3}}}}={v_{escaping}^2 \over 2}$$ which implies that $$v_{escaping}={449 m\over{s}}$$. Thus it would appear that he could never achieve the required $$2400 m\over{s}$$. Reaper Eternal (talk) 22:04, 6 March 2012 (UTC)