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August 28

recharge standard batteries?

Can batteries that are not designed to be rechargable be effectively recharged in a recharger? Bubba73 ^{You talkin' to me?} 00:56, 28 August 2013 (UTC)[reply]

No. And it might be dangerous to try. Looie496 (talk) 02:13, 28 August 2013 (UTC)[reply]

Thank you. Bubba73 ^{You talkin' to me?} 02:48, 28 August 2013 (UTC)[reply]

Resolved

The above answer is incomplete. You can't recharge them with a standard charger, but it is possible with the right equipment. We have an article on this: Recharging alkaline batteries, although the information in it is a bit limited.--Srleffler (talk) 03:42, 28 August 2013 (UTC)[reply]

I suspect both our article and the one non peer reviewed source are being a bit liberal with the word "recharge". In general non rechargable batteries produce power using non reversible chemistry (the first step is reversible but the products are absorbed irreversibly by the carbon slush. I suspect what these chargers are doing (with a claim to get 25% of the original charge or similar once only) is not recharging but recovering more of the original charge. If you have a battery in an application which requires close to full power it stops working when it still contains energy. Putting a DC through it, or leaving it in the sun for an hour, might help with this but it is not technically "recharging". It also might be dangerous--BozMo talk 06:02, 28 August 2013 (UTC)[reply]

Yes, "recharge" is not the correct technical term for this process, but I don't remember the correct one. I think there is more to it than just recovering the original charge, but it has been many years since I learned about it and I don't remember the details. --Srleffler (talk) 04:13, 29 August 2013 (UTC)[reply]

To be clear - using a standard recharger (ie one designed to recharge NiCd or Lion batteries) puts energy into a non-rechargeable alkaline battery without recharging it. With no place to go, that energy turns into heat. The battery gets very hot, and pretty soon, the liquids inside boil. That produces an enormous pressure spike and the battery explodes - splattering sharp bits of metal casing and hot, nasty chemicals in all directions! So DON'T DO THAT!

Several dubious companies and a bunch of well-meaning amateur nut-jobs have come out with so-called "rechargers" for alkaline batteries. I'm not sure what chemistry they are relying on - but they don't work very well. They can't completely recharge the battery - and the battery only survives one or two of these "recharge" cycles before going dead forever. I strongly suspect (without evidence) that User:BozMo is correct about the process merely releasing the final charge that was already inside the battery. One could imagine that by creating heat inside that's enough to cause the contents of the battery to flow more easily (but not become hot enough to boil and explode)...then, perhaps convection currents would slowly stir up the chemicals and bring material that had not yet released it's charge into the main mixture - thereby resurrecting the battery for a little while longer. That's not "recharging" - but it might seem that way. But these rechargers need some sophistication - controlling the voltage and current over time and monitoring the temperature of the batteries. That makes them expensive (or dangerous!) and it would require a heck of a lot of slightly-recharged batteries to pay for the cost of the equipment. Certainly they take a long time to do what they do - and they don't do it very well.

Bottom line is that using regular rechargers with alkaline batteries is VERY DANGEROUS and doesn't work - and the dubious-quality specialist alkaline battery "rechargers" are expensive and don't work well enough to justify buying one. So the simplest answer to this question is "No!". SteveBaker (talk) 13:50, 28 August 2013 (UTC)[reply]

Thanks, I was asking about a standard battery charger designed to work on rechargable batteries. Bubba73 ^{You talkin' to me?} 14:10, 28 August 2013 (UTC)[reply]

Yeah - DON'T DO THAT! SteveBaker (talk) 03:30, 29 August 2013 (UTC)[reply]

I wonder how many of the posters above have actually tried using a charger specially designed for recharging alkalines. I have. It works. It works well. Batteries do not explode or get too hot in the charger. Why dont you people believe what our article says?86.181.30.8 (talk) 13:59, 29 August 2013 (UTC)[reply]

Probably because it isn't a good article. There is only one source cited actually on the topic - and that is a review of a single charger on a 'photography community' website. I'm wondering whether the article should be deleted... AndyTheGrump (talk) 14:11, 29 August 2013 (UTC)[reply]

No, I wouldn't imply that batteries would explode or overheat in a properly designed alkaline battery recharger - it's when you try to charge them in a NiMH or a NiCd charger or something like that. SteveBaker (talk) 18:16, 30 August 2013 (UTC)[reply]

How much would we suffer if Earth was overcast?

Change reality the minimum possible to make it cloudy 24/7, or at least 90 percent coverage. For the purposes of this I'll consider the sun never being uncomfortably bright to stare - or invisible, to be overcast. Sagittarian Milky Way (talk) 03:10, 28 August 2013 (UTC)[reply]

What if we could choose the type (stratus, cumulonimbus etc.), or even distribute those by latitude or even sun altitude (-90 to +90 degrees) Sagittarian Milky Way (talk) 03:19, 28 August 2013 (UTC)[reply]

See Year Without a Summer, except that your scenario would be quite a bit worse. Looie496 (talk) 03:35, 28 August 2013 (UTC)[reply]

Would it be as bad as Nuclear winter? 24.23.196.85 (talk) 06:49, 28 August 2013 (UTC)[reply]

I've actually considered this Q before. If the Earth were permanently overcast, we wouldn't know about stars or planets. We would have noticed day and night, though, and possibly noticed the cycles of the moon, not by observing it directly but by noticing the slight change in the light level of the overcast sky on certain nights, and then matching that to the tides. We also wouldn't have shadows cast by the Sun, making it much more difficult to calculate the diameter of the Earth. So, astronomy and geography would suffer quite a bit. We would eventually figure out the diameter of the Earth, after sailing around it, and would eventually see the stars, after we made planes that could fly above the clouds. Then we could begin forming all the astronomical models that in our world happened over thousands of years, based on high altitude observations. So, I think we'd eventually get to the same place we are now, it would just take a bit longer.

If you take it a bit further and imagine a world always covered in ground-level fog (ignoring how plants would get enough light to grow), then flying planes visually would be impossible, as would sailing ships. We wouldn't be able to get from one disconnected continent to another, and would have little reason to think the world was round. The invention of radar would finally allow for airplanes and sea voyages, and from there we would begin the discovery process described above. StuRat (talk) 11:48, 28 August 2013 (UTC)[reply]

Well, if we're talking historically then it wouldn't really matter. "Overcast" simply means that the sky is opaque to visible light via scattering of particles (water droplets, dust, soot, etc..) suspended/dispersed in the atmosphere. If for whatever reason the atmosphere were never transparent to visible light in the first place, It's likely we'd simply have wound up with eyes that were sensitive to wavelengths where the effect of scattering is minimal. That's my $0.02 anyway. Scattering intensity depends strongly on particle size and wavelength. (+)H₃N-Protein\Chemist-CO₂(-) 12:19, 28 August 2013 (UTC)[reply]

Overcast means opaque to visible light ? That would mean a black sky. More like translucent, I think, where visible light would still be useful. StuRat (talk) 12:23, 28 August 2013 (UTC)[reply]

Translucency is basically just light scattering. A thick enough layer of suspended particles in the atmosphere would substantially reduce the transmission of visible light via scattering, and thus make the sky more opaque (no need to invoke 100% opacity here). Since there's a strong angular dependence to the intensity of scattered light, and since you can't exactly change the scattering angle in this scenario, you'd be losing a fair amount of light. It depends somewhat on the size of the particles, but anything big enough to obscure light would work here, I'd expect absorbance to be a relatively minor component in all this. (+)H₃N-Protein\Chemist-CO₂(-) 03:35, 29 August 2013 (UTC)[reply]

I believe "opaque", when unqualified, means 100% opaque: [1]. It's like when you say a door is "closed". That means 100% closed, unless you qualify it with "partially closed", "50% closed", etc. Of course, in both cases, "100%" is actually "almost 100%". That is, a photon or two can pass through almost anything, and there are gaps around any door. StuRat (talk) 06:32, 29 August 2013 (UTC)[reply]

I don't really see why I would use the colloquial definition of "opaque" when I was clearly discussing optics. (+)H₃N-Protein\Chemist-CO₂(-) 10:27, 29 August 2013 (UTC)[reply]

Knowing how and why the earth is so overcast makes a big difference - we can't ignore it.

If it's surface-level water-vapor cloud - then the increase in albedo of the planet would turn it into an ice-ball pretty quickly - the oceans would freeze and life would either die out - or never develop. However, a surface level water-vapor cloud wouldn't exist in that kind of cold either - so the sky would clear up before that happened.
If it's a high-altitude water-vapor cloud then it would cause a rapid onset of very severe global warming because water vapor is a much nastier greenhouse gas than CO2 - and then the oceans would boil and the planet would turn into another Venus.
I suppose a very carefully balanced cloud layer somewhere between those two extremes might keep the planet habitable - but it would be a highly unstable equilibrium - so it would likely flip one way or the other within not too many years (eg A large volcano might add enough material to flip it one way or the other).

Of course you can imagine clouds made of other stuff - but things get complicated to imagine under those circumstances. SteveBaker (talk) 13:31, 28 August 2013 (UTC)[reply]

Article on Pharm D

Your outline of the Doctor of Pharmacy is misleading in its outline for education, It does not require 2-4 years + 4 years, The Pharm. D is the entry level for pharmacy requiring the completion of 6 years entered directly from high school. It replaced the Bachelors of Science in Pharmacy, a five year program, with a 6 year program, It is not a graduate degree it is a professional degree. — Preceding unsigned comment added by 76.19.174.161 (talk) 05:59, 28 August 2013 (UTC)[reply]

You can make these changes to the article yourself; please include citations to reliable sources OldTimeNESter (talk) 13:09, 28 August 2013 (UTC).[reply]

What determines a planet's angle of rotation?

Exactly as the question states. I'm curious as to what factors come into play for determining the angle of a planet's rotation, and why we can see such a wide spectrum of angles, as in the case of say Earth versus Uranus.

Martyk7 (talk) 07:00, 28 August 2013 (UTC)[reply]

Second paragraph at Planet#Rotation has some information which might keep you occupied until someone comes along wit more detail. 163.202.48.125 (talk) 08:13, 28 August 2013 (UTC)[reply]

In theory they should all start out rotating in the plane of the ecliptic. However, collisions with random sub-planets would soon change the rotational axis. In the case of Earth, for example, one theory of the Moon's formation holds that a giant impact knocked the material loose from Earth which then formed the Moon. Such an impact would also knock the Earth off it's original axis. You might think such impacts would be rare, but in the early solar system, they were probably quite common. They only became rare after most of those objects either impacted somewhere, fell into the Sun, fell into a stable orbit, or left the solar system. Looking at all the craters on the Moon, you can get an idea of how many impacts there have been. The Earth would have far more impacts, but erosion here covered most of them up. StuRat (talk) 11:35, 28 August 2013 (UTC)[reply]

Thanks, that's exactly what I was looking for. Martyk7 (talk) 21:53, 29 August 2013 (UTC)[reply]

You're quite welcome. StuRat (talk) 05:43, 30 August 2013 (UTC)[reply]

Black hole question

Escape velocity is the speed at which an object can leave another's abject gravity WITHOUT further propulsion. So, if someone could go near a black hole with a rocket propulsion, especially if the black hole is a supermassive black hole with reasonable surface gravity, he could enter and get out from event horizon. So, why is ever written that is impossible to escape a black hole, while there is that possibility? I know that isn,t an easy question, but thanks for answering. Francesco. — Preceding unsigned comment added by 95.239.192.85 (talk) 10:00, 28 August 2013 (UTC)[reply]

The escape velocity from inside the event horizon is more than c. 163.202.48.126 (talk) 10:44, 28 August 2013 (UTC)[reply]

Other strange things happen at the event horizon, such as infinite gravitational red shift. This results in time slowing to a stop from the point of view of the outside world. This means that if you went to the event horizon and survived, you would never return in the life of the universe. And in case you did not get the point of 163.202.48.126, your rocket can never accelerate you to the speed of light, so you will never escape. Graeme Bartlett (talk) 10:58, 28 August 2013 (UTC)[reply]

Yes, but the rocket can, as happens with space aircraft, get me outside the event horizon where the escape velocity is less than c; if the surface gravity of a black hole is, for example 10 m/s" (as that of earth) ten meters under is howewer similar and this isn't difficult to get out; the surface gravity is infinite on the singularity, not on event horizon... — Preceding unsigned comment added by 95.239.192.85 (talk) 11:18, 28 August 2013 (UTC)[reply]

The gravity at the event horizon of a black hole is a heck of a lot more than that. However, a rocket flying outside the event horizon would be able to escape the black hole at distances where a rock with no propulsion, of the same mass, would fall in. StuRat (talk) 11:29, 28 August 2013 (UTC)[reply]

One of the many problems is (as others have pointed out) that relativistic time dilation is caused by gravitational fields - so as you approach the event horizon from outside, from your perspective, the rest of the universe would appear to go into "fast-forward" - and as the gravity increases, the rest of the universe starts to change faster and faster. At the exact boundary of the black hole, time in the rest of the universe is running infinitely fast - which means that whatever ultimate fate befalls the universe, happens instantly. You can't "escape" because from the perspective of the rest of the universe, time for you has stopped and you're frozen forever on the event horizon. SteveBaker (talk) 13:20, 28 August 2013 (UTC)[reply]

That is true only if you stay around very close to the event horizon. If you are free-falling through the horizon, the universe outside doesn't appear running infinitely fast. At least according to General Relativity (if it's true, nobody will ever be able to report to the outside, so everybody who wants to know has to fall into the black hole). Icek (talk) 14:31, 28 August 2013 (UTC)[reply]

I know, but my point is: as happens with space shuttle, that don't need earth's escape velocity of 11,2 km/s to escape earth's gravity but reaches with rocket engine higher quotas with less gravity and can escape with minor speed (generally 7-8 km/s), so a little under event horizon, far from singularity, the surface gravity is finite and so is possible WITH EXTERNAL PROPULSION get out from event horizon and reaches distance when is possible to escape. The escape velocity is a limit only without further propulsion; with constant pull each speed can be sufficient.95.239.192.85 (talk) 13:22, 28 August 2013 (UTC)[reply]

No. The event horizon is by definition, the point (actually boundary) of no return. As the article states, it is "the point at which the gravitational pull becomes so great as to make escape impossible." You could never go fast enough no matter how many rockets you had. Clarityfiend (talk) 13:35, 28 August 2013 (UTC)[reply]

Yes, but why? So, ten meters over the surface of a very supermassive black hole the surface gravity is reasonable (it can be even near that of earth if the black hole is very large) and ten meters under becames so strong to be impossible to resist it with even very strong aceleration? This seems to me very strange. The surface gravity on event horizon is finite or infinite?95.239.192.85 (talk) 13:46, 28 August 2013 (UTC)[reply]

Ten meters above the event horizon of a supermassive black hole, the effect of gravity is still extremely strong. A supermassive black hole does not at all have anything close to a "reasonable surface gravity", either a little above or a little below the event horizon. I'm not sure where you got the impression that a supermassive black hole can have surface gravity similar to that of Earth. A supermassive black hole can have a density that's less than that of Earth, but that's a different matter. Red Act (talk) 14:22, 28 August 2013 (UTC)[reply]

The gravitational acceleration can be made arbitrarily small by making the black hole large enough. Icek (talk) 14:31, 28 August 2013 (UTC)[reply]

Indeed; the concept of 'surface gravity' gets really weird when you're near a black hole's event horizon – see surface gravity#Surface gravity of a black hole – and actually varies inversely with the black hole's mass. TenOfAllTrades(talk) 14:39, 28 August 2013 (UTC)[reply]

If you approach the event horizon (from the outside of course) very slowly, braking with your rocket engine, then, as SteveBaker pointed out, you will see the universe running at a faster pace, which also means all kind of things will fall into the black hole and onto yourself within a short time (as seen from your perspective); it's more or less like you are traveling at a very large speed (close to the speed of light) relative to nearby galaxies while still being not in the immediate vicinity of large masses. And the equivalent to being at rest at a large distance from the black hole is more or less freely falling into the black hole.

In order to have a more complete picture, you have to consider space-time geometry (details are in the article Schwarzschild metric). Before anything else, you need to know what the coordinates actually mean: You can measure the circumference of a circle around the black hole at a constant distance from the center. The radial coordinate r is the radius that this circle would have in flat space. And the time coordinate is the time that an observer would measure very far away from the black hole. If you now look at the coefficients of the time coordinate differential squared (dt²) and the radial coordinate differential squared (dr²), you should notice that these coefficients switch signs at r = r_S (the Schwarzschild radius). That means that inside the event horizon, the radial direction effectively isn't a spatial direction anymore, but it's a time-like direction. I.e. it is as impossible to go to a larger r as it is impossible to travel to the past.

Icek (talk) 14:31, 28 August 2013 (UTC)[reply]

Question: So the only "direction" you can point your hpothetical rocket is the future - towards the singularity? 163.202.48.125 (talk) 14:51, 28 August 2013 (UTC)[reply]

Are you pointing rockets in a "future" direction on the surface of Earth? If you cannot do that, then you cannot do that inside a black hole either. Space there has 2 "circular" dimensions along the surface of an imagined sphere around the singularity, and 1 other dimension (corresponding to the Schwarzschild time coordinate). You can move in this 3-dimensional space. But note that the size of the 2 closed (circular) dimensions becomes smaller as time passes (proper time that is measured by your clock, as opposed to coordinate time). Icek (talk) 16:00, 28 August 2013 (UTC)[reply]

Your explanation is interesting. So the black hole can't be leaved because space-time. Howewer, to answer to other utents: what matters in escaping, if you have a rocket, isn't the speed but the acceleration you reach: if the surface gravity is 10 km/s^2 and your acceleration is 11 km/s^2 you can win gravity and escape (mantaining that acceleration, obviously) no matter how fast you go and how high could be escape velocity. For this reason i said that could be theoretically possible to escpape a black hole, especially if very large (10^10-10^12 solar masses) — Preceding unsigned comment added by 95.239.192.85 (talk) 14:55, 28 August 2013 (UTC)[reply]

No, you cannot. You need to get the energy for acceleration from somewhere. And the potential energy for lifting an object out of a black hole is greater than the mass-equivalent of the object. So even if you successively convert all your object into energy and use it to accelerate the rest, you will not be able to lift it out of the black hole. Indeed, whenever lift anything with a reaction drive out of a gravity field, you want to accelerate as fast as possible - otherwise you waste energy simply by lifting fuel. And if you drive that to the limit, you want to accelerate your ship immediately to c (you cannot do better), and we know that that is not enough to escape the black hole. --Stephan Schulz (talk) 15:38, 28 August 2013 (UTC)[reply]

Before thinking about acceleration, think about velocity (and let's stay on radial paths for the sake of simplicity). You might think of dr/dt (how much the radial coordinate of an object changes as its time coordinate changes by one (sufficiently small) unit) as the velocity, but that is only a kind of coordinate velocity. In the Schwarzschild metric, outside the event horizon, this coordinate velocity is limited to

c(1-{\frac {r_{S}}{r}})

in the same way that it is limited to c in flat spacetime.

And if you accelerate with your rocket engine near the event horizon, the acceleration you measured with the same rocket configuration in flat spacetime is now proper acceleration. Already in flat spacetime, proper acceleration a' of an object moving with respect to the observer is not equal to coordinate acceleration a (otherwise you could simply accelerate past the speed c), but it's

a'={\frac {a}{(1-{\frac {v^{2}}{c^{2}}})^{\frac {3}{2}}}}

(if the acceleration is along the same direction as the velocity v)

So I hope you can imagine (or even calculate by a few manipulations of the Schwarzschild metric equations) that an acceleration a bit larger than the gravitational acceleration isn't going to give you the speed you might have hoped for, and therefore the transition from above to below the event horizon isn't as discontinuous as it might have seemed.

Icek (talk) 16:00, 28 August 2013 (UTC)[reply]

What the OP needs to understand is that when you are inside of a black hole, even if you are moving outwards (so to speak) at the speed of light, you're are still moving towards the center. There is no path out. Dauto (talk) 16:05, 28 August 2013 (UTC)[reply]

Look at that Penrose diagram. The center of the black hole is the red line, the black hole is the triangle next to the red line and the rest of the diagram represents the whole universe outside of the black hole. Objects traveling at the speed of light follow diagonal lines at 45 degrees. If you're inside of the hole, no matter how fast you're going "outwards" you will still eventually reach the center of the hole. Dauto (talk) 16:16, 28 August 2013 (UTC)[reply]

As pointed out, the term "surface gravity" has problems in that I presume you're thinking of that term in the Newtonian sense, but the Newtonian concept of surface gravity becomes meaningless at a black hole's event horizon. As TenOfAllTrades pointed out, there does exist an alternative definition of the phrase "surface gravity", which is basically the Newtonian surface gravity renormalized by multiplying it by the gravitational redshift factor, and that renormalized value does go to zero as M increases. However, that's very different from there being something close to a Newtonian surface gravity that's close to zero close to an event horizon.

Avoiding the problematic term "surface gravity", and being super pedantic in order to be precise, the proper acceleration required to keep you at an r Schwarzchild coordinate that's ten meters greater than the black hole's Schwarzchild radius (which is at the event horizon) increases to infinity as the black hole's size increases to infinity, i.e., as the black hole's mass increases to infinity. Icek's statement at 14:31 that the acceleration can be made smaller by increasing the black hole's size is incorrect. An acceleration vastly greater than 10 m/s² is required to keep you at an r Schwarzchild coordinate that's ten meters greater than the Schwarzchild radius even with a black hole with a mass about that a star, and instead dealing with a supermassive black hole only makes the required acceleration much larger, not smaller. Red Act (talk) 16:47, 28 August 2013 (UTC)[reply]

Whoops. According to the top equation here and doing a little math, the proper acceleration required to stay a fixed δr above the event horizon only increases with increasing R_s up to R_s = 2 δr, beyond which contrary to my expectation it decreases. So I struck out my post, which apparently is incorrect. Red Act (talk) 18:54, 28 August 2013 (UTC)[reply]

So the surface gravity is infinite not only on the singularity, but even on the event horizon? Why (I believed only singularity had an infinite surface gravity)?95.239.192.85 ([[User talk: 95.239.192.85|talk]]) 16:58, 28 August 2013 (UTC)[reply]

What you said earlier was correct. The surface gravity is inversely proportional to the mass. But that doesn't mean that you can hover at the surface by accelerating outwards, and that's what Red Act was explaining. Your difficulty understanding the difference comes from the fact that you're imagining a black hole to be an object embedded in a Euclidean space (That's the most natural thing to do, but it is wrong). All it means is that if you fall into the black hole you will find yourself accelerating towards the center at that finite acceleration. If you try to hover by blasting your rocket engines, you will still find yourself approaching the center, though a bit more slowly. You cannot simply subtract one acceleration from the other since you are not in a Euclidean space. Dauto (talk) 17:32, 28 August 2013 (UTC)[reply]

It's worse than that; the longest (in proper time) worldlines there are the geodesics. Accelerating (even "outwards") gets you there faster, at least in your estimation. I'm not sure what it does in terms of "outside time" — whether, for instance, accelerating would allow you to rendezvous with a colleague who fell in shortly after you. --Tardis (talk) 13:07, 29 August 2013 (UTC)[reply]

What the OP needs to realize is that a speed limit is hard to break when you can't go any faster. You just simply can't get up to the speed to go out, even though your jets are blasting away - all they're really doing is increasing your relativistic mass and various other weird relativity parameters at that point. Wnt (talk) 18:55, 28 August 2013 (UTC)[reply]

The point of the question is that it is possible to leave a planet without ever reaching its scape velocity, so why can't one leave a B-hole without ever reaching c? Answer: The whole idea that c is the scape velocity of a B-hole is wrong to begin with. It's an oversimplification. Dauto (talk) 19:01, 28 August 2013 (UTC)[reply]

Yeah, you're right. I'm thinking of something relative to spacetime but that's not the same thing. Wnt (talk) 04:19, 31 August 2013 (UTC)[reply]

So, if I understood, the space time on the event horizon is so curved that you could only go into black hole and in every direction you move get into it. Another question: on object when reaches event horizon it's said that stops to esternal observer, but this is only an optical illusion, because black holes in reality increase their mass by absorbing other stars that even to esternal viewers as we are, fall into black holes (even to our proper time, black holes increase their mass, not only to objects that fall into).95.239.192.85 (talk) 00:51, 29 August 2013 (UTC)[reply]

Yes, all directions lead to the center, but the horizon itself doesn't have to be particularly curved. Curvature is proportional to surface gravity. Being a B-hole is a global property, not a local property. There is nothing particularly special about the horizon itself. We might be crossing some gigantic B-hole horizon right now as we speak (type) and not know it. Dauto (talk) 12:15, 29 August 2013 (UTC)[reply]

Looking up "black hole" and "unbreakable rope" got me to this [2] which says (p.105-110 or so) that there is a "maximum force" c⁴/4G, which limits the ability of a rope to hold someone suspended inside the horizon, etc. But I could use some further explanation of the details and rationale myself. Wnt (talk) 04:19, 31 August 2013 (UTC)[reply]

Enthalpy of formation

I'm trying to calculate the enthalpy of formation for molecular lithium hydride, and I'm coming up with an incorrect answer. What am I missing?

Molar enthalpy of fusion of lithium
Molar enthalpy of vaporisation of lithium
Half molar bond-dissociation enthalpy of hydrogen
Negative molar bond-dissociation enthalpy of lithium hydride

I should get ~141 kJ mol⁻¹, but my answer is >300 kJ mol⁻¹. Plasmic Physics (talk) 14:29, 28 August 2013 (UTC)[reply]

What is your source for your numbers so we can check the data? --Jayron32 14:48, 28 August 2013 (UTC)[reply]

Lithium
' '
Bond-dissociation energy
[3] (Table 2, D_e for X ¹Σ⁺) Plasmic Physics (talk) 23:09, 28 August 2013 (UTC)[reply]

It's been a long while since I've done somehing like this, but does the fact that lithium hydride is solid at STP change the answer at all? It seems like the enthalpy of fusion/vaporisation (which will have to be theoretical, since LiH decomposes before it vaporises) should "give some energy back" MChesterMC (talk) 08:29, 29 August 2013 (UTC)[reply]

I'm not trying to find the standard enthalpy of formation. Molecular LiH does actually exist. Plasmic Physics (talk) 22:24, 29 August 2013 (UTC)[reply]

Rope stretch

In rock climbing (a sport which I have recently taken up) there are two types of nylon kernmantle rope generally used. Dynamic rope has significant elasticity to act as a shock absorber and reduce peak impact forces during a fall (they will typically stretch around 30-40% of their length during a simulated fall of an 80kg load a distance 1.77x the length of the rope). "Static" rope on the other hand has much less stretch and is a better choice for abseiling (rappelling) and hauling loads since dynamic rope can end up rather annoyingly bouncy in these applications.

My question is: what is the difference between static and dynamic climbing rope which results in the difference in elasticity - is it different monomers, a different polymer chain length, a different weave pattern or something else? Thanks, Equisetum (talk | contributions) 15:11, 28 August 2013 (UTC).[reply]

It is perfectly possible to make a high-stretch dynamic rope and a low-stretch static rope out of the same materials. In the dynamic rope, the sheath is braided, and the core is filled with loosely twisted fibers, which are not themselves laid together, like this:[4], which allows them to slip against eachother, and against the sheath. After a drop, the sheath may be the only structural part remaining intact: the energy of the fall is absorbed in part by the friction of the core fibers pulling apart (this is why you don't use dynamic rope to bear human loads after it's been used to break a big fall). For a a static rope, (which may or may not have a braided sheath), the core is composed of fibers, laid into yarn, which is in turn laid into a solid, twisted cord, like this:[5], or in some cases a solid braid(see other pics in previous link). For actual rope products on the market, different fibers may be used, but the principle differences in performance of dynamic vs. static cordage are due to the structural properties of how the fibers are arranged. Of course, if you have spare sections of each, just do some dissecting! SemanticMantis (talk) 15:31, 28 August 2013 (UTC)[reply]

Update, if you read all of rope carefully, it says that dynamic kernmantle has the core fibers chopped into shorter sections, and the static kernmantle has both longer fibers, and less twist on the core. A conventionally laid or braided rope of the same fiber would have intermediate stretch, between the dynamic and static. SemanticMantis (talk) 15:37, 28 August 2013 (UTC)[reply]

Excellent answers, thank you. I should have done some dissection since I just retired my father's old dynamic rope and had some offcuts of static cord I used for re-slinging some nuts, but then, these were from different manufacturers and one was new and the other 20 years old (don't worry - it hasn't been used for over 10 years, I know that climbing ropes have limited lifespan), so I would hardly be controlling my variables very well! Equisetum (talk | contributions) 16:05, 28 August 2013 (UTC)[reply]

Homophobia

Is finding homosexual sex (or more broadly a kind of sex that does not fit one's sexuality) "icky" a learned behavior or is it an inherent part of human behavior. personally I don't find any kind of sex icky, including those kinds that don't fit my sexuality, though I don't find them attractive either. But many other people seem to genuinely disgusted by it even though they are not homophobes (or heterophobes?)--Irrational number (talk) 16:18, 28 August 2013 (UTC)[reply]

It's natural behavior, what happens is that a straight person will imagine him/herself to be in the position of the gay person and then judges that to be icky, obviously this is caused by this imagined scenario not being consistent with his/her own sexual preference. Count Iblis (talk) 16:36, 28 August 2013 (UTC)[reply]

Source for that? This is a complicated matter not helped by the fact that sexuality is something of a spectrum. 163.202.48.125 (talk) 16:41, 28 August 2013 (UTC)[reply]

Not to mention that homosexuality has been very differently treated in different societies (see e.g. the uber-macho Sacred Band of Thebes), and that at least in societies with general sexual taboos, many people often find all kind of sex "icky" - from kids to stereotypical "spinsters" to Lady Hillingdon of "Lie Back and Think of England" fame. --Stephan Schulz (talk) 17:43, 28 August 2013 (UTC)[reply]

The elephant in the room here is that when people think of straight sex, they tend to think of a penis penetrating a vagina, and when they think of gay male sex, they tend to think of a penis penetrating an anus. We know what comes out of anuses, and that's where the ickiness factor comes in. Well, just as penis-vagina contact is SO NOT the only kind of straight sex, penis-anus contact is SO NOT the only kind of gay male sex. Oh, I've had many discussions with people who say that these other activities are not really "sex", they're just foreplay or "fooling around" or whatever one does in the lead-up to actual sex. That is just so much garbage. So, we need to be really clear what we're talking about when we say "sex". -- Jack of Oz ^{[pleasantries]} 21:28, 28 August 2013 (UTC)[reply]

There's a certain amount of "ickiness" in sex in general, although "messiness" might be the better way to put it. It reminds me of an extremely old joke from Woody Allen: "Q: Is sex dirty? A: Only if you're doing it right." ←Baseball Bugs ^{What's up, Doc?} carrots→ 22:17, 28 August 2013 (UTC)[reply]

This may be of some interest: Disgust#Evolutionary_significance, also Evolution_of_morality#Disgust. If I recall, I think Richard Dawkins or similar did a program about where disgust comes from. The analogy used in it was that the thought of two ugly people having sex gives one a feeling of disgust but rationally one can see that the disgust mechanism doesn't indicate whether something is right or wrong, IRWolfie- (talk) 22:31, 28 August 2013 (UTC)[reply]

@ Jack above, a better solution than such clarity might be a boycott on answering any heterosexuals' questions on gay sex until they have tried it first. Kind of like the homework rule, above, "we'll get you over the stuck part". μηδείς (talk) 00:53, 29 August 2013 (UTC)[reply]

Yuck Medeis, did you have to choose those words? - ¡Ouch! (^{hurt me} / _{more pain}) 07:45, 29 August 2013 (UTC)[reply]

I thought you liked pain. Be careful of what you ask for. You'll get it, but it may not be exactly what you had in mind. -- Jack of Oz ^{[pleasantries]} 08:27, 29 August 2013 (UTC) [reply]

Well played... - ¡Ouch! (^{hurt me} / _{more pain}) 09:19, 30 August 2013 (UTC)[reply]

I don't have the answer, but I don't think "picturing it" explains it- even if picturing it causes some to feel disgust, it obviously doesn't in others. Moreover, lot's of men enjoy anal sex with a female partner and, yet, find the imagery of it with another male disgusting; and the same goes for other sex acts they would enjoy with a woman. Finally, I, personally, think it all seems pretty hot to picture, and to do, (either gender) and have felt this way since a young age- By numbers, I would definitely be labeled as straight, and, while this is all anecdotal, I can't imagine my perspective/experience is uncommon to the point of anomaly. In other words, "picturing it" causing disgust in some is what needs explained, not an explanation.Phoenixia1177 (talk) 03:44, 29 August 2013 (UTC)[reply]

I still don't have an answer for you, but you may find our articles: Homosexuality and psychology, Biology and sexual orientation, Societal attitudes toward homosexuality, Sexual norm, Heteronormativity, and Homophobia as a good place to start; especially the portions talking about societal attitudes. I'll be honest, this is one of those areas that's a giant morass of studies, politics, and sociology- in other words, it's gigantic and half of it's science, half of it's philosophy, half of it's pedantry, and half of it's bullshit (not exclusive halves:-) ). Because of my last sentence I'm not going to attempt any type of summation, better just to read through the articles and see if they takes you somewhere with a satisfying answer.Phoenixia1177 (talk) 09:40, 29 August 2013 (UTC)[reply]

Weather and sleep

So I live in Tehran and as you can tell it has a very polluted air. And I usually feel tired when I wake up, even when I wake up spontaneously (without an alarm clock) and I sleep in the normal time interval people sleep. And I found out that in the other cities that I've been to, although they are in very different locations of the country, I sleep very well, independent of the time interval, and I wake up feeling fresh, I wonder what is this related to... --Irrational number (talk) 16:23, 28 August 2013 (UTC)[reply]

Might be related to the fact that big cities suck? Dauto (talk) 16:28, 28 August 2013 (UTC)[reply]

Perhaps also noise perturbing sleep? Count Iblis (talk) 16:30, 28 August 2013 (UTC)[reply]

can't be it my room is quiet.--Irrational number (talk) 16:47, 28 August 2013 (UTC)[reply]

If your room isn't very dark, then that could also perturb your sleep. But if it really has to do with the polluted air, that would suggest that you may have apnea due to the air causing breathing problems that you may not notice when you are awake. Then it's best to see a doctor to get a sleep test done and test for allergies, asthma etc. Count Iblis (talk) 16:58, 28 August 2013 (UTC)[reply]

We may be straying into MedAdvice land here.. 163.202.48.125 (talk) 16:42, 28 August 2013 (UTC)[reply]

AW COME ON! let me ask it in a different way "does the weather of the place you sleep in affect the quality of your sleep in any way"? Is that better? I'm just curious about something!--Irrational number (talk) 16:47, 28 August 2013 (UTC)[reply]

From an anecdotal level (but confirmed by many), yes, absolutely. If it is too hot at night, I sleep badly. Somewhat surprisingly, heat during the day does not seem to be similarly bad (but maybe I'm just more tired because of little sleep at night ;-). --Stephan Schulz (talk) 17:45, 28 August 2013 (UTC)[reply]

The Med Advice police needs to chill out. We are just taking about a topic here. I still think the general suckyness of big cities is the culprit. Dauto (talk) 17:47, 28 August 2013 (UTC)[reply]

The line was crossed when Count Iblis suggested apnea as a medical cause - which could be construed as making a diagnosis - a clear violation of Kainaw's Criteria. SteveBaker (talk) 19:50, 28 August 2013 (UTC)[reply]

I think he said the best thing is to go see a doctor which doesn't break the rules. I think I'm gonna make a diagnosis as well: Half of the inhabitants of this forum are obsessed - You know who you are. Dauto (talk) 20:20, 28 August 2013 (UTC)[reply]

There's a big difference between:

"Go and see a doctor, who is the most appropriate source of advice about problems with the functioning of your body", and
"Go and see a doctor, because you may have apnea / cancer / a brain tumour / cholera / AIDS / the bubonic plague / an infestation of cockroaches in your guts / whatever" . -- Jack of Oz ^{[pleasantries]} 21:16, 28 August 2013 (UTC)[reply]

In the US we have lots of ads on TV that say "If you have symptom X, then you may have syndrome Y, so ask your doctor about it" (who will then prescribe their dangerous meds instead of sensibly suggesting lifestyle changes). We even have ads that say "You wouldn't want a doctor to do your job, so don't try to do his" (just take whatever crazy drugs he prescribes and shut up). I wish the makers of those ads could be convicted of practicing medicine without a license. StuRat (talk) 06:23, 29 August 2013 (UTC) [reply]

Some people have looked into a weather/sleep or air pollution/sleep connection. Plugging air pollution sleep into PubMed I get these hits:

Association between PM10 exposure and sleep of Egyptian school children ("There were statistically significant associations between particulate matter with an aerodynamic diameter <10 µm (PM10) exposure and DIMS and sleep hyperhidrosis (P?<?0.05)")
Is sleep apnea a winter disease?: meteorologic and sleep laboratory evidence collected over 1 decade ("More sleep-disordered breathing events were recorded in winter than in other seasons. This finding suggests that obstructive sleep apnea severity may be associated with other seasonal epidemiologic phenomena")
Association between exposure to secondhand smoke and sleep bruxism in children: a randomised control study ("High and moderate exposure to SHS is associated with sleep bruxism characterised by teeth grinding or clenching movements during sleep in children")

Perhaps the scientists here can tell you if these studies seem reasonable. None of them are a body of evidence, just single studies. 184.147.119.141 (talk) 00:29, 29 August 2013 (UTC)[reply]

Is there a physical barrier between the peritoneum and fallopian tube?

If not, what keeps all that stuff from just leaking out?

Thanks, Saintrain (talk) 20:41, 28 August 2013 (UTC)[reply]

Our internal organs don't just float; there is all manner of connective tissue holding things together. ←Baseball Bugs ^{What's up, Doc?} carrots→ 22:14, 28 August 2013 (UTC)[reply]

Sorry, wrong term. I meant "peritoneal cavity". (A little learning, something, something, spring.)

According to Ovary, "The ovaries are not attached to the fallopian tubes ..." and "The ovaries are uncovered in the peritoneal cavity ...", and, from Fallopian_tube#Anatomy, "The tubal ostium is the point where the tubal canal meets the peritoneal cavity ...". Since the fallopian tubes and ovaries have no physical (direct) connection, that implies that the tubes are open to the peritoneal cavity. So, why no leakage? Saintrain (talk) 22:52, 28 August 2013 (UTC)[reply]

I hope it's OK to quote non-WP sources :-), from The peritoneal environment in endometriosis,

"Fallopian tubes and ovaries are bathed in PF [peritoneal fluid]. Oocytes are

exposed to the peritoneal environment even after they are

captured by the fimbria because the Fallopian tube is a

conduit freely communicating with the peritoneal cavity.

Spermatozoa are exposed to PF factors in the Fallopian

tube before and during fertilization. The embryo undergoes

early development in the Fallopian tube. where it is also

potentially exposed to cellular and soluble components of

PF. There is evidence suggesting that at least some of the

uterine fluid may be of peritoneal origin (Casslen. 1986)."

Saintrain (talk) 23:09, 28 August 2013 (UTC)[reply]

Abdominal pregnancy can occur when an embryo floats from the fallopian tube into the abdomen and implants on internal organs. Based on this and on Saintrain's second source regarding PF in fallopian tubes, it appears that "leakage" does occur. 198.190.231.15 (talk) 16:54, 29 August 2013 (UTC)[reply]

This question always has freaked me out. In addition, I had to get a procedure where they stuck a tube only slightly thinner than a drinking straw into my abdomen to drain fluid after an operation, leaving a puncture shaped scar, yet when I asked the doctor said there was no need to suture it. μηδείς (talk) 00:50, 29 August 2013 (UTC)[reply]

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