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Wikipedia:Reference desk/Archives/Science/2010 December 16

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December 16

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What is a Hill sphere of Sun?

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What is a Hill sphere of Sun? Basically at what distance from Sun object will orbit Galaxy center rather then Sun? If you can, please use AU when answering. 70.52.184.140 (talk) 02:47, 16 December 2010 (UTC)[reply]

I thought the article on comets might help and indeed under the heading of "single period comets" was the answer The Sun's Hill sphere has an unstable maximum boundary of 230,000 AU (1.1 parsecs (3.6 light-years)).. Vespine (talk) 04:41, 16 December 2010 (UTC)[reply]
I'm not sure that's accurate. The comet article cites a 1964 Paper by Chebotarev as the source for that figure - that paper is reproduced in English here - the relevant section is section 6 "solar gravitational sphere", pp621-622 of the original journal. There Chebotarev says "For the mass of the galaxy we assume the value M=1.3 × 1011 solar masses." And indeed if one runs that through the formula given for a simple (non-eccentric) system given at Hill sphere, one gets a value for r of 225837 AU (with the value for the semimajor axis Chebotarev cites); that's consistent with the 230,000 value he gives.
But that paper is 46 years old, and it looks like the current estimates of the galaxy's mass are quite different. Looking first at Wikipedia's numbers - Milky Way says its mass is roughly 1.4×1042 kg, and the Sun article says its mass is 1.9891×1030. Going by those figures, the mass of the galaxy is 7.038×1011 solar masses, making the galaxy 6 times more massive than Chebotarev's numbers. Running that through the equation (with a slightly refined value for a, again from Sun) gives us a Hill radius of 130,261 AU (down from Chebotarev's 230,000).
But there's more. If you ask Wolfram Alpha for "mass of the galaxy" it says 6×1042 kg which it says is "based on 2009 velocity data from Mark Reid using the Very Long Baseline Array (VLBA)" (I guess that's the finding discussed here). That gives us a mass ratio of 30.2×1011, making the Hill radius about 80,000 AU - that's about a quarter of Chebotarev's number.
I've done the math on this in a hurry, so I'd be grateful if someone could check my numbers. But it does look like Chebotarev's calculation is based on a very wrong estimate of galactic mass. -- Finlay McWalterTalk 19:41, 16 December 2010 (UTC)[reply]
Moreover, ask Wolfram Alpha "distance from the sun to the centre of the galaxy" and it says 2.349×1020 m (it cites a bunch of astronomical sources). That's quite a bit lower than the values given by Wikipedia or Chebotarev; plugging that into the equation and we get an estimated Hill radius of 75,350. It's noteworthy that the estimated radius of the Oort cloud is around 50,000 AU. -- Finlay McWalterTalk 19:59, 16 December 2010 (UTC)[reply]
Wow, I'm going to bookmark you if I ever have any astronomy questions, great answer. Vespine (talk) 04:35, 17 December 2010 (UTC)[reply]
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I just wrote this poem and am curious for someone to shed light on the biological function breakdown of the things I exclaim in it...

Entry004:

The tier on which love operates at its prime; a prime that has seemingly lasted forever, and has no plans of changing anytime soon: (In no particular order of importance..)

Getting to actually know her, her secrets, what makes her tick,, the like; and loving every part about them.

Evolving my golden standard in perfect sexual chemistry, and furthermore the best sex ever even beyond what I once deemed fathomable.

She is the absolute definition of beauty, be it both in pure hotness, cutest, or most gorgeous (Jewish accent)i. And is such a good fuck, yet also the mistress of making love all with a synergistically perfect” key-lock combo”

And all of these things intersect in spacetime…

A-A-W. Cuddlyable3 (talk) 10:51, 16 December 2010 (UTC)[reply]

Current electricity

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Even though current is scalar why the directions are assigned? —Preceding unsigned comment added by 180.215.52.172 (talk) 10:56, 16 December 2010 (UTC)[reply]

One of the properties of a direct electric current is that it is accompanied by a magnetic field. The magnetic field is described by vectors and the orientation of these vectors is dependent on the direction of flow of the electric current. Therefore it is possible to identify the direction in which the electric current is flowing. Dolphin (t) 11:32, 16 December 2010 (UTC)[reply]
Scalars can be signed. Ginger Conspiracy (talk) 07:04, 17 December 2010 (UTC)[reply]
To say that perhaps a bit more clearly, current is not actually a scalar, it is a vector. In many applications it is possible to simplify things by working only with the magnitude of current and ignoring its direction, but even so it does have a direction. Looie496 (talk) 00:10, 18 December 2010 (UTC)[reply]
Current is certainly a vector. Consider applying an electric potential across an anisotropic conductor such as a piece of graphite: the direction of the current will depend on both the direction of the field and the orientation of the graphite (and in general will not be parallel with the field. --ColinFine (talk) 23:51, 19 December 2010 (UTC)[reply]

Prison Riot

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in a Prison Riot do the guards get raped —Preceding unsigned comment added by 66.66.92.152 (talk) 13:39, 16 December 2010 (UTC)[reply]

I don't know. Which particular prison riot did you have in mind? I'd imagine that in some cases, guards get assaulted. No idea on the nature of the assault. --Jayron32 14:27, 16 December 2010 (UTC)[reply]
The Wikipedia article on Etiquette does not address the issue but such an action would conform less than favourably with the Manners that are due to a kindly person who tucks one in at night. Cuddlyable3 (talk) 16:19, 16 December 2010 (UTC)[reply]
A simple search for 'prison guards raped riot' finds [1] which suggests it's a common fear among guards that it may happen in a riot (it also mentions one riot where it happened among hostages which I guess includes guards), our article New Mexico State Penitentiary riot which mentions some some guards were treated for beatings and rapes (it also mentions several prisoners were tortured, burned and killed in the riot) and [2] on the 25 (or 26 if you include the fun one) worse riots which includes the New Mexico State Penitentiary riot (the only other prison related item in that list is the 1981 Irish hunger strike, it's perhaps not the best source considering it says the Kosovo is Serbia protest 2008#Protest in Montenegro was about Serbia recognising Kosovo). As Jayron said it obviously happens, whether it happened in any particularly riot, you'd need to look in to that riot. Nil Einne (talk) 17:01, 16 December 2010 (UTC)[reply]

What do whip scorpions do with their tails? The Wikipedia article says nothing about this, which is odd, since it is a pretty distinguishing physical characteristic of the creature in question and the origin of its popular name.

This site says they use them in self-defense. (How?) This one says it is a sensory organ. (What kind?) In general there seems to be a lot less discussion of this on the web than I'd expect.

What's the truth? Can someone find a solid reference and update the article with it? --Mr.98 (talk) 14:42, 16 December 2010 (UTC)[reply]

Interesting question. All the scholarly articles I can find on defensive behavior focus on the composition and effects of the chemicals released from the glands at the base of the tail. Seems like the defensive use of the whip is a bit of a red herring, although it may help them mimic regular scorpions to scare away predators. This article is probably a good source, but I can't easily find it online; Biologie der Geißelskorpiones (Uropygi Thelyphonida) J Haupt - Memorie della Società Entomologica Italiana, 2000 SemanticMantis (talk) 16:26, 16 December 2010 (UTC)[reply]
Just because its tail has a different shape than other scorpions' doesn't mean it won't have a stinger. The smaller tail likely allows greater agility for dual use as a feeler. Ginger Conspiracy (talk) 07:01, 17 December 2010 (UTC)[reply]
Whip scorpions do not have stingers on their tails. Cursory googling will give you plenty of references. SemanticMantis (talk) 16:49, 17 December 2010 (UTC)[reply]

Boat movement problem

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If a boat is moving in upstream with velocity of 14 km/hr and goes downstream with a velocity of 40 km/hr, then what is the speed of the stream ?

I assume that this is a homework problem and you meant to say that the stream is flowing at the rate of 40 km/hr (although that's mighty fast, so maybe that's the boat and the stream is flowing at 14 km/hr). We won't do your homework for you but can help with how to solve it:
1) If both the boat and stream are moving in the same direction (downstream), then you just add the numbers together and that's how fast they are moving downstream.
2) If they are going in opposite directions then you subtract the smaller number from the larger, and that's how fast it's going, and whichever original direction number was larger, that's the direction it's going. So, if the upstream speed is bigger, then subtract the downstream speed from it and that gives you how fast it's going upstream. But, if the downstream speed is bigger, subtract the upstream speed from it and that gives you how fast it's going downstream. (This problem can also be done by adding two numbers, one of which is negative, but that's more complicated). StuRat (talk) 15:07, 16 December 2010 (UTC)[reply]

x-y=14
x+y=40
x= boat speed
y= stream

although i guess you could switch the variables —Preceding unsigned comment added by 173.166.164.91 (talk) 15:41, 16 December 2010 (UTC)[reply]

Please do your own homework.
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.
Just to clarify: if motion upstream is positive and motion downstream is negative, and x and y are as you define them above, so that x+y=14 km/hr relative to a stationary observer on the bank of the river, what would x-y be then? Is that what you are talking about, or is it something else? WikiDao(talk) 20:55, 16 December 2010 (UTC)[reply]

Paul Dirac Equations??

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I was watching a BBC4 program last night regarding equations and was drawn in by the enthusiasm of, I assume, a Dirac fan.

He wrote down an equation that I believe represented some explanation to the overall understanding of matter/anti-matter but cannot find this equation listed in the Wikipedia page on Paul Dirac.

The equation is as follows:

i,gamma,p,spinner = m,spinner

Please advise where I have mis-understood what was being relayed and if this equation exists? —Preceding unsigned comment added by 86.10.74.128 (talk) 14:51, 16 December 2010 (UTC)[reply]

The equation you're looking for is the Dirac equation, which in natural units such that and c are 1 can be expressed as
.
See also Dirac spinor. Red Act (talk) 16:48, 16 December 2010 (UTC)[reply]

A bomb’s flash, Why would it happen?

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From the NYT: "the scientists found that a bomb’s flash would blind many drivers, causing accidents and complicating evacuation. " However, wouldn't the glass block the UV, therefore, not blinding drivers? 80.58.205.34 (talk) 17:18, 16 December 2010 (UTC)[reply]

It is more than just UV that burns, it is virtually the whole spectrum. The flash can melt the retina -even if your wearing shades..Effects of nuclear explosions--Aspro (talk) 17:34, 16 December 2010 (UTC)[reply]
Also, you're only considering permanent damage, but for someone driving, even momentary/short-term vision impairment is a problem. A bright visual flash can do that (consider your visual ability right after staring directly at a flashbulb, or a Flashbang grenade). DMacks (talk) 17:38, 16 December 2010 (UTC)[reply]
According to one of his memoirs, Richard Feynman shared your reasoning and watched the Trinity test through a windshield with no goggles. --Sean 19:52, 16 December 2010 (UTC)[reply]
About 40 years ago I read that AND he also described his observation point being about 20 miles from the initiation site. Considering the intensity of light falls off as the .... of the distance... etc. A windscreen (laminated with PVA as opposed to triplex as common in 1940 auto mobiles) provides little if any protection to flash blindness emanating from a bucketful of sunshine.
According to flash blindness, a 1-megaton bomb can cause temporary blindness at a distance of up to 13 miles. The Trinity nuclear test was only 20 kilotons, so it seems plausible that Feynman would be fine observing it directly at a distance 20 miles. Of course, it was certainly a non-trivial risk to decide to look at the very first detonation since he couldn't really be sure what the effects would be. Dragons flight (talk) 21:35, 16 December 2010 (UTC)[reply]
Or what the yield would be, for that matter. But it should be remembered that 20 kt was actually a very high yield for the bomb. It was expected that it would be far less efficient than that — most scientists involved had guessed it would be only a kiloton or two at most. --Mr.98 (talk) 21:50, 16 December 2010 (UTC)[reply]
It's of note that Feynman himself reported an impressive afterimage at 20 miles. If he was driving said car, and had no idea that a bomb was about to go off, would he have kept it together or would he have crashed it? We can't know, obviously, but it's worth noting that even in the Feynman scenario (ideal from a number of standpoints), it was still a pretty potent visual effect. --Mr.98 (talk) 21:48, 16 December 2010 (UTC)[reply]
Indeed. In Surely You're Joking, Mr. Feynman!, he recounts
"Time comes, and this tremendous flash out there is so bright that I duck, and I see this purple splotch on the floor of the truck."
While he wasn't completely blinded, he saw one heck of an afterimage. Even though Feynman was expecting the blast he still reflexively turned away from it; consider the effect on a motorist moving at speed and caught by surprise. 04:48, 17 December 2010 (UTC)

Does the volume of cylinder affects Pressure

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Let consider I have a cylinder filled with compressed gas (let air), if half of gas is taken of would the pressure of remaining gas reduce. Here I mean that when the half of cylinder will be empty then the remaining gas will have some volume to expand inside the cylinder so will the pressure decrese..? —Preceding unsigned comment added by 220.225.96.217 (talk) 20:32, 16 December 2010 (UTC)[reply]

Yes. This is an example of Boyle's law, which is a special case of the ideal gas law that applies when the temperature is constant. Red Act (talk) 20:44, 16 December 2010 (UTC)[reply]
However if the gas is partially liquified due the high pressure than some of the liquid will evaporate filling the space and keeping the pressure constant, as long as there is enough liquid inside. 71.101.41.73 (talk) 03:08, 17 December 2010 (UTC)[reply]

Why fog is mostly seen in morning ?

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Why fog is mostly seen in morning ?. and disappear as the sun shines.... does it because we can't see them in darkness of night... or is there some science behind that...? —Preceding unsigned comment added by 220.225.96.217 (talk) 20:37, 16 December 2010 (UTC)[reply]

The sun heats up the earth, which allows the air to hold more moisture, so the fog (floating water droplets) evaporate. --Sean 20:44, 16 December 2010 (UTC)--Sean 20:44, 16 December 2010 (UTC)[reply]
Fog happens when the ground is warm and moist and the air is cold. That situation is more common in the morning, but can also happen at other times of day if a mass of cold air moves into an area where the ground retains heat. Looie496 (talk) 21:12, 16 December 2010 (UTC)[reply]
See Fog#Types. The type of fog that develops overnight (and you can see in the morning) is called "radiation fog" and the one that's made by moving air masses such as a Warm front is called "advection fog". Alansplodge (talk) 09:30, 17 December 2010 (UTC)[reply]
@Sean - Air does not "hold" moisture. For a full explanation (and, for the OP, a clear and very accurate description of how fog is created) see here and here for a more in-depth background. Matt Deres (talk) 23:44, 17 December 2010 (UTC)[reply]
Fog is essentially a cloud on the ground. The Sun increases the temperature of the near-surface air in the last morning so as to decrease the relative humidity of the air so the fog droplets no longer form. Ice fog also often occurs in the morning as does black ice. ~AH1(TCU) 02:33, 18 December 2010 (UTC)[reply]

Why on every Wathc they write Quartz ?

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I have seen many watch from hand watch to wall clock and seen that all of them having written Quartz on them.. why so ... is there any component which is made of Quartz. or it is some kind of patent...? —Preceding unsigned comment added by 220.225.96.217 (talk) 20:39, 16 December 2010 (UTC)[reply]

See quartz clock. --Sean 20:44, 16 December 2010 (UTC)[reply]
Thanks to cheap quartz watch brands such as Swatch, having a quartz crystal is not the novelty it used to be, so the "quartz" labelling is not always shown. Cuddlyable3 (talk) 01:12, 17 December 2010 (UTC)[reply]
Though labeling tells you that it doesn't need to be wound in any way. The counterpart on older (wound) watches was a listing of how many jewels it contained. --Mr.98 (talk) 03:22, 17 December 2010 (UTC)[reply]
Yea, it's a bit like finding a radio that boasts "stereophonic sound !" ... you know you have an antique. StuRat (talk) 05:05, 17 December 2010 (UTC)[reply]
This is an excellent little video that describes how a quartz watch functions. --Mr.98 (talk) 13:47, 17 December 2010 (UTC)[reply]
It's a pretty video but gives no idea of the circuit needed to make the crystal oscillate, just connecting a battery won't do. It would also have helped to explain why 32768 Hz is chosen, it is 215 cycles per second. Cuddlyable3 (talk) 20:54, 17 December 2010 (UTC)[reply]
Quartz watches are practically normal and have nothing to do with quality. I have a watch that cost 1 US dollar and it says "quartz" on it. --Chemicalinterest (talk) 14:34, 18 December 2010 (UTC)[reply]

Thallium(I) oxide production

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Can it be made by burning thallium powder in air or would that also make the sesquioxide? --Chemicalinterest (talk) 22:04, 16 December 2010 (UTC)[reply]

I'm not a chemist but isn't it unlikely that simply combusting it in air will result in a significant amount of the (I) oxide? I would guess there will be a lot of other oxides, including nitrogen oxides. Vespine (talk) 23:27, 16 December 2010 (UTC)[reply]
I can't find the exact answer, but the reaction of thallium with moist air produces TlOH. The standard method for production the anhydrous oxide is by decomposition of the carbonate at 700 °C. To produce the sesquioxide, you seem to need an oxidizing agent that is stronger than air (e.g. Br2), but I'm going on fairly sparse data there. Physchim62 (talk) 23:49, 16 December 2010 (UTC)[reply]
From doi:10.1006/jcht.1993.1128 Tl2O made by Tl2O3 volatilized at around 1000 K in a stream of water-vapor/oxygen, which equilibrates with TlOH under those conditions. DMacks (talk) 10:30, 17 December 2010 (UTC)[reply]
Thallium(I) carbonate, I presume, is made by reacting thallium(I) oxide with carbon dioxide... --Chemicalinterest (talk) 12:09, 17 December 2010 (UTC)[reply]