Wikipedia:Reference desk/Archives/Science/September 2005

If an adult human breaks a part of his/her tooth, does the tooth grow or always remains partially broken? If the tooth is pulled out, does a new tooth grow in adults? --anon

:As someone who has spent entirely too much money in lifetime paying Dentist bills, and now 100% false teeth, it seems to me the battle was to save what was left of eroded teeth until I was willing to part with the last of them. In children there is something called wisdom teeth, in which something grows, comes out, then is replaced with another, but in my adult experience, once you got a damaged tooth, or a cavety, the dentist can patch it up, but there is no "natural" cure. AlMac|^(talk) 07:18, 30 August 2005 (UTC)

::I'm not sure what you mean about the wisdom teeth? --Phroziac ^(talk) 16:58, September 3, 2005 (UTC)

• Children will exchange their milk teeth, but after that your teeth won't grow anymore. I speak from experience as two people in my family had broken front teeth. The dentist patched it up with fake teeth or fake partial teeth, but once damaged, adult teeth can not recover. In fact, once you've exchanged your teeth as a kid, your new teeth won't grow anymore. They're already adult size. - Mgm|^(talk) 08:37, August 30, 2005 (UTC)
• Humans only get two sets of teeth - their milk teeth and the adult set, though the wisdom teeth at the back may not appear until you're in your twenties and (in my case) have to come out straight away. If you damage your adult teeth, that's it, you're stuck weith them - they don't grow any more. -- Arwel 12:24, 31 August 2005 (UTC)
• Which I can, unfortunatly, confirm from my own exprerience. A small piece of my lateral incisor broke of after I fell flat on my face. More than a decade later it still hasn't grown back, although it has become smoother. --R.Koot 13:10, 31 August 2005 (UTC)

• By the way, why are they called milk teeth? --Phroziac ^(talk) 16:58, September 3, 2005 (UTC)

who discovered rational numbers?

:The ancient Egyptians knew quite a lot about them, though the general idea may have crossed somebody else's mind before that. David Sneek 07:21, 30 August 2005 (UTC)

:See rational numbers. --R.Koot 11:45, 30 August 2005 (UTC)

My sister's computer is becoming a real pain. While it worked before, suddenly items she tries to download won't show in the traffic screen after clicking download. According to the program she's connected (searches work fine and there's a disconnected option to click in the menu) so what's going on? - Mgm|^(talk) 08:40, August 30, 2005 (UTC)

:Are other people able to download files from your sister while she's connected to the network? It might be that her ISP is blocking the Kazaa service on their end, especially if she's on a college campus or something similar. Garrett Albright 16:33, 30 August 2005 (UTC)

• Haven't been able to check, because I don't know how to make sure someone gets on the same network. Anyway, I doubt our ISP is blocking the service, because she used the same ISP the day before when it worked fine and it's at home, not a campus of any kind. - Mgm|^(talk) 21:21, August 30, 2005 (UTC)

Have you installed any new software, a firewall perhaps? If all else fails and you're sure that your internet connection is working fine, you may want to try reinstalling it. If you do this, be careful that it doesn't delete all of your shared files during the reinstallation pro

What do doctors add in the medicine that contains antibodies in a syringe?

I might have misunderstood your question... but if you mean excipients (agents that are added to improve stability of the active ingredient or to maintain it in solution) then you can find quite a lot of information on the electronic Medicines compendium if you know the name of the vaccine. The website is http://emc.medicines.org.uk and once you've typed in the name of the vaccine, choose the link with the letters SPC (which stand for Summary of Product Characteristics) next to it, and browse the excipients section (I think it is always section 6.1). This works for UK medicines, I don't know what other databases there are, but Google might have the answer. Hope that helps, Jo Brodie, 16:49 BST, August 30, 2005.

This is a peculiar question and I suspect represents a misunderstanding or a non-Western practice. It is extremely rare for American or western European doctors to give injections of antibodies in a syringe. Antibodies in the form of gamma globulin are given by iv infusion for many problems but this doesn't involve a syringe. I cannot off the top of my head think of any examples except old fashioned passive immunity treatments for rabies, tetanus, and snakebite antivenin. There are a number of newer monoclonal antibodies used as treatments for specific uncommon conditions, but most of these are still experimental research protocols and most are given by infusion rather than injection. There were some quack cancer treatments involving injections in the past which claimed to induce a patient to make antibodies against the tumor, but these were pretty uncommon. Can you give us more context? What disease is being treated, and in what country, and is the doctor a mainstream MD or alternative practitioner? alteripse 16:19, 30 August 2005 (UTC)

The questioner really needs to give us more to go on. Another possibility is that they are referring to the use of radiolabelled antibodies in nuclear medicine. - Nunh-huh 20:41, 30 August 2005 (UTC)

:Assuming the antibodies is a mistake, and concentrating on the "What doctors add" bit: I believe one thing added to (all?) injections of medicine is a buffer, to prevent blood pH from changing. -- DrBob 20:49, 30 August 2005 (UTC)

::To the extent that any diluent is a buffer, it's so to avoid discomfort from the injection of highly basic or acidic solutions, which are locally irritating. It's a rare injection that would change blood pH....50 cc of sodium bicarbonate, usually in multiples, gets the job done for a while, but not your everyday run-of-the-mill injection. - Nunh-huh 22:45, 30 August 2005 (UTC) Absolutely right. alteripse 01:17, 31 August 2005 (UTC)

:::Huh, I guess my A-level chemistry lied to me, then. -- DrBob 03:17, 31 August 2005 (UTC) Disillusionment can be tough. Sorry if it was your first time. alteripse 10:47, 31 August 2005 (UTC)

In The Dilbert Future, author Scott Adams says, "Some physicists theorize that reality is like frames of an animated movie, with infinite universes existing at once." Any idea which physicists they are or what their theory is called. Multiverse article didn't help me. Jay 18:42, 30 August 2005 (UTC)

:Probably not the physicist he was refering to, but if you're looking for that type of metaphore, I suggest you try Michio Kaku, he's pretty good for a pop culture physicist--152.163.101.12 21:28, 30 August 2005 (UTC)

One thing which I've always wondered. If you were to build a steel ring with a diameter of, say, 100 metres or 1 kilometre greater than that of the Earth, and somehow place it around the entire Earth, would it more or less hover in mid-air with no support of any kind? — JIP | Talk 19:50, 30 August 2005 (UTC)

:Someone's been reading a bit too much Arthur C Clarke, personally i thought 3001 was a bit of a let down, but I guess you liked it well enough--152.163.101.12 21:25, 30 August 2005 (UTC)

:Not really. The Earth is pretty rough at that scale - you have mountains, seas, that sort of thing - which are very inhomogenous; it'd be like dealing with a lot of weird mascons. (Consider the gravitational effects of "ring next to a mountain" vs. "ring over central Pacific). It'd probably "crash to the ground" somewhere along the length - or, indeed, buckle under its own weight and do so all over the place. Shimgray 19:55, 30 August 2005 (UTC)

:It wouldn't be very stable unless you spun it, but you could make it a geosynchronous orbit. You might want to look at Ringworld and Dyson sphere. Bovlb 19:59:37, 2005-08-30 (UTC)

::At geosynch you might manage stable, but not at a kilometer high - local mass differences are just too significant, then. Shimgray 20:33, 30 August 2005 (UTC)

:You might also find an interest in the space elevator concept. How is that supposed to stay where it is supposed to be? AlMac|^(talk) 06:08, 31 August 2005 (UTC)

::The plan for that is to be above geosynchronous orbit (much much more than 1 km above the surface). Technically it would work at geosynchronous orbit, but the designers want the elevator structure to be under some tension. The ring structure would "hover" at geosynchronous orbit, except that our geoid isn't perfect and we've dumped lots of satellites up there, so I don't think it'd be particularly stable, but — as mentioned above — spinning it up would help. — Laura Scudder | Talk 15:49, 31 August 2005 (UTC)

::I believe the general concept is to counterweight an elevator so that the center of gravity is at the geosynch point. As Laurascudder noted, though, we'd also have to do some major orbital cleanup. IIRC, Kim Stanley Robinson discusses a lot of this in his book Red Mars — Lomn | Talk / RfC 15:54:06, 2005-08-31 (UTC)

Just curious, but is "luminous toxin" (the poison that slowly killed the character in the 1950 movie D.O.A.) a real poison? According to the movie, the poison is real (from the movie: "``The medical facts in this motion picture are authentic. Luminous toxin is a descriptive term for an actual poison.'' — Technical Adviser, Edward F. Dunne, M.D.") However, after I have searched the web and Wikipedia for "luminous toxin" and only found references to it in articles related to the film (no scientific/medical articles), I am starting to wonder whether "luminous toxin" is real or not. Thanks. 61.94.149.174 06:33, 31 August 2005 (UTC)

: A "descriptive term for an actual poison" means thay are describing the toxin rather than giving you its name. Medical articles will refer to it by its actual name. I could describe milk as "a white liquid complete baby food" but everyone else will still call it milk. Theresa Knott (a tenth stroke) 19:30, 31 August 2005 (UTC)

: Expect a 40's film to be very loose with scientific merit. Very roughly, you could describe phosphosous as a luminous toxin. White phosphorous, specifically, glows and is a systemic poison, although mostly known for bone damage. This was probably close enough for the purposes of a sensational movie script. However, having several days of relative health before dying (despite the best medical intervention) is possibly more charactistic of overdose from paracetamol or deadly Amanita mushrooms, both of which cause liver failure that results in death after an initial recovery 131.172.99.15 (talk) 12:44, 23 December 2008 (UTC)snaxalotl

: Without a reliable citation, phosphorous or one of its compounds is just a guess. I have marked the phosphorous passage 'citation needed'. I doubt that either of the authors of D.O.A. knew of a specific chemical with the properties described in the film. It's probably literary invention. Without inside information, we cannot even be sure there really was a medical consultant on the film. David Spector (talk) 16:50, 27 January 2011 (UTC)

The "luminous toxin" was iridium. Watch DOA! - that's not correct. You misunderstood the plot line.

Explain why in a group of 35 children at a school, a randomly selected individual has a high probability of being the oldest son or daughter in his or her family.

--anonym

:Maybe because there are a lot of one / two child famillies nowadays? --βjweþþ (talk) 07:54, 31 August 2005 (UTC)

Presumably this is a math assignment. If the average number of children per family are less than 2 (as is true in most of the western world), any randomly chosen child is more likely to be an eldest than a non-eldest.

Average children per family (CperF) can be expressed as N1+N2+N3+N4+N5+N6...+N20 divided by the NF, where N1 is total number of eldest children, N2 is total number of second children, N3 is total number of third children, etc, and NF is the total number of families. There is a value of CperF below which more than half of any group of children are eldest (N1). Is this what you are supposed to compute? It would be somewhat above 2 because preschoolers (age under 5 years) would be selectively excluded from your classroom sample.

Unless King Herod or the Tenth Plague of Egypt has visited the community lately, there will always be more eldest children than second, more second children than third, etc (i.e., N1>N2>N3...N20), so that even in a society in which there are on average many children per family (high CperF), any randomly chosen child is more likely to be an eldest than a second, third, fourth, etc. alteripse 11:16, 31 August 2005 (UTC)

: Also if the scholl is a kindergarten, the youngest son/daughter might not have been born yet. --R.Koot 12:22, 31 August 2005 (UTC)

:: Actually, this is not relevant. Whether a younger sibling was already born or not, it won't change the fact that the elder is the elder (even if s/he is the only child). Ornil 16:53, 31 August 2005 (UTC)

:: I think it is relevant, since having a child in kindergarten correlates with having young children in general (i.e. having a young family) and therefore correlates with not having had all of your children yet; i.e. the distribution will be skewed toward having one (or few) child MORE THAN the average distribution131.172.99.15 (talk) 12:51, 23 December 2008 (UTC)snaxalotl

A magazine clearinghouse holds a sweepstakes contest to sell subscriptions. If you return the winning number, you win $1,000,000 (USD). You have a 1-in-20-million chance of winning, but your only cost to enter the contest is a first-class stamp to mail the entry. Use the current price of a first-class stamp to calculate the expected net winnings if you enter this contest. Is it worth entering the sweepstakes?

--anonym

:Yes. No. Maybe. Depends whose stamp it is. I think it's probably better to be doing your homework than entering the sweepstakes. Notinasnaid 07:58, 31 August 2005 (UTC)

• If the price of a first-class stamp is $1.29, then you can expect to lose $\backslash approx-\$1.24$.

I want to ask a question about power and torque, more specifically about rally car engines.

Reportedly, Peugeot 206 WRC's 2-liter turbocharged engine generates 635 Nm of torque at 4000 rpm. if i'm not mistaken, this makes 667 hp at the same rpm. but the car's maximum power output is reported to be 300 HP at 5250 rpm (and 300 HP is the maximum legal power output in the championship). How can the engine develop such a high amount of torque considering that the power output is not that high? Isn't there an inconsistency between the figures? Am I just miscalculating or how is this possible?

Thanks,

--Ozkaplan

:According to the article on torque, if $P$ is power in kilowatts, $T$ is torque in Newton metres, and R = rotational speed in revolutions per minute, if you manipulate the constants:

:$P\; =\; \backslash frac\{T\; \backslash times\; R\backslash times\; 2\; \backslash pi\}\{60000\}$

:If you plug in T = 635 and R = 4000 into that I get approximately 266 kilowatts, which is about 357 horsepower. Considerably over the limit, still. As to how it's done, let's just say that creative interpretations of physics, mathematics, and indeed logic pertaining to rulebooks have been a stock in trade of automobile racing for a very long time. In the 1980's, the maximum level of turbo boost in Formula One was supposed to be restricted by the use of a standard popoff valve, which opens up to bleed off pressurised air if the pressure gets too high. Teams apparently got around this by fitting huge turbos to their qualifying motors which pumped so much air the popoff valve couldn't get rid of it all...--Robert Merkel 12:18, 31 August 2005 (UTC)

what's the speed of processor for a computer system?

Speed is measured in Hertz, or cycles per second. Modern systems run in the Gigahertz range, though one cycle does not necessarily equal one instruction. See Central processing unit and clock rate for more info. — Lomn | Talk / RfC 15:36:50, 2005-08-31 (UTC)

:Analysing the performance of a computer system is a very, very complex topic, if you want to tackle it in all its glory. However, these days, for most purposes, for most people, the speed of the processor in your PC doesn't really matter - you will barely be able to tell the difference in CPU speed between the slowest and fastest x86-compatible PC Intel and AMD make. The thing that causes the most perceived delay in my use of a computer is almost always my Internet connection. --Robert Merkel 22:45, 31 August 2005 (UTC)

I checked a minor manufacturing topic; it got me looking into Manufacturing as a whole. It seems to me that it is a very big topic that is fragmented and difficult to navigate. I am not sure how to start reviewing it without being a vandal! The categories lists also seem in need of review. I might start by creating a "Manufacturing overview" page similar to this temporary page. Joe1011010 19:41, 31 August 2005 (UTC)

:I think you should go ahead with a rewrite if you think you can improve the article(s). BUT, in a high level topic normally we would not just say "The beginings of manufacturing is covered in the Industrial Revolution." Rather, it is better to have a condensed overview of the industrial revolution in the manufacuring article, along with an italisized statement indicating that more detail can be found in the article on the main article on the industrial revolution. Happy editing ike9898 17:22, September 6, 2005 (UTC)out, does a new tooth grow in adults? --anon

As someone who has spent entirely too much money in lifetime paying Dentist bills, and now 100% false teeth, it seems to me the battle was to save what was left of eroded teeth until I was willing to part with the last of them. In children there is something called wisdom teeth, in which something grows, comes out, then is replaced with another, but in my adult experience, once you got a damaged tooth, or a cavety, the dentist can patch it up, but there is no "natural" cure. AlMac|(talk) 07:18, 30 August 2005 (UTC)

I'm not sure what you mean about the wisdom teeth? --Phroziac (talk) 16:58, September 3, 2005 (UTC)

Children will exchange their milk teeth, but after that your teeth won't grow anymore. I speak from experience as two people in my family had broken front teeth. The dentist patched it up with fake teeth or fake partial teeth, but once damaged, adult teeth can not recover. In fact, once you've exchanged your teeth as a kid, your new teeth won't grow anymore. They're already adult size. - Mgm|(talk) 08:37, August 30, 2005 (UTC)

Humans only get two sets of teeth - their milk teeth and the adult set, though the wisdom teeth at the back may not appear until you're in your twenties and (in my case) have to come out straight away. If you damage your adult teeth, that's it, you're stuck weith them - they don't grow any more. -- Arwel 12:24, 31 August 2005 (UTC)

Which I can, unfortunatly, confirm from my own exprerience. A small piece of my lateral incisor broke of after I fell flat on my face. More than a decade later it still hasn't grown back, although it has become smoother. --R.Koot 13:10, 31 August 2005 (UTC)

By the way, why are they called milk teeth? --Phroziac (talk) 16:58, September 3, 2005 (UTC)

mathematics

Edit

who discovered rational numbers?

The ancient Egyptians knew quite a lot about them, though the general idea may have crossed somebody else's mind before that. David Sneek 07:21, 30 August 2005 (UTC)

See rational numbers. --R.Koot 11:45, 30 August 2005 (UTC)

My sister's computer is becoming a real pain. While it worked before, suddenly items she tries to download won't show in the traffic screen after clicking download. According to the program she's connected (searches work fine and there's a disconnected option to click in the menu) so what's going on? - Mgm|(talk) 08:40, August 30, 2005 (UTC)

Are other people able to download files from your sister while she's connected to the network? It might be that her ISP is blocking the Kazaa service on their end, especially if she's on a college campus or something similar. Garrett Albright 16:33, 30 August 2005 (UTC)

Haven't been able to check, because I don't know how to make sure someone gets on the same network. Anyway, I doubt our ISP is blocking the service, because she used the same ISP the day before when it worked fine and it's at home, not a campus of any kind. - Mgm|(talk) 21:21, August 30, 2005 (UTC)

Have you installed any new software, a firewall perhaps? If all else fails and you're sure that your internet connection is working fine, you may want to try reinstalling it. If you do this, be careful that it doesn't delete all of your shared files during the reinstallation pro

Medical Information

Edit

What do doctors add in the medicine that contains antibodies in a syringe?

I might have misunderstood your question... but if you mean excipients (agents that are added to improve stability of the active ingredient or to maintain it in solution) then you can find quite a lot of information on the electronic Medicines compendium if you know the name of the vaccine. The website is http://emc.medicines.org.uk and once you've typed in the name of the vaccine, choose the link with the letters SPC (which stand for Summary of Product Characteristics) next to it, and browse the excipients section (I think it is always section 6.1). This works for UK medicines, I don't know what other databases there are, but Google might have the answer. Hope that helps, Jo Brodie, 16:49 BST, August 30, 2005.

This is a peculiar question and I suspect represents a misunderstanding or a non-Western practice. It is extremely rare for American or western European doctors to give injections of antibodies in a syringe. Antibodies in the form of gamma globulin are given by iv infusion for many problems but this doesn't involve a syringe. I cannot off the top of my head think of any examples except old fashioned passive immunity treatments for rabies, tetanus, and snakebite antivenin. There are a number of newer monoclonal antibodies used as treatments for specific uncommon conditions, but most of these are still experimental research protocols and most are given by infusion rather than injection. There were some quack cancer treatments involving injections in the past which claimed to induce a patient to make antibodies against the tumor, but these were pretty uncommon. Can you give us more context? What disease is being treated, and in what country, and is the doctor a mainstream MD or alternative practitioner? alteripse 16:19, 30 August 2005 (UTC)

The questioner really needs to give us more to go on. Another possibility is that they are referring to the use of radiolabelled antibodies in nuclear medicine. - Nunh-huh 20:41, 30 August 2005 (UTC)

Assuming the antibodies is a mistake, and concentrating on the "What doctors add" bit: I believe one thing added to (all?) injections of medicine is a buffer, to prevent blood pH from changing. -- DrBob 20:49, 30 August 2005 (UTC)

To the extent that any diluent is a buffer, it's so to avoid discomfort from the injection of highly basic or acidic solutions, which are locally irritating. It's a rare injection that would change blood pH....50 cc of sodium bicarbonate, usually in multiples, gets the job done for a while, but not your everyday run-of-the-mill injection. - Nunh-huh 22:45, 30 August 2005 (UTC) Absolutely right. alteripse 01:17, 31 August 2005 (UTC)

Huh, I guess my A-level chemistry lied to me, then. -- DrBob 03:17, 31 August 2005 (UTC) Disillusionment can be tough. Sorry if it was your first time. alteripse 10:47, 31 August 2005 (UTC)

universe as animation frames

Edit

In The Dilbert Future, author Scott Adams says, "Some physicists theorize that reality is like frames of an animated movie, with infinite universes existing at once." Any idea which physicists they are or what their theory is called. Multiverse article didn't help me. Jay 18:42, 30 August 2005 (UTC)

Probably not the physicist he was refering to, but if you're looking for that type of metaphore, I suggest you try Michio Kaku, he's pretty good for a pop culture physicist--152.163.101.12 21:28, 30 August 2005 (UTC)

Ring around the Earth

Edit

One thing which I've always wondered. If you were to build a steel ring with a diameter of, say, 100 metres or 1 kilometre greater than that of the Earth, and somehow place it around the entire Earth, would it more or less hover in mid-air with no support of any kind? — JIP | Talk 19:50, 30 August 2005 (UTC)

Someone's been reading a bit too much Arthur C Clarke, personally i thought 3001 was a bit of a let down, but I guess you liked it well enough--152.163.101.12 21:25, 30 August 2005 (UTC)

Not really. The Earth is pretty rough at that scale - you have mountains, seas, that sort of thing - which are very inhomogenous; it'd be like dealing with a lot of weird mascons. (Consider the gravitational effects of "ring next to a mountain" vs. "ring over central Pacific). It'd probably "crash to the ground" somewhere along the length - or, indeed, buckle under its own weight and do so all over the place. Shimgray 19:55, 30 August 2005 (UTC)

It wouldn't be very stable unless you spun it, but you could make it a geosynchronous orbit. You might want to look at Ringworld and Dyson sphere. Bovlb 19:59:37, 2005-08-30 (UTC)

At geosynch you might manage stable, but not at a kilometer high - local mass differences are just too significant, then. Shimgray 20:33, 30 August 2005 (UTC)

You might also find an interest in the space elevator concept. How is that supposed to stay where it is supposed to be? AlMac|(talk) 06:08, 31 August 2005 (UTC)

The plan for that is to be above geosynchronous orbit (much much more than 1 km above the surface). Technically it would work at geosynchronous orbit, but the designers want the elevator structure to be under some tension. The ring structure would "hover" at geosynchronous orbit, except that our geoid isn't perfect and we've dumped lots of satellites up there, so I don't think it'd be particularly stable, but — as mentioned above — spinning it up would help. — Laura Scudder | Talk 15:49, 31 August 2005 (UTC)

I believe the general concept is to counterweight an elevator so that the center of gravity is at the geosynch point. As Laurascudder noted, though, we'd also have to do some major orbital cleanup. IIRC, Kim Stanley Robinson discusses a lot of this in his book Red Mars — Lomn | Talk / RfC 15:54:06, 2005-08-31 (UTC)

"Luminous toxin"

Edit

Just curious, but is "luminous toxin" (the poison that slowly killed the character in the 1950 movie D.O.A.) a real poison? According to the movie, the poison is real (from the movie: "``The medical facts in this motion picture are authentic. Luminous toxin is a descriptive term for an actual poison.'' — Technical Adviser, Edward F. Dunne, M.D.") However, after I have searched the web and Wikipedia for "luminous toxin" and only found references to it in articles related to the film (no scientific/medical articles), I am starting to wonder whether "luminous toxin" is real or not. Thanks. 61.94.149.174 06:33, 31 August 2005 (UTC)

A "descriptive term for an actual poison" means thay are describing the toxin rather than giving you its name. Medical articles will refer to it by its actual name. I could describe milk as "a white liquid complete baby food" but everyone else will still call it milk. Theresa Knott (a tenth stroke) 19:30, 31 August 2005 (UTC)

Expect a 40's film to be very loose with scientific merit. Very roughly, you could describe phosphosous as a luminous toxin. White phosphorous, specifically, glows and is a systemic poison, although mostly known for bone damage. This was probably close enough for the purposes of a sensational movie script. However, having several days of relative health before dying (despite the best medical intervention) is possibly more charactistic of overdose from paracetamol or deadly Amanita mushrooms, both of which cause liver failure that results in death after an initial recovery 131.172.99.15 (talk) 12:44, 23 December 2008 (UTC)snaxalotl

Without a reliable citation, phosphorous or one of its compounds is just a guess. I have marked the phosphorous passage 'citation needed'. I doubt that either of the authors of D.O.A. knew of a specific chemical with the properties described in the film. It's probably literary invention. Without inside information, we cannot even be sure there really was a medical consultant on the film. David Spector (talk) 16:50, 27 January 2011 (UTC)

The "luminous toxin" was iridium. Watch DOA! - that's not correct. You misunderstood the plot line.

Probability: The “Oldest Son or Daughter Phenomenon”

Edit

Explain why in a group of 35 children at a school, a randomly selected individual has a high probability of being the oldest son or daughter in his or her family.

--anonym

Maybe because there are a lot of one / two child famillies nowadays? --βjweþþ (talk) 07:54, 31 August 2005 (UTC)

Presumably this is a math assignment. If the average number of children per family are less than 2 (as is true in most of the western world), any randomly chosen child is more likely to be an eldest than a non-eldest.

Average children per family (CperF) can be expressed as N1+N2+N3+N4+N5+N6...+N20 divided by the NF, where N1 is total number of eldest children, N2 is total number of second children, N3 is total number of third children, etc, and NF is the total number of families. There is a value of CperF below which more than half of any group of children are eldest (N1). Is this what you are supposed to compute? It would be somewhat above 2 because preschoolers (age under 5 years) would be selectively excluded from your classroom sample.

Unless King Herod or the Tenth Plague of Egypt has visited the community lately, there will always be more eldest children than second, more second children than third, etc (i.e., N1>N2>N3...N20), so that even in a society in which there are on average many children per family (high CperF), any randomly chosen child is more likely to be an eldest than a second, third, fourth, etc. alteripse 11:16, 31 August 2005 (UTC)

Also if the scholl is a kindergarten, the youngest son/daughter might not have been born yet. --R.Koot 12:22, 31 August 2005 (UTC)

Actually, this is not relevant. Whether a younger sibling was already born or not, it won't change the fact that the elder is the elder (even if s/he is the only child). Ornil 16:53, 31 August 2005 (UTC)

I think it is relevant, since having a child in kindergarten correlates with having young children in general (i.e. having a young family) and therefore correlates with not having had all of your children yet; i.e. the distribution will be skewed toward having one (or few) child MORE THAN the average distribution131.172.99.15 (talk) 12:51, 23 December 2008 (UTC)snaxalotl

The Expected Value of a Sweepstakes Contest

Edit

A magazine clearinghouse holds a sweepstakes contest to sell subscriptions. If you return the winning number, you win $1,000,000 (USD). You have a 1-in-20-million chance of winning, but your only cost to enter the contest is a first-class stamp to mail the entry. Use the current price of a first-class stamp to calculate the expected net winnings if you enter this contest. Is it worth entering the sweepstakes?

--anonym

Yes. No. Maybe. Depends whose stamp it is. I think it's probably better to be doing your homework than entering the sweepstakes. Notinasnaid 07:58, 31 August 2005 (UTC)

If the price of a first-class stamp is $1.29, then you can expect to lose

≈

−

$

1.24

{\displaystyle \approx -\$1.24}.

Power and torque output (of WRC cars)

Edit

I want to ask a question about power and torque, more specifically about rally car engines. Reportedly, Peugeot 206 WRC's 2-liter turbocharged engine generates 635 Nm of torque at 4000 rpm. if i'm not mistaken, this makes 667 hp at the same rpm. but the car's maximum power output is reported to be 300 HP at 5250 rpm (and 300 HP is the maximum legal power output in the championship). How can the engine develop such a high amount of torque considering that the power output is not that high? Isn't there an inconsistency between the figures? Am I just miscalculating or how is this possible? Thanks, --Ozkaplan

According to the article on torque, if

P

P is power in kilowatts,

T

T is torque in Newton metres, and R = rotational speed in revolutions per minute, if you manipulate the constants:

P

=

T

×

R

×

2

π

60000

{\displaystyle P={\frac {T\times R\times 2\pi }{60000}}}

If you plug in T = 635 and R = 4000 into that I get approximately 266 kilowatts, which is about 357 horsepower. Considerably over the limit, still. As to how it's done, let's just say that creative interpretations of physics, mathematics, and indeed logic pertaining to rulebooks have been a stock in trade of automobile racing for a very long time. In the 1980's, the maximum level of turbo boost in Formula One was supposed to be restricted by the use of a standard popoff valve, which opens up to bleed off pressurised air if the pressure gets too high. Teams apparently got around this by fitting huge turbos to their qualifying motors which pumped so much air the popoff valve couldn't get rid of it all...--Robert Merkel 12:18, 31 August 2005 (UTC)

computers

Edit

what's the speed of processor for a computer system?

Speed is measured in Hertz, or cycles per second. Modern systems run in the Gigahertz range, though one cycle does not necessarily equal one instruction. See Central processing unit and clock rate for more info. — Lomn | Talk / RfC 15:36:50, 2005-08-31 (UTC)

Analysing the performance of a computer system is a very, very complex topic, if you want to tackle it in all its glory. However, these days, for most purposes, for most people, the speed of the processor in your PC doesn't really matter - you will barely be able to tell the difference in CPU speed between the slowest and fastest x86-compatible PC Intel and AMD make. The thing that causes the most perceived delay in my use of a computer is almost always my Internet connection. --Robert Merkel 22:45, 31 August 2005 (UTC)

Review of Manufacturing topic / categories

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I checked a minor manufacturing topic; it got me looking into Manufacturing as a whole. It seems to me that it is a very big topic that is fragmented and difficult to navigate. I am not sure how to start reviewing it without being a vandal! The categories lists also seem in need of review. I might start by creating a "Manufacturing overview" page similar to this temporary page. Joe1011010 19:41, 31 August 2005 (UTC)

I think you should go ahead with a rewrite if you think you can improve the article(s). BUT, in a high level topic normally we would not just say "The beginings of manufacturing is covered in the Industrial Revolution." Rather, it is better to have a condensed overview of the industrial revolution in the manufacuring article, along with an italisized statement indicating that more detail can be found in the article on the main article on the industrial revolution. Happy editing ike9898 17:22, September 6, 2005 (UTC)

biosphere

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Ñ How does energy and nutrients move through the biosphere? How are these two different?

I suggest you start by reading the Biosphere article and then do you're homework yourself - believe it or not, this will help you in the long run (and, no I don't think I would have believed it when I was in school!). Thryduulf 20:50, 31 August 2005 (UTC)

Truth Drug

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how do u test or what kind of test can you take to verify that you have been given a truth drug with out consent?

Well you would notice a sedative effect. Like being drunk. See Truth drug for more info Theresa Knott (a tenth stroke) 20:46, 31 August 2005 (UTC)

Also, a blood test for drugs (the kind pathologists give accident victims, for instance) taken afterward would also reveal their presence. I don't know how long the presence of the different types of potential truth drugs is detectable after their admission, and specifically the "window" between when they wear off sufficiently to give you enough self-control to voluntarily get a blood sample taken, and when they are no longer detectable. I suspect it's a matter of hours rather than days, but I'm certainly no expert in the area. --Robert Merkel 22:59, 31 August 2005 (UTC)

S-box S5 on Substition box page

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First, thank you for your service.

On http://en.wikipedia.org/wiki/S-box, I believe the binary value 1010 is missing as a column heading and as an output.

I think it should be the column heading for column 11, which is currently labeled 1100.

According to http://www.tropsoft.com/strongenc/des.htm, 1010 should be in column 6 (0101), row 1 (00); column 12 (1011), row 2 (01); column 5 (0100), row 3 (10); and column 13 (1100), row 4 (11). At Wikipedia, anybody can edit, with or without an account. If you spot a logical error, go to the page in question, hit the edit this page link at the top, and go to it! For added niceness, you can then proceed to the discussion link and add a link to the source so that we've got it available for review. — Lomn | Talk / RfC 20:55:30, 2005-08-31 (UTC)

Ñ How does energy and nutrients move through the biosphere? How are these two different?

:I suggest you start by reading the Biosphere article and then do you're homework yourself - believe it or not, this will help you in the long run (and, no I don't think I would have believed it when I was in school!). Thryduulf 20:50, 31 August 2005 (UTC)

how do u test or what kind of test can you take to verify that you have been given a truth drug with out consent?

Well you would notice a sedative effect. Like being drunk. See Truth drug for more info Theresa Knott (a tenth stroke) 20:46, 31 August 2005 (UTC)

:Also, a blood test for drugs (the kind pathologists give accident victims, for instance) taken afterward would also reveal their presence. I don't know how long the presence of the different types of potential truth drugs is detectable after their admission, and specifically the "window" between when they wear off sufficiently to give you enough self-control to voluntarily get a blood sample taken, and when they are no longer detectable. I suspect it's a matter of hours rather than days, but I'm certainly no expert in the area. --Robert Merkel 22:59, 31 August 2005 (UTC)

First, thank you for your service.

On http://en.wikipedia.org/wiki/S-box, I believe the binary value 1010 is missing as a column heading and as an output.

I think it should be the column heading for column 11, which is currently labeled 1100.

According to http://www.tropsoft.com/strongenc/des.htm, 1010 should be in column 6 (0101), row 1 (00); column 12 (1011), row 2 (01); column 5 (0100), row 3 (10); and column 13 (1100), row 4 (11).

At Wikipedia, anybody can edit, with or without an account. If you spot a logical error, go to the page in question, hit the edit this page link at the top, and go to it! For added niceness, you can then proceed to the discussion link and add a link to the source so that we've got it available for review. — Lomn | Talk / RfC 20:55:30, 2005-08-31 (UTC)

Most statistics books seem to be written by people who lack basic descriptive skills or have yet to have agood grasp of English writing. Here we see a statistic textbook explaining cummulative frequency, a complex name for an extremely brain-dead operation, and explains it with great complexity:

"A cummulative frequency distribution can be created from a frequency distribution by adding an additional column called "Cummulative Frequency." For each score value the cummulative frequency for that score value is the frequency up to and including the frequency for that value."

Is there any books, or other sources anyone can recommend to me that will not try to complicate this very simple subject?

:Let's paraphrase:

:# Take a graph. A special graph called a "frequency distribution".

:# Let's take, say, the graph of all the edits done on the reference desk...(see below)

Image:RD edit history.PNG

:# Each bar represents the frequency of edits done per month.

:# Now, draw another graph, with the same horizontal axis. (I just replaced the month names with numbers here instead)

:# This time, each month's height corresponds to the total number of edits done up to that month. When you do this, you get something that looks something like this:

Image:Cumulative RD edits.PNG

:This second graph is the "cumulative frequency distribution". Each bar represents the "total up to that date." Sometimes, it's called the "cumulative distribution function" if it's normalized. Hope this helps! --HappyCamper 03:09, 1 September 2005 (UTC)

Is it a good idea to leave orthopedic hardware (9 screws and a metal plate) in the proximal humerus after the bone healed and maximum cure attained?--anon

Alexander

:That'd be a question for your surgeon.

:I can tell you that I know someone who had screws left in his bones for more than 30 years, and the only problem was that they had to be removed when the area needed another operation. After that long they were a bit hard to get out, and the extra hoops (waiting for it to heal up a bit before the next, more involved operation) dragged the whole process out.

:Another friend has told me that all the hardware in his legs has made running painful as the metal does not absorb impact as well as bone. I'm sure there are other potential problems that depend on exactly what is left in where.

:Your doctor will know your situation much better than us here, so ask him to go over the pros and cons with you. — Laura Scudder | Talk 04:26, 1 September 2005 (UTC)

::From our personal experience, the hardware of which you speak (titanium) and in the same location of your concern has had no adverse effects after five plus years. In fact, removal would present an unnecessary risk. I'm speaking to adult and not pediatric bone growth concerns. The only side effect for Heidi seems to be at airport security. ;-) hydnjo talk 21:05, 2 September 2005 (UTC)

I live in Bend OR. I recently witnessed 2 slugs doing their duty, ( I think!) But all I could see coming out of the aftermath were more slugs.I saw protruding tentacles, slime, ect., but no egg sack. It looked like live birth. They wrapped around each other, explored, ( as good sex should be), but then there was another slug...and another. It was beautiful, but disturbing. Did we witness hatching, or what????Do they come back to the scene of the crime? It was an intimate moment, and it looked tender ( who thought slugs could be tender), But what the >!@?":.....

:The slug article does have a section on their reproduction (Slug#Reproduction_and_life_cycle), and it does say that they wrap around each other like you describe. However they apparently lay eggs in the ground, and don't have live births. Thryduulf 09:00, 1 September 2005 (UTC)

Hi there-- recently I have been finding out about gravity and a question keeps itching me. If you were to drain half of the magma from the earths core would the remaining liquid rearrange itself into a perfect sphere centred around the point at which gravity for this planet is generated?

In case that was slightly confusing what i mean to ask is that gravity is a single point in the exact centre of our planet so if there was just an open void around that point and something was placed on that point would it just float exactly still as forces are acting equally uopn it in every directon?

:The gravity of our planet is "generated" in each and every part of it. When you add up all these minute gravitation pulls you get what we experience as gravity.

:Assuming no other objects exist apart from the earth, and the earth was made hollow (and it didn't crumble), and the earth was a perfect sphere, inside the earth there would be no gravitational field or pull. As you get closer to one side of the sphere you get closer to the particles and therefore their pull on you will be stronger, but there'll be less particles in that direction of you. This evens eachother out. As far as I know, the first person to come up with this was Isaac Newton in his Principia.

:So, to answer your question, the magma would do whatever it damn well pleased, i.e. there would be no gravitational force working on it, though there could be other forces. --fvw* 10:51, September 1, 2005 (UTC)

::See also Hollow Earth. Thryduulf 11:02, 1 September 2005 (UTC)

Thanks for the answer but it leads me to another question. If every part of the earth produces a gravitational pull and these minute pulls combine to form what appears to be one force known as gravity why is it that we are not pulled slightly towards huge clusters of particles like say a mountian and also why is it that all of these forces seem to pull towards one point in the centre of the earth?

:We are pulled slightly towards mountains, but AIUI the force is so weak that it is only measurable by scientific instruments. It takes the mass of something on the scale of a planetary body to produce a force strong enough for us to feel, and this is why we appear to be pulled towards one central point. Thryduulf 15:26, 1 September 2005 (UTC)

:(edit conflict with Thryduulf) In fact, they don't pull you exactly towards the center of the Earth. Measuring local deviations in gravity is a huge part of geophysics and is a way to get information about the density of material under the surface. See gravity anomaly and geoid. — Laura Scudder | Talk 15:33, 1 September 2005 (UTC)

Oh and your comment about getting closer to one side of the sphere and the force growing stronger but the lack of particles meaning they even each other out. What is that supposed to mean? These particles that are fewer as you move closer to one side, are they supposed to be air particles because thats the only thing i can imagine moving through as easily as you say. And even each other out!!?? If the forces evened each other out wouldn't you be stuck in your position unable to move unless you can produce a force strong enough to move in a direction?

:Basically, if you are to be pulled downwards by gravity, then the particles exerting the gravity need to be below you. When you are in the centre of the hollow sphere there is more of the sphere below you than there are when you are closer to the edge (as there is more of the sphere beside and above you). If all the forces evened each other out you would be stuck as you describe, however the force required to overcome them wouldn't be difficult - for example, consider how easy it is to move under the force of gravity from a body the size of planet earth. Thryduulf 15:26, 1 September 2005 (UTC)

:What he was saying is that you're closer to a small amount of the Earth's rock, but farther away from a large portion of the Earth's rock. So less rock is pulling on you in one direction (but there's a stronger pull per bit of rock because you're closer), while more rock is pulling on you in the other (but there's a weaker pull per bit of rock because you're farther away) and they end up cancelling eachother out. This accounts for why you don't feel any net gravitational force inside a hollow sphere. A mathematical way of proving this uses Gauss's law. — Laura Scudder | Talk 15:33, 1 September 2005 (UTC)

One more thing, if 'particles' of whatever it is are responsible for gravity then is a black hole just an unspeakable number of them clustered together?

:See teh Gravity and Black hole articles. A better answer is beyond my understanding! Thryduulf 15:26, 1 September 2005 (UTC)

:Yes. A black hole is what you get when so much matter is in one place that it all collapses under its own weight, into a singularity - a point so small it cannot be measured. The gravitational attraction of this is, as you'd expect, pretty big, since it requires a lot of matter - the absolute minimum mass to form a black hole is given by the Chandrasekhar limit, which is about one and a half times the mass of the sun. Shimgray 16:06, 1 September 2005 (UTC)

Thanks guys

i dont now

Image:Yellowsky_31_August_2005.jpg

Hello, everyone. I am writing to ask about something that I saw last night. After a rather grand storm, in which all light faded, suddenly not only the sky above, but all the space outside, became a disorientating, deep chemical yellow (see the picture to the right). Walking outside was like walking into yellowness. After about twenty minutes, during which time the yellow became darker, the light shut out, and we were in total darkness. I have never before witnessed such a thing. Do you know what this occurance is called, and what causes it? I would appreciate any help. IINAG 13:08, 1 September 2005 (UTC)

:Well, in Texas we just say it means there's going to be a tornado with the storm, but I have of course no scientific backing for that and it doesn't help with the why. — Laura Scudder | Talk 15:38, 1 September 2005 (UTC)

::Hello, Laura; thank you for your response. I had never knew before about there being a yellow sky with a tornado. The only thing is that I live right in the middle of England, in an area not noted for tornados (however, there was a small tornado in a nearby city, Birmingham, although this was a freak incident.) That has me wondering whether the same reason(s) that make(s) the sky yellow before a tornado acted when the air yellowed yesterday. IINAG 16:56, 1 September 2005 (UTC)

:::I'm trying to think of a plausible explaination for the yellow. It may be due to the storm kicking up an unusual amount of dust, and it being illuminated indirectly, by light that had already been depleted of blue by Rayleigh scattering. -- DrBob 17:37, 1 September 2005 (UTC)

::::It must be associated with some change in weather. I lived in Kansas for thirteen years and once in a long while something like you described would happen. The sky would rapidly adopt a very strange yellowish color. Often it was also strangely warm, and as Laura mentioned a decent indicator of an approaching tornado.

:::Just to clarify: yellow sky -> tornado, but tornado !-> yellow sky. In other words, as the former Kansas resident says, this is rarer than tornados so it must be related to a rather particular set of storm conditions. I found [http://www.usatoday.com/weather/resources/basics/green-thunderstorms.htm this article], which tries to explain it. — Laura Scudder | Talk 14:57, 2 September 2005 (UTC)

:::::Thank you all for all of your help; that article is quite useful. The thing that confused me was that, ostensibly, there was no change between the weather before the yellowing and after it. Before it, it was dark, and it was raining heavily; after it, it was dark, and was raining heavily. IINAG 00:56, 3 September 2005 (UTC)

:I'd like to point out that I live in Michigan, and saw this once. It wasn't quite as yellow as the picture you showed though! It was a bad storm, but I have seen worse. We did not have a tornado, but I have heard of that before. --Phroziac ^(talk) 17:12, September 3, 2005 (UTC)

I occasionally play a little with Linux on my GameCube. For that, I have a 1GB harddrive image for use with a Network Block Device server.

However, I occasionally wish to manipulate this image without having to boot the Cube into Linux - injecting files, etc. Does anyone know if there are any tools that will let me do this? It doesn't matter if the image is mountable as a drive, or if I need to use a seperate program that mounts the image internally. --Pidgeot (t) (c) (e) 13:35, 1 September 2005 (UTC)

: Search for "ext2 windows" on google. The second links looks good, but I'm sure there are several other programs if you don;t like that one. --R.Koot 15:31, 1 September 2005 (UTC)

::Actually, I already have the program at the second link installed, but it only works with physical harddisks - however, Explore2Fs, the first link, did the trick - for some reason, I never thought of the image as a regular ext2 disk. >_<

::So thanks for the wake-up call, I needed it ^_^;;; --Pidgeot (t) (c) (e) 15:44, 1 September 2005 (UTC)

Is there any simple (free) way of transforming a word document to a pdf? I tried converting to OpenOffice, but their pdf converter makes the document ugly. Why doesn't Word have its own converter?

Thanks! --Alice

:Adobe sells a PDF plugin for Word - it's not bad. The problem is that it's expensive - PDF is a proprietary file format, unlike (say) HTML, so Adobe take what the market will bear. There are a variety of free converters out there, though - do a search on doc to pdf or word to pdf - but they'll take some fiddling to work.

:Also, it's quite possible the flaw lies with Word and not OpenOffice; Word uses a lot of odd proprietary formatting tricks, and often doesn't convert very well to other formats. So you may be unable to transfer some of the formatting regardless. It may be simpler to save to an RTF file, then open that through OpenOffice. Shimgray 16:02, 1 September 2005 (UTC)

::It wouldn't surprise me if it *is* OpenOffice that's to blame - in my experience, it's very bad at PDFs, even with documents never opened in Word.

::I suggest downloading pdf995 ([http://pdf995.com]), a printer driver. It lets you export any printable document to PDF by printing it through a special printer. You'll have to live with it popping up a "registration reminder" everytime you make a PDF with the free version, but it's really no big issue, and you can just keep on using the free version as long as you want, there are no limitations. --Pidgeot (t) (c) (e) 16:11, 1 September 2005 (UTC)

: PDF actually is an open format. Have you tried OOo 1.1? If found that its PDF converted does a much better job than OOo 1.9/2.0. --R.Koot 16:25, 1 September 2005 (UTC)

• There's also a program called CutePDF, but the people who told me about that weren't all that happy about it. Still worth a try maybe. - Mgm|^(talk) 17:39, September 1, 2005 (UTC)

Hi all, thanks for your replies.

I tried Pidgeot's suggestion and it seemed to work well. However, it changes the formatting o my document. I've got a 40+ page paper, with chapter headings at the tops of pages and References at the end. There appear to be a different number of lines per page in the pdf, which makes my headings appear in varying places down the page, and the references after 1 1/2 blank pages. How can I get it so that the formatting is the same as in my word document (and yes, I'll learn latex for the next paper!).

Thanks, --Alice

:Alice, one free (but somewhat inconveient) way around this is to install a postscript printer driver (you can download one for free from Adobe, IIRC), print your document to a postscript file, and use Ghostscript to convert the postscript to PDF - it comes with a utility called ps2pdf for just this purpose. Your mileage may vary...--Robert Merkel 23:26, 1 September 2005 (UTC)

:If pdf995 changed the formatting of your document, it may those changes would have been there anyway. If you were printing from OOo, that could be why. Use print preview (in Word!) to make sure it isn't a formatting error somewhere in your document.

:You can also try the print->PostScript->PDF solution mentioned above, but since that's basically what PDF995 already does, it might not help. --Pidgeot (t) (c) (e) 04:48, 2 September 2005 (UTC)

Speaking of pdf, I have scanned some pages of a document and saved it as a png file (resolution 800px). I want to stitch all these pages as a single pdf file. How do I do this? =Nichalp «Talk»= 06:34, September 2, 2005 (UTC)

:There are a few solutions to this, none of them necessarily easy. If you have a full copy of Adobe Acrobat available, it woudl be pretty easy, but if you are asking on here you probably don't have one. If you are using an Mac running OSX, you could "print" them as PDFs (see below) individually and then use Automator, I think, to add them into one big file. The other option I know of involves using the PDF drivers which come with PHP to stitch them together, but this takes a fair amount of coding knowledge (I've done it with TIFs before). --Fastfission 19:28, 2 September 2005 (UTC)

::No, I don't have access to a Mac :(. I use Win and have a limited knowledge in Linux. What if I embed the images in each page and use OO to 'print' the pdf's? I won't lose the formatting as it is a png afterall. =Nichalp «Talk»= 05:51, September 3, 2005 (UTC)

Alice: If you are working on, or have access to, an Apple Macintosh running Mac OS X, and which also has a copy of Word (perhaps your local library or uny computer lab?), it's easy: Open the file up with Word on the Mac, select "Print" from the "File" menu, then click the "Save as PDF" button. The PDF will look more or less exactly like how the document would look if you printed it from Word. Viola! Yet another reason to love the Macintosh. Garrett Albright 06:53, 2 September 2005 (UTC)

My favorite utility so far for PDF creation is [http://www.primopdf.com/ PrimoPDF]. It's a wrapper around GhostScript to make it act as a printer driver, and it's all open source with no registration reminders (unlike CutePDF). --David Wahler (talk) 17:52, 7 September 2005 (UTC)

What are some accurate methods you could use to determine the identity of a sample of a white solid?

--anon.

:your title is missleading, this isn't physical chemistry, this sounds like a general chem lab, at best--64.12.117.12 02:10, 4 September 2005 (UTC) You didn't read the title carefully, 64. Physical science is not the same as p chem. In the US, physical science is often an early secondary school subject that is a less rigorous combination of chemistry and physics for those who will not be taking them individually. Our questioner will likely never know what p chem is. alteripse 04:22, 4 September 2005 (UTC)

• Solubility in water and other solvents
• pH in solution and reactions to acid or alkali
• Microscopy to look at crystal shapes
• The sequence of procedures of qualitative analysis (organic or inorganic)
• X-ray diffraction analysis
• Mass spectrometry
• Reaction of animal trained to react to drugs or explosives
• Feed it to a guinea pig or other, even better, a disposable, non-cute and non-furry, animal
• Taste it, like Alice did
• Circumstances of how it came into your possession (location, container, amount, accompanying objects, labels, etc)
• Ask the owner or provider of the sample

alteripse 03:44, 2 September 2005 (UTC)

I'd go for taste if it wasn't for the potentially deadly results. If you disolve it in water and it the water can now transfer electricity it's a salt. If you smell cloride it's a cloride salt. You can then add other water solutable salts to test whcih to ions the salt was composed of (look in the chapter Salts in your chemistry book for the exact procedure). --R.Koot 03:55, 2 September 2005 (UTC)

I was once told that Humphry Davy tested for taste of every new substance he discovered. I was surprised he survived, given that he isolated Sodium and Potassium among others. DJ Clayworth 19:41, 8 September 2005 (UTC)

Also flame test and borax bead test. These will help identify the substance, if only to show what it isn't, since some

substances do not give helpful results. Melting point is another useful tool, as well as density.

Hi,

I know this might not be technical enough to warrant your attention but do you happen to know the exact speed of sound in MPH, KPH and Knots? I know its around 750MPH but i need to be sure in order for my calculations to be correct. Ryan

:The speed of sound is not a constant. It highly depends on the material that carries it and the temperature of that substance. --R.Koot 10:06, 2 September 2005 (UTC)

:See speed of sound for some more information. --R.Koot 10:07, 2 September 2005 (UTC)

In that case can you tell me which sequence of variables they use to determine the top speed of planes ie. If a company states it's plane can go Mach 2 then which speed are they referring to as it could be anywhere between 1200mph and 1400mph?

:I would bet that the Mach number is a more accurate measure of the limit as a plane's top speed will also change with conditions. — Laura Scudder | Talk 15:41, 2 September 2005 (UTC)

::As stated in the Mach number article "Mach number (Ma) is defined as a ratio of speed to the speed of sound in the medium in case." It would seem that the reference speed of sound is that in which the plane is flying. hydnjo talk 19:37, 2 September 2005 (UTC)

::There are a few directions you can go with this (since the speed of sound, even if you fix the medium as "air", will vary):

::*speed of sound at a standardized altitude (say, sea level)

::*speed of sound at a nominal cruising altitude, not necessarily standard across aircraft types

::*speed of sound at the best possible, albeit extremely unlikely, conditions

::From a marketing standpoint, guess which is most likely to be the case :) — Lomn | Talk / RfC 21:14:14, 2005-09-02 (UTC)

Just to complicate things, note that a plane (a jet, as prop plane are always subsonic, I think) will go faster in rarefied

air, because of reduced friction, but sound will tavel slower in that medium. There is a law of diminshing returns here, as

rarefied air will provide less oxygen to the jet engine and it will produce less power.

Hi,

I recently read an article on a similar site that stated; if you were on a train travelling at 1000mph (and the ride was so smooth you couldn't notice) and you fired a bullet off the front of the train (in the exact direction it was travelling) then the bullet would be travelling at 2000mph relative to a stationary point on the ground and at 1000mph realtive to your position (moving at 1000mph). This I can believe, but it also states that if you fired a bullet under the same circumstances off the back of the train (this is all theory, I'm not taking into account aerodynamics or anthing) that the bullet would be travelling at 2000mph relative to your position on the moving train but it would be travelling at 0mph in relation to a stationary point on the ground. This suggests that the bullet would float in that air on then just fall to the ground. Is that correct because it seems very unlikely?

: Yes it is true, however... The bullet would not float in the air, but immediatly start to fall to the ground. If the bullet and the train are travelling at the same speeds, but in opposite directions, it will fall on the ground, exactly were you shot it. You will be quite a distance away however, because you are traveling at such a great speed.

: You can test this yourself with a heavy object. Hold object, start running, then rlease it (don't throw it) it should travel along you for a brief moment, and will not fall on the floor were you released it. Do it agian, but trhrow the object in the opposite direction of which you are running, with a bit of tuning you can make the object land at the spot were you threw it. --R.Koot 11:33, 2 September 2005 (UTC)

: I completely agree with you. This can be explained on the basis of the fact that motion itself is a relatie term. It depends on the frame of reference when you percieve an object to be in motion. By frame of reference i mean your position with respect to the object in motion (like the train and the stationary point). Most of the time the apparent speed of the object can be calculated using simple addition of vectors. --StratOnLSD 11:58, September 2, 2005 (UTC)

: I think the air behind this 1000 mph train would be somewhat disturbed by the passage of the train, and more than able to whip a bullet around all over the place. So the actual result wouldn't be as simple as "floating in the air". Notinasnaid 12:03, 2 September 2005 (UTC)

: yes, it would just drop to the ground (apart from being sucked along a bit in the train's wake, but that would not be very much anyway), but it would not travel "2000mph relative to your position on the moving train"; only 1000 mph. It is a 1000mph bullet.

Thank you to R.Koot and StratOnLSD (interesting name) for those answers, and to Notinasnaid; I believe if you read my question you would understand I was disregarding any other forces, variables and factors not covered in my question and if technicality is your specialty (which it appears to be) then you shouldn't have answered a question about a train moving at 1000mph.

:Also, if you do three things simultaneously: fire the bullet forwards, fire the bullet backwards (both exactly horizontally), and drop the bullet from your hand, all three bullets will hit the ground at exactly the same time.

:Another interesting result (again assuming no air resistance or other complicating factors), if you shoot the bullet directly upwards from the moving train, it will come back down at precisely the same point on the train. -- DrBob 16:38, 2 September 2005 (UTC)

::Which has got to make one wonder why shooting bullets up into the air is such a common form of celebration in some cultures. Has no one ever been indirectly shot by a falling bullet? hydnjo talk 19:02, 2 September 2005 (UTC)

:::I'd think it's happened at some point. Similar injuries occured during WW2 as a result of shrapnel from anti-aircraft barrages eventually having to fall to the ground -- since those were usually over large cities, the chance of injury was a good bit higher. — Lomn | Talk / RfC 21:10:39, 2005-09-02 (UTC)

:::I don't have the details to hand, but IIRC there is at least one recorded case of an aircraft being hit by its own bullets - a modern jet, on a gunnery range, fired some ammunition and then went into a shallow dive, picking up speed. By a combination of a miscalculation and some very bad luck, it accelerated underneath its own falling shells... Shimgray 03:20, 3 September 2005 (UTC)

:::Yes, this definitely happens. I read an article about someone dying from gunshots into the air in a wedding somewhere in the mideast. Luckily most of the cultures that do that are in more sparsely populated areas. One specific example, my Aunt's store had a bullet come down at a very steep angle through the window in the front of the store. It shattered the window and a number of glass items in displays. I can't remember how far away they estimated the bullet was from, but likely within a mile or so. The other addition to DrBob's point, is that not only does the bullet come down in the same place if fired straigh up with no wind, but even with air resistence, it comes down at essentially the same speed it left the barrel with. fast enough to kill. - Taxman ^Talk 18:22, September 3, 2005 (UTC)

:::: I very much doubt this. The Mythbusters proved you can't be killed by a penny dropped from a building, so I doubt you can be killed by a bullet falling from the air. It will be travveling at a much slower speed than when it was shot at you, due to air friction limiting the bullet to a certain speed. --R.Koot 19:52, 3 September 2005 (UTC)

::::::They explained it pretty well below. The deaths part, and the exact speed coming down I did not look up, but the part about my aunt's window was unequivocably confirmed by finding the bullet that did the damage, and being able to discern the path it took. - Taxman ^Talk 16:10, September 6, 2005 (UTC)

:::::It's also, however, more aerodynamic and heavier. IIRC, the Mythbusters conclusion was that pennies lacked sufficient mass more than they lacked velocity. — Lomn | Talk / RfC 01:46:37, 2005-09-04 (UTC)

::::Point 1: bullets are relatively aerodynamic, but air resistance still slows them down hugely and quickly. Take a look at the data table in external ballistics - even though this bullet has a high ballistic coefficient, it sill looses one third of its initial velocity in about two thirds of a second. This is an average decelleration of about 400 m/s^2 i.e. about 40g. Point 2: the terminal velocity of a falling bullet is between 60 and 100 m/s (depends which source you look at - probably depends greatly upon type of bullet, whether it tumbles etc.). Fast, but nowhere close to its muzzle velocity. Point 3: even so, are people killed by falling bullets ? Yes, for example it happens regularly in Los Angeles, according to the sources quoted in [http://www.straightdope.com/classics/a950414b.html this article]. Gandalf61 11:29, September 4, 2005 (UTC)

Hello!

I'm in need of power formulas associated with both delta and wye connected AC generators.

Can anyone help or provide links?

Thanks,

Rick Gilbert

Presuming a balanced three-phase resistive load:

DELTA

Vdelta = Vline

Iline = √3 Idelta

Ptotal = 3 Vdelta Idelta = √3 Vline Iline

WYE

Vwye = √3 Vline

Iwye = Iline

Ptotal = 3 Vwye Iwye = √3 Vline Iline

Yoke

I got stumped working on the Elbow macaroni article. They make macaroni by extruding dough through a circular die with a pin blocking the center of the circle. I see a number of references that say that a "groove" in the pin causes the pasta to curve as it is extruded. How? What does this groove look like? Why doesn't it form a ridge or valley on the finished product? Bunchofgrapes 17:11, 2 September 2005 (UTC)

• I haven't found the answer yet, but I did just find an assertion that Thomas Jefferson invented a macaroni extruder!--jpgordon∇∆∇∆ 02:31, 4 September 2005 (UTC)

It has to do with relative rates of extrusion. Pasta machines work by forcing pasta dough through a die, which simplistically speaking is just a negative silhouette of the desired shape. But if more dough is forced through one part of the die than another part, the extrusion there will happen faster, and the extruded pasta will curl.

A picture will make this much clearer.

Here is the front view of an elbow macaroni die:

Image:Elbow_macaroni_die_front.jpg

This is about 30cm in diameter, and as you can see it extrudes 5 tubes of pasta at once.

But now let's look at the back:

Image:Elbow_macaroni_die_back.jpg

The ridges you see divide the total output of the pasta machine

(that is, the stream of pasta dough which this die is the target of)

into ten "capture regions".

(There's a rather nice analogy here to watershed.)

Two regions flow into each extruded macaroni tube,

but significantly, one of them (the outer one) is considerably larger than the other.

Therefore, the outer edge of each tube will extrude faster than the inner, and each one will therefore curl inwards.

I haven't seen one, but I presume that the die for curly-edge lasagne consists of the long straight slit you would expect, but with some extra "capture region" at each end of the slit (corresponding to each edge of the lasagne ribbon). In this case, the extra dough volume is too localized to cause a curl, but it has to go somewhere. The edges end up being "too long" for the main body of the ribbon, so the edges end up rippling to consume the extra material.

Steve Summit (talk) 04:36, 6 September 2005 (UTC)

How were portions of New Orleans built below sea level? Did these areas sink? Did they drain and/or excavate after building levees? Are below-sea-level-portions of the city built on drained parts of The Mississippi River or Lake Pontchartrain?

Any information would be much appreciated.

Thank you,

Neil Higgins

:The answer is complex. Start with the Wikipedia article New Orleans, Louisiana and then take a look at Hurricane Katrina, Effect of Hurricane Katrina on New Orleans and Damage to infrastructure by Hurricane Katrina. I've found a lot of historical reference within these articles that address your questions. hydnjo talk 19:47, 2 September 2005 (UTC)

A slot machine has three wheels: Each wheel has 11 positions – a bar and the digits 0, 1, 2,…, 9. When the handle is pulled, the three wheels spin independently before coming to rest. Find the probability that the wheels stop on at least one bar.

I have two answers. I think the first one is right, but I do not understand why the second one might be wrong.

P(E₁) = $1-\backslash frac\{10^3\}\{11^3\}=\backslash frac\{331\}\{1331\}$,

where I subtract the probability of getting no bar three consecutive times (10/11)³ from 1.

P(E₂)= $\backslash frac\{1\}\{1\}+\backslash frac\{1\}\{11^2\}+\backslash frac\{1\}\{11^3\}=\backslash frac\{133\}\{1331\}$, where I add the probability of getting one bar, two bars, and three bars.

Can you explain this discrepancy? Why do these two answers not conform to each other?

--anonym

:The second method you used, while the idea of adding up the probablity of getting one, two, and three bars will give you the correct answer, you have not done so correctly, and it is much simpler to take the probablity of getting no bars and subtract it from 1. Your first method is correct. Y0u (Y0ur talk page) (Y0ur contributions) 21:36, September 2, 2005 (UTC)

To expand on Y0u's answer. The probability of getting one bar is

$\backslash frac\{1\}\{11\}\backslash times\backslash frac\{10^2\}\{11^2\}\backslash times\; 3\; =\; \backslash frac\{300\}\{11^3\}$

i.e. the probabliity of a bar on the 1st multiplied by a not bar on the other two multiplied by three (because the bar could be on any of the three not just the first one).

Likewise the probability of two bars is

$\backslash frac\{1^2\}\{11^2\}\backslash times\backslash frac\{10\}\{11\}\backslash times\; 3\; =\backslash frac\{30\}\{11^3\}$ Theresa Knott (a tenth stroke) 22:21, 2 September 2005 (UTC)

I recently saw a commercial for Old Navy where folks were picking Old Navy products from a field of some kind of gourd-like plants, handsome and erect with big leaves. Does anyone know what kind of plants those are? Attractor 00:08, 3 September 2005 (UTC)

:$12 Jean plants!!!! {{=)|wink}} — Ilγαηερ (Tαlκ) 01:12, 4 September 2005 (UTC)

Does anyone publish detailed scene-by-scene film synopses that fill in plot gaps that you can't get by listening alone? Attractor 00:08, 3 September 2005 (UTC)

• Well, for the really good films, you can find complete walkthroughs at http://www.filmsite.org -- it's worth bookmarking that site in any case. --jpgordon∇∆∇∆ 01:53, 3 September 2005 (UTC)

:There's also [http://www.themoviespoiler.com/ The Movie Spoiler], which might be useful. --Sum0 18:32, 15 September 2005 (UTC)

A safe containing $1,000,000 is locked with a combination lock. You pay $1 for one guess at the six-digit combination. If you open the lock, you get to keep the million dollars. What is your expectation?

Thus, $E\; =\backslash frac\{10^6\}\{720\}-\backslash frac\{719\}\{720\}\backslash approx\$1,387.89$.

Am I wrong?

--anon

$E=(10^6-1)\backslash frac\{1\}\{P(10^6,6)\}+(-1)(1-\backslash frac\{1\}\{P(10^6,6)\})=-\$1$.

--anonym

If there are 1000000 possible combinations (assuming 6 decimal digits), then $1000000/1000000 - $1, and if you get to try again, $1000000/999999 - $1 and then $1000000/999998 - $1 the next two times. Κσυπ Cyp   06:09, 3 September 2005 (UTC)

In the World Series, the top teams in the National League and the American League play a best-of-seven series; that is, they play until one team has won four games. (No tie is allowed, so this results in a maximum of seven games.) Suppose the teams are evenly matched, so that the probability that either team win a given game is$\backslash frac\{1\}\{2\}$.

I would like to model the World Series with a program on my graphing calculator (TI-83), where “heads” represents a win by Team A and “tails” a win by Team B. Then I want to use the program to estimate the probability that an evenly matched series will end in four, five, six, and seven games. Even though I did a little Visual Basic programming about five years ago, I am clueless.

--anonym

: the TI-83 interface is probably too primitive for you to do that, but if you really want to try, you might try looking up the rnd(0,1) command, random number generator, if I remember right.. nothing like Vb really, more like QB--64.12.117.12 02:08, 4 September 2005 (UTC)

• I know that rand generates . All the programming commands are in the catalog menu or the PRGM menu. The same instructions should be executable on all Texas Instruments calculators, id est, they are the same. --anonym
• Not being a TI user myself, I can't really help - but it may be helpful to know the way of calculating a random number when rand generates a number between 0 and 1 is to multiply the number by the upper end of your range (so if you wanted numbers from 1 to 10, RAND*10) and then round to 0 decimal points. Additionally, if you want to increase the lower end, try eg. RAND*5 + 2 for random numbers between 2 and 7.
• If you want integer values though, the TI-83(+)[possibly the regular version too] can do that directly. Use randInt(near rand). Syntax is randInt(1,10) to generate an integer from 1 through 5(inclusive).
• If you just want to know the probability of the series ending in 4,5,6, or 7 games, it's easier to do it mathematically.
• 4:(1/2)^4*2=1/8
• 5:(1/2)^5*4*2=1/4
• 6:(1/2)^6*10*2=5/16
• 7:(1/2)^7*20*2=5/16
• if I'm not mistaken. Tell me if you want a more in-depth explanation. The probability of any win-loss combo is (1/2)^(num. games). Then you multiply by the number of possible combos for each situation. You multiply by two at the end because the quantity before that is the probability of one of the teams winning in that number of games but you want to find the probability of either of them doing it. For example, there are only two ways a series can end in four games. Team A wins four and a row or Team B wins four in a row. The probability of either is 1/16. Hence, the probability of a four game series is 1/8, the sum. However, this is all probably a huge waste of time because your teacher is requiring you to do the modelling. If that's the case, I'll provide a few hints. First: To decide who wins a game, generate a random number with rand. If it's less than .5, give it to the National league. If it's greater than or equal to .5, give it to the American leaguers. I.E.

If(rand<=.5)

Then

Disp "National win!"

Else

Disp "American win!"

End

If, Then, Else, and End are in the CTL(control) menu while programming. <= is in the Test menu. Press 2nd, Math, then select it. If you need more help, tell me and I'll see what I can do. Superm401 | Talk 23:03, 13 September 2005 (UTC)

• I was going to give up on this problem and wait until I learn about programming and then do it if I ever get a chance in the coming years. I thoroughly understand the mathematics behind it and I spent over an hour constructing the table below it. Here is my nonfunctional guess.

:BASEBALL

:0-->A:0-->B

:For(N,1,4)

:Rand--->X

:Int(2X)--->Y

:A+(1-y)-->B [(1-y) is either 0 or 1.]

:End

Again, I am clueless about programming and this is giving me hints is not likely to be fruitful. Yet, I am meek and am not demanding an answer. If you can, then please write the program. Also, I neither attend any school nor is this a homework. I am an old adult who is doing this for satisfaction. This programming problem is from an elementary mathematics textbook. --anonym

In the World Series, the top teams in the National League and the American League play a best-of-seven series; that is, they play until one team has won four games. (No tie is allowed, so this results in a maximum of seven games.) Suppose the teams are evenly matched, so that the probability that either team win a given game is $\backslash frac\{1\}\{2\}$.

What is the probability that the series will end in four games? Five games? Six games? Seven games?

--anonym

And how far with this homework have you got all by yourself? Theresa Knott (a tenth stroke) 16:32, 3 September 2005 (UTC)

• I do not know how to start. --anonym.

Does P(4 games) = $\backslash frac\{C(4,2)\}\{C(7,2)\}=\backslash frac\{6\}\{21\}=\backslash frac\{2\}\{7\}?$

--anonym

Nope. For P(4 games) you need to have one of the two teams win all four games in a row. (There is no other way it can happen). Call the teams A and B. So I reckon (and I'd like people to check my reasoning please) that

$P(4\; \backslash \; games)\; =\; \backslash Big\backslash \{\; P(A\_1)\backslash times\; P(A\_2)\backslash times\; P(A\_3)\backslash times\; P(A\_4)\; \backslash Big\backslash \}\; +\; \backslash Big\backslash \{\; P(B\_1)\backslash times\; P(B\_2)\backslash times\; P(B\_3)\backslash times\; P(B\_4)\; \backslash Big\backslash \}$

Where $P(A\_1)$ means the probability that side A wins game 1 and so on.

Now all these probablilities = 1/2 so that gives

P(4 games) = ?

It's harder as you go to 5 games because you need A to win 4 and lose one and there are 5 different ways this can happen (do you see why?) Theresa Knott (a tenth stroke) 22:34, 3 September 2005 (UTC)

:Actually, there are only 4 ways for this to happen... You don't play game 5 if you win the first 4. Similar conditions must be imposed on the 6- and 7-game scenarios. — Lomn | Talk / RfC 22:36:46, 2005-09-03 (UTC)

:Oh yes you are right! Theresa Knott (a tenth stroke) 22:39, 3 September 2005 (UTC)

::Technically, only the 5- and 6-game scenarios need such conditions, as P(7)=1-P(6)-P(5)-P(4). --Pidgeot (t) (c) (e) 22:42, 3 September 2005 (UTC)

:::True. The conditions do exist, but there's no need to run the calculations involving those conditions. — Lomn | Talk / RfC 23:04:23, 2005-09-03 (UTC)

: To the anon - can you see why that's true? I have to say your teacher set a pretty difficult HW but if you study it enough you'll learn a lot. Theresa Knott (a tenth stroke) 22:45, 3 September 2005 (UTC)

• We shall consider that team A wins. We can double the probabilities for either team winning. Moreover, we will denote the wins of A win W and the losses of A with L. We also should remember that the last game must be a win to finalize the World Series.

• (The total probability for each game)=2×(probability for A). Note that the sum of all the total probabilities is one. My table on the Edit page does not show, no matter what I do.

The last game MUST be a win

Number of games No. of Sequences Probability

--------------- --------------- -------------

4 WWW|W = 1 (1/2)^4 = 1/16

5 LWWW|W = 4 4 x (1/2)^5 = 1/8

6 LLWWW|W = 10 10 x (1/2)^6 = 5/32

7 LLLWWW|W = 20 20 x (1/2)^7 = 5/52

The total probabilities will be double the probabilities shown, So

Series ends after 4 games = 1/8

" " 5 " = 1/4

" " 6 " = 5/16

" " 7 " = 5/16

-------------

Total = 1.00 (so this checks)

--anonym

What is the HTML equivalent of {nowiki} {/nowiki}? --Commander Keane 10:44, September 3, 2005 (UTC)

:One way to do it is using Pidgeot (t) (c) (e) 10:54, 3 September 2005 (UTC)

::Unfortunately, I didn't remember correctly. The other tag is Pidgeot (t) (c) (e) 11:24, 3 September 2005 (UTC)

::Note that PRE is intended to treat whitespace literally. It does not suppress the interpretation of HTML tags and entities. There is a small set of inline elements that are permitted within PRE, although browsers seem to tolerate almost anything. See [http://www.htmlhelp.com/reference/html40/block/pre.html this website] for more details. Bovlb 14:44:37, 2005-09-03 (UTC)

You could use < ; and > ; tags like this <html tag > Theresa Knott (a tenth stroke) 16:29, 3 September 2005 (UTC)

:Keane, could you perhaps give us a little bit more explicit idea of what you're trying to do? Garrett Albright 17:20, 3 September 2005 (UTC)

::I was trying to ask about the use of "&nbsp;" here, which I acheived by swapping the usualy "&" with "&". I've done it now, and I'll try to be more specific next time I ask a question. Thank you. --Commander Keane 17:36, September 3, 2005 (UTC)

:::You have to use the "&" trick even with nowiki, it's a real pain until you discover it... Physchim62 20:38, 4 September 2005 (UTC)

It's (yes, it's very obscure, unless you are a XHTML user, where it's used all the time with