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When did the state of New Jersey started regulating psychotherapy? I found this online:1945 "The state of Connecticut passed licensure legislation for psychologists, becoming the first state to recognize psychology as a protected practice oriented profession." (source).
--JWSchmidt 04:40, 31 July 2006 (UTC)[reply]

Looks like it was around 1968. .pdf file of New Jersey law - Nunh-huh 04:51, 31 July 2006 (UTC)[reply]

Thanks. Is it safe to assume that before ~1968 the New Jersey State Board of Medical Examiners would have been involved in State regulation of psychologists and anyone practicing psychotherapy or was psychotherapy basically not regulated by the State? --JWSchmidt 05:06, 31 July 2006 (UTC)[reply]

I don't think the Board of Medical Examiners played any role, as the practice of psychotherapy seems to have been unregulated before that time. The NJBME would have (still does) regulated psychiatrists, of course, but that was the practice of medicine. - Nunh-huh 05:25, 31 July 2006 (UTC)[reply]

Wikipedia articles about scientology heavily cite the website of Operation Clambake. One page at their website says, "In 1951, the New Jersey State Board of Medical Examiners filed charges against HDRF for teaching a branch of medicine without a license". I'm trying to figure out if this makes sense. The idea that a "Board of Medical Examiners filed charges" seems odd. --JWSchmidt 06:11, 31 July 2006 (UTC)[reply]

The NJSBME certainly brings charges against those who practice medicine without a license. They list the names of those deemed guilty monthly. "Filed" wouldn't be my verb choice, but it's pretty much synonymous. - Nunh-huh 06:28, 31 July 2006 (UTC)[reply]

Thanks for all of your help. This is starting to make sense. I found this which suggests that the attorney general brought the case to court. --JWSchmidt 06:41, 31 July 2006 (UTC)[reply]

No problem. Perhaps the board only handles civil penalties, and the attorney general handles cases referred from the Board for criminal prosecution. - Nunh-huh 06:44, 31 July 2006 (UTC)[reply]

Is there a general name for a reaction in which a large molecule is split by the addition of a small molecule? It's the reverse of condensation, and hydrolysis is a special case. Solvolysis isn't it because the small molecule that's added isn't necessarily the solvent. Any ideas? —Keenan Pepper 07:01, 31 July 2006 (UTC)[reply]

"cleavage" is used a bit in the literature, i can't think of anything more scientific, but i might be forgetting something obvious. Xcomradex 11:14, 31 July 2006 (UTC)[reply]

I like cleavage.

Don't all men ? :-) StuRat 22:48, 31 July 2006 (UTC)[reply]

Depends on the cleavage and the men. DirkvdM 09:50, 1 August 2006 (UTC)[reply]

In nuclear reactions, a large nucleus is split into two smaller atoms by collision with a small nucleus (or neutron). This is called nuclear fission but I doubt it is what you are looking for. Perhaps, decomposition? Nimur 16:37, 31 July 2006 (UTC)[reply]

He asked about molecules, not atoms. DirkvdM 09:50, 1 August 2006 (UTC)[reply]

I'm pretty sure it's called "cleavage"... It's used all over the place in reports, specifically I recall, my old chemistry teacher's PhD work on "Rapid syntheses of difluorinated polyols using cleavable carbamates". Other stuff I've come across using 'cleavage' is "Base-catalyzed cleavage and homologation of polyhedral oligosilsesquioxanes". ≈Eh-Steve 04:26, 2 August 2006 (UTC)

Bah... What an awful article to have red-linked... Someone should sort that out... :D ≈Eh-Steve 04:30, 2 August 2006 (UTC)

s black hole a partial black body?? if no please explain how--59.178.4.239 09:26, 31 July 2006 (UTC)[reply]

No. A blackbody (usually written as one word) is an object which reflects absolutely no electromagnetic radiation that falls on it. A black hole is a perfect blackbody.

Unless you believe in Hawking radiation.... TenOfAllTrades(talk) 13:10, 31 July 2006 (UTC)[reply]

Hawking radiation is emitted from (as opposed to reflected by) black holes. Blackbodies may emit radiation (stars are almost perfect blackbodies) they just don't reflect any.

Why doesn't aluminium foil get hot in the oven? I know it is an insulator and effective in reflecting heat, but why? Would other metals (say copper) have a similar effect? BenC7 09:51, 31 July 2006 (UTC)[reply]

It does get hot, but because it is so thin and has a large surface area compared to its mass, it loses most of its heat almost as soon as it comes out of the oven.--Shantavira 09:58, 31 July 2006 (UTC)[reply]

It is important to note the difference between Temperature and Heat. Though the temperature may rise, there is not much heat energy stored in the foil because it is so low-mass. Also, although it reflects heat, I believe Aluminum would usually be classified as a conductor. This has a double-effect: the heat is easily conducted away whether it is warming up or cooling off! Nimur 16:35, 31 July 2006 (UTC)[reply]

I'll note in response to Nimur's comment that while aluminum is a pretty good conductor of heat, it does an excellent job of blocking radiant heat (infrared radiation). Also, since the foil traps the air around the wrapped-up food, it reduces the convective cooling. TenOfAllTrades(talk) 17:48, 31 July 2006 (UTC)[reply]

Well spoken, Ten. Maybe we should link to Convection, Conduction, and Radiation - which are the three mechanisms of heat transfer. Understanding the interplay between these will help the original questioner get a sense of the way food actually gets warm inside an oven. Nimur 20:46, 31 July 2006 (UTC)[reply]

Yes, I understand why food gets hot in an oven... So why are metals good at reflecting radiant heat? I assume it has something to do with the delocalization of electrons, which makes it hard for them to be pushed into an excited state...? BenC7 01:06, 1 August 2006 (UTC)[reply]

See plasma frequency. Basically, if the EM wave is "slow" enough (has a frequency below the plasma frequency of the metal), the electrons in the metal (which are delocalized, as you said, and mostly unbound) can match its pace perfectly. Then all the incident energy goes into moving the electrons, which then (because they are accelerating) re-radiate the energy immediately. At higher frequencies, the wave "sloshes" the electrons instead and is absorbed within the metal. Many common metals have plasma frequencies in the UV, which makes them excellent reflectors of visible light and IR. --Tardis 16:49, 1 August 2006 (UTC)[reply]

Thanks. Nicely explained. BenC7 03:38, 2 August 2006 (UTC)[reply]

I began thinking about this while driving down the road one day. Say I'm accelerating at a rate of 5 mph/sec and I come to a hill which drops my rate of acceleration to 2 mph/sec. While my rate of acceleration is decreasing, am I decelerating or still accelerating? Or both?

I suppose that you mean to say that you are travelling at a velocity of 5 mph and when you encouner a hill, your velocity drops to 2 mph. That means that you have decelerated. If your velocity is still decreasing while you are climbing uphill, then you are still decelerating. We can say that the deceleration is constant if the rate of decrease in velocity is constant (say the velocity drops by 1 mph every second).

If this is not the answer you are looking for, you may need to reframe the question.--Wikicheng 13:02, 31 July 2006 (UTC)[reply]

You are still accelerating because your acceleration is still greater than 0. You just accelerate more slowly when you are on the hill. --Yanwen 13:56, 31 July 2006 (UTC)[reply]

Yanwen is right. Your rate of acceleration is decreasing, but your velocity is still increasing. As an aside, as you get to the hill, and your rate of acceleration drops, your jerk is decreasing - jerk is another word for "rate of acceleration". All these quantities should become blissfully clear if you take a basic calculus course - all these quantities are different derivatives of your position. Your position is the "zeroth" derivative of position, your velocity is the first derivative of position, acceleration is the second derivative of position, jerk is the third, etc. And consequently acceleration is the first derivative of velocity, and jerk is the second, etc etc. --Bmk 14:39, 31 July 2006 (UTC)[reply]

That's close, but not perfect. If the hill is ideal and encountering it is instantaneous, the acceleration drops instantly to 2 mph/s, so there is infinite jerk for 0 time at that point and no jerk elsewhere (the jerk has the form of a Dirac delta function). Part of what's confusing is the phrase "rate of acceleration" which can be interpreted as "rate of velocity change (a.k.a. acceleration)" or "rate of [change of] acceleration, a.k.a. jerk". The second interpretation is a bit odd, but there's confusion nonetheless (with "rate of speed" and such too). I'd recommend using "amount" or "rate of change of" (as appropriate) instead of "rate" where there are already speed-like variables and time-derivatives everywhere. --Tardis 23:12, 31 July 2006 (UTC)[reply]

Other possibility: if the 2 mph/s value is at the top of the hill, to the entirety of which the reduction in acceleration is attributed, then there has been a negative jerk during the trip up the hill (although we can't say anything about the precise nature or distribution of the jerk without more detailed information). But this isn't what a hill would really do (given constant driving power); it would be closer to the instant change I suggested a minute ago, where when on the hill the acceleration had one constant value and when off the hill it had another, also constant, value. In other words, with a different (odd) interpretation, we can have a negative jerk, but nothing has indicated a decreasing jerk as Bmk suggested. --Tardis 23:17, 31 July 2006 (UTC)[reply]

Good point - that wasn't what I meant to say - I meant to say the jerk is negative, not decreasing. Thanks --Bmk 03:22, 1 August 2006 (UTC)[reply]

mph/s? Is that unit actually used? If so, it might deserve a mention (and conversion) in the acceleration article. It isn't as weird as kWh/yr, though, because that mixes up three units of time, second, hour and year (the second is in the watt, which is J/s). DirkvdM 10:02, 1 August 2006 (UTC)[reply]

What is the most visible part of the sun? a)Corona, b)Chromosphere, c)photosphere, d) prominence

If you read our article on the sun (<== click on the link), you will quickly find out that the most visible part of the sun is the....aack! (*#&$#@$#( Gasp.... the gods of the reference desk are smiting me for almost answering a homework question! I'm sure you'll find the answer pretty quickly. (PS: I would suggest using your browser's text search function and search for "visible" once you get to the sun article) --Bmk 14:49, 31 July 2006 (UTC)[reply]

This does sound rather like a homework question. The answer should be pretty obvious if you paid attention to the different layers of the Sun.

...or paid attention in class. - Cybergoth 03:00, 1 August 2006 (UTC)[reply]

The outside part? --Ginkgo100 ^{talk · contribs · e@} 14:56, 31 July 2006 (UTC)[reply]

The shiny bits. – ClockworkSoul 20:05, 31 July 2006 (UTC)[reply]

The surface? --Bowlhover 03:35, 1 August 2006 (UTC)[reply]

Page 3? -- Rockpocket 05:43, 1 August 2006 (UTC)[reply]

You could also take a hint from the word photon. DirkvdM 10:07, 1 August 2006 (UTC)[reply]

Try not to put answer choices next time. --Proficient 19:08, 1 August 2006 (UTC)[reply]

Is the North Star the major reference point in the celestial sphere?

Depends where you live. Not in Australia. Notinasnaid 15:14, 31 July 2006 (UTC)[reply]

Yup. See North Star and Pole Star.--Shantavira 15:25, 31 July 2006 (UTC)[reply]

Nope. Polaris is 44 minutes (0.74 degrees) away from the celestial north pole, so it can't be used to pinpoint exactly where the celestial north pole is. The "first point of Aries" is the reference point for right ascension: it's the point in the sky (relative to the stars) where the Sun appears to be, at the March equinox. (To be more specific: at some point during March, the Sun will appear to be directly overhead at some point along Earth's equator. The position of the Sun at this moment, relative to the stars, is right ascension 0.) As for declination, the declination of a star is simply the latitude at which the star appears to be at a 90-degree altitude. (Of course the star can't be directly overhead along the entire line of latitude, but it will appear overhead at some location along the latitude line.) --Bowlhover 22:59, 31 July 2006 (UTC)[reply]

The North Star is Polaris.

It hasn't always been, and it will not always be. And Polaris is not a perfect "north star". --Bowlhover 22:59, 31 July 2006 (UTC)[reply]

GPS? DirkvdM 10:12, 1 August 2006 (UTC)[reply]

I'm confused. You say that both Jabir ibn Hayya and al-Razi discovered sulfuric acid. Which one was it?

I'm not sure who "you" is, but the wikipedia article on Sulfuric Acid states that Jabir ibn Hayyan is credited with the discovery, but al-Razi studied its properties and production. I don't see a contradiction - hope that clears things up. And btw, it's usually helpful to link to articles that you are referring to so everyone knows what you mean - do this by enclosing an article name in double square brackets. --Bmk 15:41, 31 July 2006 (UTC)[reply]

Quoting from the article on al-Razi: "Razi is credited with the discovery of sulfuric acid". --Lambiam^Talk 17:27, 31 July 2006 (UTC)[reply]

Ah - thanks. That seems like a difficult contradiction to resolve - probably will take someone more familiar with the topic and the sources. There may be no good answer, but the articles should at least agree with each other. --Bmk 17:38, 31 July 2006 (UTC)[reply]

What is the correct term for a person who is neither male nor female? One one hand, Wikipedia's articles state that "hermaphrodite" and "pseudo-hermaphrodite" are antiquated and offensive, and have been replaced by "intersexual." On the other hand, I have seen numerous uses of these words in the media. Including the article by Jared Diamond in Discover magazine. JianLi 16:13, 31 July 2006 (UTC)[reply]

Hermaphrodite would mean both male and female, neither would be asexual. Hermaphrodite is only somewhat offensive to humans, but the term is used in biology all the time. I have never heard intersexual. Is that like transexual?

According to the relevant articles, asexuality refers to the condition of having no sexual interests or desires, while intersexuality describes the condition of having ambiguous genitalia. The latter sounds like what the questioner was asking. Intersexuality notes that advocates for intersexual people do not like the terms hermaphrodite and pseudohermaphrodite, which suggests they should be restricted to describing non-human animals. --Ginkgo100 ^{talk · contribs · e@} 21:25, 31 July 2006 (UTC)[reply]

Of course asexuals have no sexual desire! They can't have sex! Asexually reproducing biota include bacteria, sponges, and cnidarians.

There are two meanings of the word; one refers to organisms that reproduce asexually, the other to humans who don't feel sexual desire (see Asexuality). --Allen 04:00, 1 August 2006 (UTC)[reply]

I think it's "transgender"

Transgender refers to individuals who practice behaviors usually associated with the opposite gender or who identify with the opposite gender. Asexual, in the context I was using it, also refers to a human sexual identity. Of course asexual also refers to a type of reproduction not involving the exchange of genetic material used by many non-human organisms. I apologize if I was not explicit enough with context. --Ginkgo100 ^{talk · contribs · e@} 19:49, 2 August 2006 (UTC)[reply]

I'm considering buying myself a new digital camera, as my old one seems to have finally died on me. Quick question - are the newer ones still as battery-hungry? My previous camera (bought about seven years ago) would go through 4xAA batteries in a couple of hours, less with rechargables - I hardly ever used it because it was so damn impractical. --Kurt Shaped Box 18:40, 31 July 2006 (UTC)[reply]

I don't know about cameras being less power hungry, but in my experience modern NiMH rechargeables should easily outlast ordinary disposable AA cells. Mind you, the highest-capacity ones tend to be costly, and the capacity ratings are usually "ideal" ones: a "2400 mAh" battery does not usually last 1.5 times as long as a "1600 mAh" one in practice, and will usually cost more than 1.5 times as much. Still, even the low-end ones often outperform disposable cells in camera use, since they deliver a more stable voltage over time. A good battery charger is also important — it doesn't have to be an expensive brand-name one, but it should be ΔV-controlled ("intelligent") and should have independent charging circuits for each cell. I've had good personal experiences with the cheap store-branded "Rapid Charger" and cells from Biltema; similar products, or possibly even the exact same ones modulo branding, can probably be found in local stores elsewhere. —Ilmari Karonen (talk) 19:36, 31 July 2006 (UTC)[reply]

I have a German brand, Concord 5345z (5 Megapixels), which comes with a recharger for it's two rechargeable AA Nickel-Metal Hydride batteries. I would call it a battery hog, yes, as the two batteries only last about a day if you are taking pics constantly. However, as long as I remember to recharge them every night, a one day charge seems workable. One bad side, though, is that uploading the pics to my laptop also depletes the charge. I've learned not to dawdle when doing so. I used to upload one pic, then edit it, then do the next, etc., which ran my batteries down. If I just upload them all at once, then turn off my camera and recharge the batteries while I edit the pics, it works out much better. StuRat 22:37, 31 July 2006 (UTC)[reply]

I don't know if power consumption has changed, but the major factor here is the display. A bigger one will (I assume) consume more power, but more important is how much you leave it on. If you are likely to run out of juice, turn it off when you don't need it. That way you'll still have the camera standing by without having to worry too much about the battery running down. Make sure the camera has a separate and handy button for this.

Also, whatever the power consumption might be, it is always good to have two batteries, so you can use one while the other one is being charged. Or better still, you don't need to worry about when to charge because if you take two batteries and the one in the camera runs down you can use the other. You're unlikely to go through more than one battery in one day, so in the evening you can charge the one that ran down.

So in the price of the camera you should include the price of two batteries - they differ quite a bit in price, especially if there are (decent) alternatives by other manufacturers (such as Hama (if you buy two of those you'll have Hamas :) )) DirkvdM 11:59, 1 August 2006 (UTC)[reply]

Lets say I take a shiny metal object, one that is polished and reflects light well, and heat it until it glows red-hot. Does the object continue to reflect light while it is glowing, (I assume the intensity of the radiated light would be many times greater than that of the reflected light, obscuring the latter), or are radiation and reflection mutually exclusive behaviors? Assume that the shiny metal is heated in an oxygen-poor environtment to avoid tarnishing oxidation. I was unable to find the answer in blackbody or associated articles.Tuckerekcut 19:03, 31 July 2006 (UTC)[reply]

Yes, it will continue to reflect light. And if you could build a precise-enough instrument to measure the intensity of light, (or any other part of the EM spectrum), you would be able to see that the two processes add linearly. Nimur 20:47, 31 July 2006 (UTC)[reply]

Actually, this isn't really correct. If the object is hot enough to be visibly glowing, then the spectrum of excitations responsible for that glow will also make it receptive to absorbing visible light, at which point the light shined on to it would be partially contributing to heating the object. Since energy must be conserved, the total emitted light would still be expected to vary linearly with the light shined on it (neglecting other modes of energy dissipation), but the spectrum would be changed and not have the wavelength preserving qualities normally associated with reflections. Dragons flight 21:43, 31 July 2006 (UTC)[reply]

in the same vein as the previous post, if the light being reflected is of a high enough frequency to be out of the blackbody radiation for the objects temperature, then it will reflect. think about a regular mirror at 298K, it is still pumping out black body radiation, yet it still reflects light, because visible light is of a much higher frequency than the blackbody radiation at room temperature. Xcomradex 08:34, 1 August 2006 (UTC)[reply]

Note also that when you heat a metal until it grows red-hot, its chemical characteristics might also change in some cases, which could affect how it reflects light. – b_jonas 13:07, 1 August 2006 (UTC)[reply]

Thanks, all. I'll try it in a glovebag ( [nearly] oxygen-free) when I get back to school. I'll see if I can get the palladium catalysts red hot then try to bounce a laser off the surface. I'll post my findings around here somewhere...Tuckerekcut 21:07, 1 August 2006 (UTC)[reply]

Hello. How does string theory solve the renormalization problem of gravity? And if it doesn't, why is it considered a possible T.O.E?AmateurThinker 22:48, 31 July 2006 (UTC)[reply]

At a guess I would suggest that it solves the problem by saying gravitons are massless. If not (assuming it does solve it) perhaps gravitons having a non-zero spatial extent would help. I have heard a suggestion that spacetime itself is effectively comprised of gravitions, rather like a sheet of chainmail. I cannot remember where I heard this (I may even have just dreamt it up) so I cannot comment on the credibility of this theory. If it were so then gravity could be treated as a field and the problem would be almost solved.

Several articles I've read have mentioned such impairments, but did not mention whether they were reversible upon receiving an adequate amount of sleep, or whether such impairments were residual. Anyone?

I hope not, because I'm really tired right now :) actually I'm pretty sure not. However check the article on Sleep debt. 71.199.123.24 01:04, 1 August 2006 (UTC)[reply]

of course this depends on how serious a situation you are talking about. Several days without any sleep at all can result in permanent madness and is actually a form of torture (a 'mild' form of which is used in Guantanamo Bay by continually leaving the lights on). Mind you, I'm talking about no sleep at all. Normally, one will have at least some microsleep (dozing off for a sec), which makes a big difference. DirkvdM 12:14, 1 August 2006 (UTC)[reply]

Long-term health effects of sleep deprivation are unknown, however the cognitive function impairments all reverse after a few hours of sleep no matter how long the deprivation.

What is the name of the disorder/allergy where red skin is induced by scratching?

Dermatographism, as seen in Darier's sign which is seen in diseases such as mastocytoma. InvictaHOG 23:37, 31 July 2006 (UTC)[reply]

Hives and/or contact dermatitis. It's not mentioned in wikipedia, but there is a form when rubbing/scratching the skin rasies red welts (bump). I saw a picture where someone wrote the word hello on someones back, just by lightly scratching them! Also see Eczema, and maybe general Dermatitis. 71.199.123.24 00:59, 1 August 2006 (UTC)[reply]

Itch? DirkvdM 12:15, 1 August 2006 (UTC)[reply]

That would be a symptom of a disorder/allergy, not the disorder/allergy itself. JackofOz 14:14, 1 August 2006 (UTC)[reply]