Wikipedia:Reference desk/Archives/Science/2006 December 14

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December 14

Why do faster moving fluids have lower pressure?

I understand the concept that as fluids (i.e. water/air) flow faster, their pressure drops. But what I don't understand is why. Please explain? --AstoVidatu 03:16, 14 December 2006 (UTC)

Bernoulli's principle may be helpful. TenOfAllTrades(talk) 03:33, 14 December 2006 (UTC)

A fluid rushing through a pipe is composed of molecules, and those molecules are all bouncing around, off of each other, and off of the walls of the pipe. How often they smack into the walls of the pipe results in pressure. For a given flow rate, they bounce into the walls at a certain rate which we then measure as the pressure. Now, if the flow of the fluid through the pipe is increased, think of what happens to the molecules... they are traveling faster down the pipe, and so fewer of them will smack into the pipe in any one particular place. Fewer molecules smacking into the wall of the pipe = lower pressure. At least that's how I remember it being explained to me in high school physics. 71.112.115.246 05:27, 14 December 2006 (UTC):P

Wind and toilets

I live on the fifth floor of a ten-floor apartment building. When the wind speed gets up around 30 miles per hour, the water level in the toilet bowl is an inch lower than it is in calm weather. Why is this? --67.185.172.158 04:18, 14 December 2006 (UTC)

Plumbing drains, including toilets, have in addition to the pipe which carries waste down to the sewer, an up vent which goes up to the roof. Its purpose is to prevent the draining water from siphoning the water out out of the U shaped trap, which is there to prevent sewer gasses from entering the living area. One way in which a high wind level could cause a drop in the toilet bowl level is if the air rushing past the vent on the roof lowers the pressure. If the air pressure in the vent stack were lower than the air pressure in the bathroom, the water level in the bowl would drop. One inch of water is not a huge pressure difference. Edison 04:39, 14 December 2006 (UTC)

Hmmmm... why does this comment appear above mine when the timestamp is ten minutes later? Anyway, that's what I was thinking: Bernoulli's principle in action. —Keenan Pepper 04:45, 14 December 2006 (UTC)

Where is independent of when. An accomodation to minimize deletion. Adaptron 08:07, 14 December 2006 (UTC)

Hmmmm... could this be related to the previous question? I honestly don't know. —Keenan Pepper 04:29, 14 December 2006 (UTC)

(Expanding on Edison's comments...) The water in the trap of your toilet bowl acts as a manometer between the air pressure inside your house and the air pressure in the plumbing vent stack. When the wind whips across the vent stack, the venturi effect lowers the pressure in the stack and the water in the toilet-bowl manometer shifts towards the vent stack. Unfortunately, unlike in a true manometer, it then sloshes over the upper edge of the trap and falls down the drain, permanently lowering the water level in your toilet bowl, at least until you flush the toilet and refill the trap again. If you catch this phenomenon in action, you may actually observe waves and motion in the water in the bowl.

Atlant 12:50, 14 December 2006 (UTC)

Digital camera shutters

Is it correct to say that all digital single-lens reflex cameras have mechanical shutters but no live-preview digital cameras do, or is the situation more complicated than that? In cameras without a mechanical shutter, how exactly is the "shutter speed" (integration time) controlled? —Keenan Pepper 04:28, 14 December 2006 (UTC)

Not quite.. The digital camera doesn't need a shutter - the ccd can recieve light all the time - and not be overexposed like a film. In fact a digital slr doesn't have a shutter proper as a normal slr - both will have the mirror, but only the non digital version has a focal plane shutter. So no digital camera needs a shutter - but a digital slr will have the moving mirror (which is a bit like a shutter).

'Shutter speed time' - the ccd is an array of light sensitive components - light falling on an element of the array will produce a voltage/current/resistance change - The longer this current is measured the more accurately it is known. So digital cameras do have a close analogy of shutter speed - the exposure time - short exposures will have more noise, longer exposures will have less noise but be prone to motion blur. Hope that answers all your questions.

See also digital single-lens reflex camera#Digital SLR versus SLR-like and compact cameras, note that some SLR like cameras do away with the mirror as well. 83.100.174.70 11:34, 14 December 2006 (UTC)

The description in this review of a DSLR says that with full frame transfer CCDs, "a mechanical shutter is absolutely required to control the start / stop measurement of light", which seems to directly contradict you. —Keenan Pepper 20:49, 14 December 2006 (UTC)

"Your typical consumer digital camera uses what is called an Interline Transfer CCD, put simply the CCD can itself control the start / stop of when it measures light falling on it, otherwise known as an electronic shutter..." I'd suggest that this camera is aimed at people with more money than sense or who want to 'live in the past' why quote something saying " seems to directly contradict you" when actually reading the page confirms what I've been saying???87.102.19.95 22:58, 14 December 2006 (UTC)

You said "no digital camera needs a shutter". It says that some digital cameras do need shutters — the exact logical negation. I asked a question, you answered incorrectly, get over it. —Keenan Pepper 03:45, 15 December 2006 (UTC)

Actually they don't need shutters - the transfer of info. from the ccd is done electronically - using an electronic switch - there's no reason why at the start of an exposure the ccd could be blanked - and then at the end the resultant set of electrical signals be transfered over. Even in the case given above, so as I said "no digital camera needs a shutter" . And learn some manners as well.87.102.8.6 14:44, 15 December 2006 (UTC)

As the article says, that is only for full frame ccds. not all dSLRs use full frame ccds, eg. the Pentax *ist DS doesn't. Xcomradex 21:27, 14 December 2006 (UTC)

But does it have a mechanical shutter? —Keenan Pepper 22:14, 14 December 2006 (UTC)

Google says yes. Why does it have a mechanical shutter if it doesn't need one? —Keenan Pepper 22:16, 14 December 2006 (UTC)

Suggest it's unneccessary but no doubt some advertising wanker has convinced people that "mechanical shutter = good quality" therefore a perferctly good digital camera is crippled because some idiot will buy it - and at 5 times the price of the equivalent product no doubt - does that sound like a realistic appraisal of 'the way things work'.. eg. Why buy a £5000 rolex when a £1 digital watch keeps better time? (<end of rant>)87.102.19.95 23:11, 14 December 2006 (UTC)

What evidence are you basing this on? Why should I belive you? —Keenan Pepper 03:45, 15 December 2006 (UTC)

Truth:

• SOME DSCs (CHEAP ones mostly) have NO shutter.
• MANY OTHER DSCs (MID- OR ABOVE) have shutters.

How to observe it:

1. Set your DSC to manual exposure (not every DSC supports this function)
2. Set to the widest aperture setting.
3. Set a reasonably long exposure time. Maybe more than 1 second.
4. Turn your DSC around so you can see the inside of its lens mechanism.
5. If you cannot see the shutter mechanism, you may adjust the mechanical zoom (usually wide angle).
6. If you still cannot see it, get a lamp or a flashlight.
7. If you have a wide-angle adapter, you may want to use it to magnify the mechanism.
8. When you think you see the mechanism, take a picture.
9. SEE! SEE! SEE! SEE! SEE! SEE! SEE! SEE!
10. If you cannot see it, adjust the zoom setting and take another picture.
11. SEE! SEE! SEE! SEE! SEE! SEE! SEE! SEE!

Under the MANUAL mode, you may find your fastest shutter setting be controlled by the aperture setting. Take my Canon PowerShot S3 IS for example, the fastest shutter speed is 1/3200 s under f/8.0 and 1/1600 under f/2.7 (aperture wide). This is a tell-tale sign of mechanical shutter.

If your camera supports long exposure, you may find it calibrates itself after a long exposure. Take my cam as an example, if I do a 15 s long exposure, it would:

1. Open the shutter and expose the CCD for 15 s. (Viewfinder ON)
2. Close the shutter. (Viewfinder OFF)
3. Save the picture to a buffer.
4. Expose the CCD for another 15 s while the shutter is closed.
5. Save the noise "picture" to the buffer as well.
6. Calculate an improved picture with the noise data.
7. Save the improved picture to the memory card.
8. Open the shutter. (Viewfinder ON)

The level of noise is subject to many real-time conditions such as the temperature and the age of your circuits. You cannot measure it in the factory and use the fixed value later. I only have one DSC so I cannot take a movie of it. Maybe you guys can take a shutter-open-shutter-closed movie. A camera with a larger and better CCD may not need to calibrate for the noise. My camera does not have a B or T exposure setting. It is simply not very practical to use a noisy chip to take a really long exposure. -- Toytoy 15:42, 15 December 2006 (UTC)

chest muscles

can the dumbells be used to work the chest muscles?

YES. Check the links for more information. I suggest particularly the PDF file that's the 5th or so link down. Anchoress 06:53, 14 December 2006 (UTC)

Dumbell flat, incline or decline bench press (like barbell benchpresses but using dumbells) and dumbell flyes are the most common exercises. Some would recommend pullovers as well. As Anchoress says, you need to look up how to do them. --jjron 12:52, 14 December 2006 (UTC)

The chest muscle which is visibly enlarged in body builders is Pectoralis Major. This is a powerful adductor, flexor, and internal rotator of the shoulder; hence any activity which produces this movement under strain will "work" the muscle:

Adduction - stand as if imitating the letter "T" with your arms out. Move hands forwards, bringing them in front of your chest, arms fully extended. This, bringing the upper limbs towards the midline, is called adduction Flexion - stand up straight, as if at military drill. Bring your arm up, forwards, so you see your hand in front of your shoulder, arm fully extended and horizontal. Int Rotation - stand up straight, upper arm against side, lower arm flexed at 90 degrees and hand out sideways. Move your hand round, using the elbw as a pivot, until your hand is across your lower chest

William Giaque

"He developed a magnetic refrigeration device of his own design in order to achieve this outcome, getting closer to absolute zero than many scientists had thought possible. This trailblazing work, apart from proving one of the fundamental laws of nature led to stronger steel, better gasoline and more efficient processes in a range of industries."

um....anyone know why this led to stronger steel and better gasoline??

As for the steel, see Basic oxygen process. It's likely that some oil refining process likes to have pure oxygen as well.

Atlant 13:12, 14 December 2006 (UTC)

given that the oxygen is as far as I know not made by a magnetic process, and oil refining does not need any o2 - unless you want explosions.. I'd say this is bullshit, an example of scientific discoveries being totally overplayed in terms of their industrial significance. However maybe I'm missing something - but I would suggest an ill informed journalist wrote this..(it's William Francis Giauque)83.100.174.70 13:24, 14 December 2006 (UTC)

Environment

How badly have we screwed ourselves? What will the world be like at the next turn of the century (guesses allowed)Sashafklein 07:11, 14 December 2006 (UTC)

lol, you have quite the hubris to believe that hairless apes can significantly affect the biosphere. Read Lewis Thomas' " Lives of a Cell".

If you already believe that humans have ruined the enivoronment, I shan't try to convince you otherwise. Simply understand that nature changes, the world changes, not necessarily for the worse. It is impossible to accept that anything is static without suffering a mental breakdown or becoming a crusader for various causes that are outside of your control. —The preceding unsigned comment was added by 70.225.160.106 (talk • contribs).

That's one way of looking at it. Proponents of this are: The Bush administration, Michael Crichton, and anyone in the oil or car business. The other way of looking at it is that we're up to our knees in...trouble, and still digging. Proponents of this are: Anyone who's ever studied science. A bit of hyperbole, but that's the gist of the situation. yandman 08:35, 14 December 2006 (UTC)

Unfortunately, the billions of people in the developing world need food, shelter, and water. And unfortunately for the eco-crusaders, this sometimes comes at the cost of the environment. Sorry, but people need to live, and the environment and species diversity comes second place.

How do you expect to convince a Brazilian lumberjack to quit his job and stop feeding his family because it bothers the conscience of some yuppie in the USA?

Sure, I'm being a little polemical, but these are things that need to be considered. —The preceding unsigned comment was added by 70.225.160.106 (talk • contribs).

Isn't it just wonderful that there are still poor people around whose indigence we rich people can use to justify our own excesses! --Rallette 11:49, 14 December 2006 (UTC)

Recommended reading on this topic includes our articles on global warming, climate change, attribution of recent climate change and economics of global warming. While there is a fair amount of (especially political) debate on this subject, the scientific consensus seems to be that while we're not completely "screwed" yet, things are not looking particularly rosy, and will only get worse unless something is done. — QuantumEleven 12:09, 14 December 2006 (UTC)

And in response to the unsigned contributions above, and also because it's very relevant, add to that list overpopulation. --jjron 12:34, 14 December 2006 (UTC)

I am of the opinion that all harm done to the environment is manifested just as it is. That is to say, we need not worry that what we've done in the past has already doomed us in the future. The world is what it is. I don't buy 'doomsaying' at all, though pundits like Al Gore make a load of money doing it. Cheers. Vranak 16:50, 14 December 2006 (UTC)

Just to butt in, we're not very good at guessing how it will be in 100 years. I wouldn't even guess at all. The margin of error is extremely large, and we haven't guessed how it will be hundred years in the future many times before to collect data on that. Climate systems are although cyclic, they are chaotic and nonlinear. We're still trying to figure out what happened in the past and present. X [Mac Davis] (DESK|How's my driving?) 22:10, 14 December 2006 (UTC)

Next century? Meteorologists can't tell me the weather next Tuesday!

Well my general sense is that a change in the environment is not necessarily a catastrophe, as long as it does not occur too rapidly. The Earth's environment changes all the time, and (thus far) life forms have managed adapt. But the problem seems to be that we're incurring a relatively rapid change that is comparable to a natural global catastrophe. Nature isn't able to adapt quickly enough, at least among the larger creatures. So we're putting the planet out of equilibrium and it will take a long time for a new balance to be reached. The ultimate cost we may end up paying is a mass die-off of humanity. But some humans will probably survive. We have some pretty clever members among our species, and we're highly adaptable. — RJH (talk) 17:48, 20 December 2006 (UTC)

canonical partition function

As V increases at constant T, why does En decrease?

units

what kind of value would be in units of K*cm (within P-chem)? These are not HW questions I'm attempting to reverse engineer my pchem midterm test questions from my graded test. Thanks.

The constant in Wien's displacement law is 0.290 cm K. —Keenan Pepper 03:51, 15 December 2006 (UTC)

milk shake

how come whenever I eat a milk shake I get really dehydrated afterwards (1-3 hours after)?

Can you clarify what you mean by 'dehydrated'? Do you just mean you get thirsty or something else? From our dehydration article "Medically, dehydration is a serious and potentially life-threatening condition in which the body contains an insufficient volume of water for normal functioning." I find it hard to believe a milkshake is doing this to you! --jjron 12:27, 14 December 2006 (UTC)

jees why does everyone here have to be such a Technical Tanya? dehydrated, really thirsty, yellow piss, etc. Great Living Luci, how do you live in society if you take everyone seriously? (maybe I'll post that one in Misc..)

Maybe he means he gets the runs. --Russoc4 15:39, 14 December 2006 (UTC)

We're not well equipped to give medical advice, sorry. I suggest referring such questions to your doctor. Ned Wilbury 16:38, 14 December 2006 (UTC)

I agree about the medical advice. And ask for a blood test; dehydration after consuming sugar is a symptom of diabetes. Anchoress 17:00, 14 December 2006 (UTC)

Thirst (rather than dehydration) an hour or two afterwards does indeed suggest transient hyperglycemia and may be an early stage of diabetes. Dehydration cannot happen in two hours and is not a subjective sensation, but assuming he means thirst, anything that raises your plasma osmolality will make you feel immediately thirsty, and glucose is the prime candidate. A glucose tolerance test will confirm or refute it. alteripse 01:40, 15 December 2006 (UTC)

May I ask where you are getting your milkshakes from? Vranak 23:41, 14 December 2006 (UTC)

Simple, most milkshakes are quite salty. Consuming salty things makes you thirsty. A Burger King king-sized vanilla shake contains 640 mg of sodium. StuRat 03:47, 15 December 2006 (UTC)

There will also be a vast difference in how a Burger King milkshake makes you feel, and how a lovely homemade milkshake made in a modest agricultural town with good grass and good cows will make you feel! Vranak 16:07, 15 December 2006 (UTC)

Absolutely. StuRat 00:34, 17 December 2006 (UTC)

Conservation of Energy and Rolling a ball down a ramp

Hello. I was wondering if anyone would be able to help me with a query about my physics coursework? What we have to do is talk about the conservation of energy, and how energy is never created or destroyed, and simply changes form, and then to prove this theory correct we had to roll a ball down a ramp at different heights, and time how long it took for the ball to reach the bottom. Thats my planning and observing stages completed, but then for the analysis we have to link our experiment up to the theory of the conservation of energy, and I have absolutely no idea how it's relevent, am I missing an equation or anything? I would greatly appreciate any help! Thanks. --88.108.21.142 07:38, 14 December 2006 (UTC)

OK. First, the experiment's wrong. But we'll get to that later, and don't worry, it's your teacher's fault. But telling him why he's wrong will surely give you extra points... You know how to calculate "gravitational potential" (or something like that) energy? Go and fetch the formula. You know how to calculate kinetic energy? Go and fetch that one, too. Hopefully you know the weight of the ball. Now, did you measure the speed of the ball at the end of the ramp, or did you measure the slope of the ramp? (If you did neither, you're in trouble...). yandman 08:23, 14 December 2006 (UTC)

If you measured the speed of the ball, you just need to use all the formulas you've collected. If you've got the slope, you can find the projection of the gravitational force on the ramp (time to get out your sin and cos..), and then you're going to use Newton's laws to find out what you need.

And the experiment is wrong because it's not taking into account the fact that the ball is rolling. It takes energy to make something spin, and you're only measuring its speed, not its spin. Therefore, you're not taking into account all the energy the ball has at the end of the ramp, and if you do the experiment accurately enough, you'll see that the "final energy" is lower than the starting energy. Added to that, there's the friction losses, but they're negligible. yandman 08:29, 14 December 2006 (UTC)

energy which goes to rolling is also likely negligable compared to the measurement techniques being used. If you want brownie, look up the formula that goes into rolling, and put an estimate on friction. Actually fuck that, derive the formula for the energy of rolling —Preceding unsigned comment added by 70.225.160.106 (talk • contribs) 08:32, 14 December 2006

but computing the rotational energy requires you to find the moment of inertia of the ball, which can be computed easily from the mass and radius if the ball is uniform density or hollow or something like that, but otherwise may be hard to measure --Spoon! 09:49, 14 December 2006 (UTC)

IGNORE this number - I think I made a mistake..((For a solid ball rolling at speed v it's 0.3mv², i think.)) See rotational energy and rotational inertia83.100.174.70 14:20, 14 December 2006 (UTC)

I'm guessing that at the level you're at (A level?) the energy due to rotational inertia is ignored.(In any case it's sensible to solve the simplified case first and then include other variables..)

If I've understood correctly you need a link showing the relevance of conservation of energy to your experiment.. This is simple. The total energy should remain the same. In this case it means that the sum of kinetic energy (1/2mv²) and potential energy (mg x height) remains constant.. So if the ball drops y metres, it has lost mgy joules of potential energy (mass times gravity = force, force times distance = energy/work) so that energy lost should be compensated by in increase in kinetic energy ie 1/2mv²=mgh so v²=2gh. So you need the speed of the ball at a given height - you can get the speed from the slope of a graph of distance versus time. Did you measure the time to different points or just to the bottom?83.100.174.70 12:51, 14 December 2006 (UTC)

I'll give you the rest of the clues.. Distance = vt +1/2at² - you know the time, the initial velocity v should have been 0, a is the acceleration (=g sin f, where f = slope of the 'ramp' an angle), the final velocity is v=at, a as before, you measured t, so you can get the final velocity, and hence the final kinetic energy 1/2mv², the change in potential energy is mg x height. If you're still stuck ask again..83.100.174.70 13:16, 14 December 2006 (UTC)

e.coli

where can i get the perfect ultra structure of Escherichia coli bacterium?--hima 11:20, 14 December 2006 (UTC)

Try checking the external links at e coli. X [Mac Davis] (DESK|How's my driving?) 18:23, 14 December 2006 (UTC)

Transmission of communication signal along with high voltage transmission line

Is it possible to transmit communication signal along with high voltage transmission line by varying frequency of communication signal

...................nivas.... " nivasfellow @ yahoo.co.in"

Yes. It's been tried and tested. yandman 11:33, 14 December 2006 (UTC)

See power line communication. Weregerbil 11:50, 14 December 2006 (UTC)

By "varying frequency of communication signal" do you mean amplitude modulation at other than the power frequency, or do you mean frequency modulation ? Various systems have been used for many decades as described in the article cited above. Electric utilities use amplitude modulated signals or unmodulated carrier transmitted over power lines for telemetry and for protective relaying. They have also used them for radiotelephone communications. One problem with using these systems for communicatind the state of the system back to the main dispatch center is that if the lines are knocked down, shorted to ground, or they overload and fail, is that the information transmission is lost with the power oine, leaving the dispatchers less able to communicate with generating stations and substations. Wavetraps have to be installed wherever transformers are connected to the powerline, which are band reject filters to isolate the radio signal from being shorted out by the transformer windings. They have to be tuned to the frequency of the carrier current signal. Frequencies below the AM broadcast band are typically used. Low power college AM radio services have been piped to dorms using carrier current at normal AM broadcast frequencies; the building wiring carries the signal to each room, without a broadcast license. Edison 15:16, 15 December 2006 (UTC)

Science question on Miscellaneous Desk

Just a note to point out a science question on the Miscellaneous Desk:

• Wikipedia:Reference_desk/Miscellaneous#Free_fall_from_space

If any regulars here could help over there, that would be great. Carcharoth 11:54, 14 December 2006 (UTC)

Maxwell–Boltzmann statistics

In Maxwell–Boltzmann statistics#A derivation of the Maxwell-Boltzmann distribution can anyone explain what "even though the ad hoc correction for Boltzmann counting is ignored, the results remain valid." means ie what is the ad hoc correction?

Also in the derivation the modal distribution ie the distribution giving greatest value of W is found by ".. taking the derivative with respect to Ni, and setting the result to zero and solving for Ni yields the Maxwell-Boltzmann population numbers", but the mode is not the average - isn't the distribution supposed to represent the average distribution based on the assumptions. i.e. it should give a mean distribution and not the mode.83.100.174.70 12:39, 14 December 2006 (UTC)

I don't know what the ad hoc correction is referring to, I've never heard of it. It sounds vague. As for your second question, the MB distribution can be interpreted as a probability distribution, so you can get all sorts of things like the mean, and the mode. It tells you much more than the average. The procedure you described does indeed give you the mode. The mode corresponds to the single peak of the curve, so if you differentiate this, the new function will be zero at precisely that point. --HappyCamper 15:55, 14 December 2006 (UTC)

I was thinking that as the derived equation (on the wikipedia page) gives an energy distribution based on the mode - isn't this wrong for a statistical treatment - in terms of statistics the mode is definately not the average (in most/many cases). The mean(averaged) distribution is in fact a noticably different shape..(and a different equation). Interpreting the modal distribution as a probability distribution is wrong, eg if I toss two dice - take the modal value (ie 7) - I can not use this as a probability..83.100.174.70 16:08, 14 December 2006 (UTC)

You know, I think the articles need to be fixed up a bit. I also I think I've misinterpreted your question. Have you taken a look at Maxwell–Boltzmann distribution? That was what I was thinking of, in particular, the equations at the bottom of the page. As far as I can tell, N_i itself is not the average, nor the mode in your original question. I'm sorry, the best I can do for now is to direct you to another source. If you have McQuarrie and Simon - (physical chemistry, a molecular approach) on hand, see chapter 17 pg. 693 - I think the idea you want to get out of this, is that (I will quote): "The probability that a system in an ensemble is in the state j with energy E_j(N,V) is proportional to $e^{-E_{j}(N,V)/k_{B}T}$". That is the crux of a lot of results that follow in "classical" statistical mechanics. --HappyCamper 16:50, 14 December 2006 (UTC)

Haven't got that specific book but have seen many similar.. My point is that the above (both) take an approximation to the actual average distribution - by taking the modal distribution (the one with the highest W - number of ways to obtain it). Unfortunately this discards all other distributions that contribute to the actual average distribution. This fact is often overlooked but I see no reason why it should not be mentioned as a 'flaw' in the derived equation in wikipedia even though some reputable books omit to mention it.83.100.174.70 17:06, 14 December 2006 (UTC)

Looking at Maxwell–Boltzmann distribution#Distribution of speeds raises even more problems as I see it.. if probability(E_n) = fn(E_n) {where fn(x)=ae^-bx} then for a noble gas (psuedo ideal gas) 1/2mv²=E therefore v=sqrt (2E_n/m) . - but as fn(E_n) is an inversly exponential graph (as given) then the speed distribution will also drop off consistently as E_n increases - therefor not having a bump as shown in the diagram or supplied equation. seeMaxwell–Boltzmann distribution#Distribution of speeds. So what on earth is going on???83.100.174.70 17:06, 14 December 2006 (UTC)

Ah, okay, I think you are using "modal" differently from how I am using it. The approximation is justified in part, by the extremely large numbers involved, but you are correct to say that there are a number of subtleties associated with this process. f(E_n) is not the same as f(v) - there is a change of variables that takes place in that derivation. --HappyCamper 19:02, 14 December 2006 (UTC)

The derivation finds the distribution that has the greatest number of ways to obtain it ie W is maximum, this distribution is the modal one (see mode), the mean, or average distribution would take all the possible distributions (weighted by the number of ways to obtain them ie multiply by W as defined in the article).

Unfortunately the approximation is not justified (if you wish to use the distribution as a measure of probabilities) - since the average distribution is very different from the modal distribution.83.100.174.70 19:17, 14 December 2006 (UTC)

The change of variables of course makes a difference - it the case of molecular speeds the x axis is scaled as the square root eg a plot with x~energy, becomes x~sqrt(energy) - this would cause a change in measured slope at a given point - but does not cause a change in form of the graph - eg from inverse exponential (e^-kE) (note no hump) to the form given in Maxwell–Boltzmann distribution#Distribution of speeds (the graph now has a hump, why??); something is very wrong here.83.100.174.70 19:22, 14 December 2006 (UTC)

Trick chemistry question.

Hey! I have a chemistry question here that I think maybe a trick question. It asks: Find the number of oxygen atoms in 25g of CO₂. Is it 2? Or is it (after stoichemetry) 6.8 x 10²³.
The way i thought of it is that there are two Oxygen atoms in the CO₂ compound. But there are A LOT of Oxygen molecules in the 25g of it. Thanks for any feedback! --Agester 14:35, 14 December 2006 (UTC)

There are no oxygen molecules in 25g of Co2, there are only Co2 molecules. It's not a trick question. Just divide 25g by (2*molar mass of oxygen + molar mass of carbon), multiply by a mole, multiply by two. Is that how you got 6.8E23? yandman 14:43, 14 December 2006 (UTC)

Yeah, there's two oxygen atoms in each molecule of CO₂, but there's lots of CO₂ molecules in 25g of the stuff. As Yandman says, do the maths. --jjron 14:55, 14 December 2006 (UTC)

Thanks for the help on that question! I thought it'd be the 6.8E23. However, I'm stuck on another tricky chem question. I don't feel the need to spam this desk with several topics of chem question but i'm really really stuck. The density of a 1.95 Molar solution of KBr in water is 1.16 g/mL. Find the molality of this solution. The formula weight of KBr is 119.0 AMU.

What I tried was assuming it's one liter as your volume. Therefore you'd have 1.95 moles. Convert that 1.95 moles into grams which came out to be 232.05g (or .23205 kg) and then tried to get the molality now by dividing moles over kg and got 8.4 Mol/kg. Which isn't a choice here.
Second attempt i tried to assume okay. I have one liter. Lets use that for the density. 1160g/1000mL. And tried to divide my moles into that and got 1.68 mol/kg which isn't a choice either.
What am I doing wrong? I know the second attempt was badly incorrect from the start but it was worth a stab. --Agester 16:09, 14 December 2006 (UTC)

I would not expect the question to be constructed in this way...here would be my approach:

1. First find the molar mass of CO₂ - that would be about 12 + 16 + 16 = 44 g / mol
2. Now, find the number of moles of CO₂ you would have. That would be (25 g )/(44 g / mol) = 25/44 mol
3. Now, for each mole of CO₂, there are 2 moles of O atoms. So that means, there are 25/44 * 2 = 25/22 mols of O atoms
4. Now, for each mole there are an Avogadro's number of particles. So that gives you (25/22 mols) * (6.022 *10²³) = 6.843 *10²³ which is precisely the answer that you got originally!

The core concept behind this question does not focus on densities, or whatnot. The question is really focusing on whether you understand what a "mole" is and whether you understand the relationships between the ratios of atoms that are in CO₂ - you do not need to be concerned with the actual connectivity that exists between the atoms - that comes much later. --HappyCamper 16:56, 14 December 2006 (UTC)

I read the question several times and was unable to think, like a robot in Star Trek given two comands at once. Bu what I get from the question is that the concept of molarity is hard to get for a non chemist.

200 people jump out of a plane. The density of air is zz pounds/cubicfoot all people fall in to water with a density of 1 ton/cubicmeter. How many people are there in the water. --Stone 17:10, 14 December 2006 (UTC)

No the questioner wants to convert molarity to molality. The number you need is that water is 1 kg per liter. So a 1.95 molar solution divided by the mass of solution (1 kg) will get the answer (assuming you have a dilute solution at STP) Rmhermen 18:44, 14 December 2006 (UTC)

So if I interpreted that right. Does that mean the Molality of the solution is 1.95 Mol/kg all along? After all if the assumed amount of water is 1 kg/L wouldn't that make the volume 1L mass 1kg therefore:
1 molar = 1.95 mole/ 1L
1.95 mole / 1 kg = 1.95 mol/kg
(and I assume it's STP too because the question doesn't list any other information on Temp or pressure.) However, what does the density have to do with this then? Just useless information? --Agester 21:08, 14 December 2006 (UTC)

The confounding quantity here is of course the volume change associated with adding the KBr to water. We know that for a volume $V$ of solute (with mass $\rho V$), we create a mixture of volume $V'$ by adding a mass $MV'\mu$ of the solvent (M=molarity; μ=molar mass), so that the mass of the solution is $\rho V+MV'\mu$. We still know neither $V$ nor $V'$, but we additionally know $\rho '={\frac {\rho V+MV'\mu }{V'}}$. From that we can derive the ratio ${\frac {V'}{V}}$ — obviously their individual values are irrelevant — and then calculate $m:={\frac {MV'}{\rho V}}$ (m=molality). It all has to do with the volume change and (confusingly) with the different systems (before and after mixing) in reference to which molarity and molality are defined. Does that help? --Tardis 22:33, 14 December 2006 (UTC)

I'm not quite to sure what you mean Tardis. What is p' (or p prime?). Most of your work there is a little confusing to me. Like i'm not sure what you mean by V' either. In addition, How do we conclude to this mass of our solute? Unless we assume we have 1L of water and say we have 1.95 moles and then use the Molar mass to find out how many grams. But even then, a similar question would be what is V' and how does the density play in effect?
also! I managed to find our volume of KBr by assuming our volume of water is 1L. therefore we have 1.95 moles of KBr. With density in mind (1.16g/mL) i worked out 232.05g / V = 1.16 g/mL and volume eventually came out to 200 mL of KBr. I'm not sure if that helps us anymore or was just extra work but I tried throwing it around and stuff but no help for me. --Agester 01:12, 15 December 2006 (UTC)

Um... ρ and $\rho '$ are densities (as mentioned at the article, although for some reason it claims that uppercase rho is used there); the prime on one of them and on $V'$ means that the variable applies to the solution and not to the solvent alone (that is, it applies to the latter state of the system; see the prime's meaning with regards to physics in its article). The mass of solute is derived from the definition of molarity (which is defined in terms of volume of solution); I left it in terms of $V'$ precisely to avoid assuming any particular volume of water. Your calculation of the volume of KBr is flawed because you are dividing the mass of the solute by the density of the solution: the volume you get is the volume of solution which would have mass equal to the mass of KBr in a different amount of solution! If you can be more specific about what you don't understand (if anything, still) about the equations I wrote, that will help: it is extremely helpful to formalize the problem such that one does not make mistakes like operating on unrelated variables and being confused by numerical coincidences. --Tardis 03:03, 15 December 2006 (UTC)

tonsilitis

is it possible to get tonsilitis even if you have had ur tonsils taken out.

sorry this question is up twice, i didnt think it went up b4 - skye

No problem - I've removed the repeat question. Strictly speaking, tonsilitis is an inflamation of the tonsils, so no, you can't have tonsilitis after your tonsils have been removed. But one of the symptoms of tonsilitis is a sore throat (pharyngitis), which has many other causes as well, so you might feel as though you have tonsilitis even though it is really something else. Usual disclaimers ... I have no medical training, we don't provide medical advice and you should see your doctor if in any doubt. Gandalf61 15:16, 14 December 2006 (UTC)

ok so, if my friend has tonsilitis i could get the sore throat from the tonsilitis but not the actual tonsilitis cause i have no tonsils. does that make sense?

I think the question comes down to this -- can tonsils grow back?

I know other organs can -- so why not tonsils? I suppose it's possible. And thus, getting tonsilitus after having ones' tonsils removed is not unfathomable. Vranak 16:54, 14 December 2006 (UTC)

thanks... skye

Sore throats and tonsillitis, whether viral or bacterial infections can be contagious. I have known strep throat to be passed from person to person. Many sore throats are dangerous and the bacteria causing them can be highly transmissable. Other sore throats are mononucleosis also highly contagious. A sore throat may be scarlet fever. Untreated strep can lead to rheumatic fever. Please call your doctor, since Ref Desk is not a good source for medical advice. Edison 17:45, 14 December 2006 (UTC)

Tonsils can grow back. (I've seen them!) By that I mean that they grow back if they were not completely excised the first time. Mmoneypenny 19:36, 15 December 2006 (UTC)

physics junior cert experiment

how would you investigate the relationship between the length of a metallic conductor and its resistance? a brief description wold be a huge help...thanks.83.71.61.203 16:51, 14 December 2006 (UTC)

Take a wire - span it like clothesline that you have both ends at one side and a u-turn at the other - measure U and I of a battery at different place and draw a chart!--Stone 17:01, 14 December 2006 (UTC)

See Electrical resistance and Ohm's Law. With a typical ohmmeter or digital multimeter, it may take a long length of very fine wire to get clear results. A spool of very small wire, like # 28 or #30 wire, obtainable at places like Radio Shack, or will have more easily measured resistance than a short length of say #12 wire or even #22 doorbell wire. Resistance varies with wire diameter and with what metal the wire is made of. You should touch the ohmmeter leads together and note the lead resistance, probably a few tenths of an ohm, and subtract that from resistance of the wire. Make sure the wire is bare and shiny where you connect the leads, either by just touching them to the wire or with alligator clips (measure and deduct the resistance of the alligator clip leads as well). Then measure the resistance of, say 1 foot, 50 feet, 100 feet, etc up to the total length. Note the room temperature and record it as part of your data, since resistance varies slightly with temperature. Graph your data and see the form of the relationship between length and resistance, with wire diameter and temperature held constant. You could also use separate ammeter and voltmeter to get a better hands on idea of what you are measuring. Edison 18:00, 14 December 2006 (UTC)

If you don't have an ohm-meter, you can use a voltmeter, ammeter and Ohm's law. The basic principle is to vary length while keeping material wire is made from, cross-sectional area, and temperature constant. 80.169.64.22 17:14, 15 December 2006 (UTC)

a maths proof

there are six people in a room. it is then certain that eiter:at least 3 people know each other OR at least 3 people dont know each other. how is it certain?

Does it count to say when all 6 people meet in a room all of them know each other then? So it's the "4 or more" choice. X [Mac Davis] (DESK|How's my driving?) 18:21, 14 December 2006 (UTC)

Answered in the Maths RD. Please do not post more than twice. -- DLL ^{.. T} 18:36, 14 December 2006 (UTC)

Metals and Metals

Can there be a chemical bond between metals? What is that type of bond called?

Metal metal bonds are common in many to core complexes. There are single double ore triple bonded metals in some complexes.--Stone 17:15, 14 December 2006 (UTC)

Cyclopentadienyliron dicarbonyl dimer is one!--Stone

See also Metallic bond. DMacks 19:15, 14 December 2006 (UTC)

Also see cold weld. StuRat 03:31, 15 December 2006 (UTC)

Dentistry as a separate course

Why is dentistry separatly taught from medicine? Why is there no separate course for neurology, etc...Mr.K. 17:52, 14 December 2006 (UTC)

Tradition! The dentist is historically derived from a poorly trainned Barber, while a doctor is coming from a university. This diversion is still there and as a combination would mean reduced funding nobody is integrating both!--Stone 17:56, 14 December 2006 (UTC)

Surgeons also started out as Barber surgeon in past centuries, and in some countries are called "Mister" rather than "Doctor." In the U.S. dentists are called "Doctor" as are surgeons. Dentists before the late 19th century were generally "tooth-drawers" with little training, but today get extensive medical training related to their specialty in their 8 years of college, then receive a Doctor of Dental Surgery or Doctor of Dental Medicine degree, then have to pass a state board exam (in the U.S.). They may then take several years of additional training in a speciality. Dentists in the U.K. have to take a shorter training course, per Dentistry, and get a Bachelor of Dental Surgery. Edison 18:11, 14 December 2006 (UTC)

Aren't the teeth pretty well independant? If you studied neurology or hematology, you'd have to know all about the body. With teeth, you just have to know the mouth and jaw, and try not to cause infections.X [Mac Davis] (DESK|How's my driving?) 01:08, 15 December 2006 (UTC)

Barber surgeon was seen as not medicine. To cut of arms and legs is more like drawing teeth! Nowerdays the seperation is still there, but not that dentists are poorly trained! I hope nobody understood it this way! --Stone 18:37, 14 December 2006 (UTC)

IANADentist, but the statement that the teeth are pretty well independent needs a wee response. Dentists (in the UK) spend 5 years at dental school, then a year as a vocational trainee before being allowed to practice independently. If they want to specialise in say Orthodontics they will spend an additional 2-3 years at a teaching hospital then do a 3 year Masters Degree in Orthodontics followed by an additional 2 years to become an Orthodontic consultant. Saying that the teeth are independent is like saying that the blood or nerves are independent. Dentists need to know about head, neck and throat cancers (since doctors very rarely look into people's mouths (as you'll know if you've ever been admitted to hospital.)) They will also need to know about congenital craniofacial abnormalities and the effects of various systemic diseases on the mouth, e.g. B12/folate deficiency, sarcoidosis etc. Mmoneypenny 19:34, 15 December 2006 (UTC)

What is repsonsible for eye drying that comes with sleepiness?

Is this melatonin? Melatonin is secreted by the retina as the article states.

I believe that there's at least half a dozen different glands that secrete different oils across the surface of the eye, to keep it nice and lubricated as you blink. If there's a problem with these glands, or your body in general (which supplies nutrients to these glands) -- or if the room you're in has a lot of dust in it, then you'll get dry-eye. Sleepiness is associated with dry-eye, but it is not the cause per se. Vranak 18:32, 14 December 2006 (UTC)

I think I read that this comes from blinking less unconciously, to try and get you to close your eyes for good so they are wet. When I don't wet my eyes by closing and er... squeezing my glands, they can get pretty dry and feel like fiberglass. eek! And this is in every climate (not just saying weather). X [Mac Davis] (DESK|How's my driving?) 20:10, 14 December 2006 (UTC)

I asked a similar question recently, which I think is what you are referring to. I don't think anyone really knew the answer. BenC7 08:59, 15 December 2006 (UTC)

Thansk for that one. I wonder where to look for an answer!! X [Mac Davis] (DESK|How's my driving?) 19:26, 16 December 2006 (UTC)

Chicken breasts

With a choice between deboned and deskinned chicken breasts versus chicken breasts with skin and bones isn't it healthier to get the chicken breast with skin and bones since the body needs some fat too? 71.100.6.152 20:31, 14 December 2006 (UTC)

Are you eating an *entirely* fat-free diet? --24.249.108.133 21:15, 14 December 2006 (UTC)

Nope. I am overweight by 30 plus pounds though, have borderline high cholesterol and high blood pressure. Even using artificial sweeteners exclusively for well over a year (except for yogurt) I have an inability to come within or near 20 pounds of being overweight no matter what foods are in my diet or if I eat so little that I stay hungry. I can not swallow tuna or other meats unless lubricated with some oil or condiments or other foods like vegetables so if an "entirely" fat-free diet would be just as healthy I would do it. 71.100.6.152 21:59, 14 December 2006 (UTC)

Health is a funny thing. In a 1567 monograph by Swiss physician Paracelsus, it was observed that "All things are poison and none are without poison," a phrase which is usually said "The dose makes the poison." Everybody and their diets are a little bit different and it can be harder to generalize than people say. We don't know you very personally, although I'm betting people will say to eat the deskinned one, since you have enough fat in your diet other places. X [Mac Davis] ::(DESK|How's my driving?) 22:18, 14 December 2006 (UTC)

The bones don't matter as you likely won't eat them anyway. The skin, however, and the fat right under the skin, are high in bad fat and bad cholesterol. You will get more than enough fat and cholesterol from lean, skinless chicken breasts. Ideally, you should get good fat and cholesterol from things like avocados, salmon, and vegetable oils (avoid partially hydrogenated oils, however). Avoid all trans-fats completely. StuRat 01:21, 15 December 2006 (UTC)

You haven't asked for weight loss advice, but I'm going to give you some. Some people can't lose weight by controlling their diets, I'm one of them. Whether it's high protein, low calorie, low fat, I always plateau and no matter how much I reduce my food intake I don't lose weight. I have to exercise. Maybe that's true for you too. I suggest that you go and get a checkup and talk to your doctor about starting (or augmenting) an exercise plan. I should also say that - for me - aerobics are fine, but I don't lose weight with just aerobic exercise (walking, running, etc). However, when I do weight training, I drop weight like, well, weight. Get some sensible guidance from people in the know and get moving. Anchoress 09:17, 15 December 2006 (UTC)

"The body needs some fat too" is correct, but do not deceive yourself, there is ample fat in lean chicken breast. The fat is in muscle cells, as well as in fat cells. In the case of chicken, there is much less fat "hidden" inside the muscle than is the case with "marbled" meat, like beef. This I have known for a long time, but I have not checked it myself, so I did some calculations based on data from the USDA website. I was surprised:

     Chicken Breast With and Without Fat and Skin 
     (Yield from a 1 kg Chicken Prepared for Cooking.)
                                    Fat(g)     Energy(kJ)
     Meat with skin and fat          10.15       1078
     Meat (skin and fat removed)      3.89        810

     Chicken Breast (fat and skin removed)
            Compared to Beef Filet of Tenderloin (all visible fat removed)
                (Quantity of each is 100g)
                 Fat(g)        Energy(kJ)
     Chicken      3.03           632
     Beef        11.12           912

So there is no risk of suffering from fat deficiency if you eat meat without visible fat - 3% fat for the leanest chicken, 11% for the leanest beef. Something you can look up for yourself: Lean pork has less fat than lean beef.

One final note: "Staying hungry" is not a criterion for deciding whether your diet is strict enough for you to be able to lose weight. If you look at the Hunger article (unfortunately at present not much more than a tag) you can click on the links there to get a better picture of the physiology of hunger. Inter alia, Leptin and Ghrelin are fashionable topics in this field. Well, now I need a peanut butter sandwich... --Seejyb 21:45, 15 December 2006 (UTC)

Wanted to point out that even if your diet doesn't appear to help you lose weight, you're still likely to be far healthier if you adopt a healthier diet. Point being, just because you don't notice a difference in weight from eating chicken with the skin-on doesn't mean it's a good idea Nil Einne 13:56, 21 December 2006 (UTC)

LOL I love peanut butter too but even though it is supposed to have more protein than calf's liver it has an awful lot of fat too. On peanut butter I gain weight almost as if I'm eating pure fat. If I could find a way not to be hungry then that would help a lot in reducing the number of calories I consume which is the problem as far as I've been told rather than either fat or protein. Its just that fat has a lot more calories than protein. Tried sipping constantly on sodas but the doc said not to do that anymore since I was putting myself at risk of contracting diabetes. 71.100.6.152 13:57, 17 December 2006 (UTC)

Absolutely zero not really that cold?

Absolute zero has always facinated me in high school chemistry. The possibility of all molecular action stopping at a "mere" -273 below zero. Considering how hot things can get -- indeed is there even a maximum for heat? -- it seems we are living on the awfully cool side of the temperature universe. --24.249.108.133 21:12, 14 December 2006 (UTC)

There is, in fact, a theoretical maximum temperature for the universe. Review Heat death of the universe, which is a very interesting concept. You might also want to read Big Freeze, which is about the cold death of the universe.

For any individual mass, however, there is no upper bound for "heat," as energy in a particle with mass can grow without bound as the particle approaches the speed of light. Hipocrite - «Talk» 21:21, 14 December 2006 (UTC)

But there is a theoretical upper limit for temperature. Ironically, it isn't what you expect: it's -0 K. See Negative temperature. Titoxd^(?!?) 22:04, 14 December 2006 (UTC)

Also interesting is Orders of magnitude (temperature) -- Sandman30s 11:41, 15 December 2006 (UTC)

Yes, it does seem we are living on the cold side. It is easy for me to heat something by way more than 273 Celsius, and impossible to cool it more than 273 Celsius starting from zero Celsius. Edison 15:21, 15 December 2006 (UTC)

bullets from spitfires and hurricanes

During the Battle of Britain, did bullets from spitfires and hurricanes shooting down Germans accidentally kill anyone on the ground? Paul Silverman 22:03, 14 December 2006 (UTC)

Seeing how they would carry ~8 .30cal machine guns, which yield a lot of rounds, there's probably a damn good chance they did. I'm not sure how easy it would be to find actual reports, however. Saying the RAF is strafing its own people wouldn't be good for morale, so the civilian deaths were probably chalked up to enemy fire.--138.29.51.251 01:42, 15 December 2006 (UTC)

latent heat

What is going on inside a molecule during a phase change that causes or allows energy to be absorbed yet without producing an increase in temperature? Adaptron 22:50, 14 December 2006 (UTC)

I haven't done heat science in a while, though i'm starting to learn thermochemistry now...but if I remember correctly the molecules are gaining potential energy during the change, and not actual kinetic energy that you would see as the temperature incereases? I'm not sure, but check out the wikipedia articles of state changes and different types of energy. 74.102.89.241 00:30, 15 December 2006 (UTC)

The energy goes to either making or destroying the bonds between the molecules. For example with water, when either boiling or melting the energy goes towards breaking the hydrogen bonds between the water molecules. -anonymous6494 02:16, 15 December 2006 (UTC)

Basically, in a substance, the atoms that make it up are pulled toward eachtother by intermolecular forces (Hydrogen, Ionic, and Covalent bonds, London dispersion forces, etc.). As the atom gains energy it moves faster and faster, "bouncing around" until it finally gains enough energy to overcome the intermolecular forces, and breaks free. However, one must remember that when boiling a liquid, or during any phase change, not all of the atoms/molecules have enough energy to break free and undergo phase change(See Maxwell-Boltzmann Distribution on Wikipedia, or here). For any temperature, molecules are moving at different speeds. As a result, only a given amount of the substance has enough energy for phase change (only those moving over a speed of "x" change - the number of the molecules with that speed increases with temperature). That's why when you boil water the whole pot of it doesn't go "poof" into water vapor, and the phase change takes place over the time (i.e. the bubbles of H₂O gas released from the pot). --AstoVidatu 02:30, 15 December 2006 (UTC)

Why then do certain substances such as carbon dioxide skip the liquid phase entirely and go "poof" i.e., go from being a solid directly to a gas? Adaptron 05:41, 15 December 2006 (UTC)

Because the pressure's too low for carbon dioxide to do so. Have a look at this. David 19:24, 15 December 2006 (UTC)

See also Gibbs free energy. In short, despite the fact that less energy would be required to turn the CO₂ into a liquid instead of a gas, leaving more energy to become heat and therefore entropy, the increased entropy (disorder) of the gas phase makes it preferable. At a high enough pressure, it requires so much energy to give the gas a volume appropriate to its temperature and mass that it becomes more profitable (in terms of entropy) to move to the liquid phase and use the energy to provide disordering thermal motion. Does that help? --Tardis 21:06, 15 December 2006 (UTC)

So then if "disordering thermal motion" is a variable like volume, pressure and temperature what is the equation that includes it as a variable? Adaptron 10:07, 17 December 2006 (UTC)