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December 13

drywall mud

re-post, got archived

I Was looking at the MSDS sheet for a powdered drywall mud that you mix with water and hardens in 20 min. what was strange to me is that it has formaldehyde in it, presumably as a biocide. I don't see any functional purpose for this as it's a powder and it dries in 20 min. can anyone shed some light on this? The link to it is below --Jrbsays (talk) 11:26, 1 December 2011 (UTC)

http://www.usg.com/rc/msds/joint-compounds/sheetrock/durabond/sheetrock-durabond-20-joint-compond-msds-en-61205014.pdf

I'm certainly not an expert on building materials, but the the text above the carcigenicity table in that data sheet says, "All substances listed are associated with the nature of the raw materials used in the manufacture of this product and are not independent components of the product formulation," which clearly implies that they're not intentionally using formaldehyde as an ingredient in the product. My guess is that both acetaldehyde and formaldehyde may be given off by the "vinyl alcohol polymer" that is an ingredient in the product. Deor (talk) 13:13, 1 December 2011 (UTC)

It's not the EVA, I don't think, otherwise the MSDS would mention the formaldehyde as a decomposition product, no? --jpgordon::==( o ) 15:11, 1 December 2011 (UTC)

It might get damp from time to time from condensation, especially on an exterior wall, such as when it's cold outside and you've just taken a shower nearby or boiled water on the stove. A roof leak or window left open during rain could also dampen it occasionally. StuRat (talk) 16:26, 1 December 2011 (UTC)

"BGC Multipurpose Joint Compound Data Sheet" says "trace amounts of residual vinyl acetate monomers, acetaldehyde and formaldehyde may be associated with the production of the emulsion polymer". So it's a by-product of the manufacture. --Heron (talk) 19:20, 1 December 2011 (UTC)

I don't think it's a leftover product, because other MSDS sheets from the same company on a different type of drywall mud do not list formaldehyde as an ingredient. That MSDS sheet can be found below

http://www.usg.com/rc/msds/joint-compounds/sheetrock/sheetrock-all-purpose-joint-compound-msds-en-61320001.pdf

In addition other brands of drywall mud such as "dap" brand do not list formaldehyde on their MSDS sheet.--Jrbsays (talk) 05:25, 2 December 2011 (UTC) — Preceding unsigned comment added by Jrbsays (talk • contribs)

Demixing of gases of different density

Suppose I put inside a gas cylinder 0.25 atomic percent of xenon and the remainder helium. Everything is thoroughly mixed and compressed to 40 bar. Now I let the bottle sit at 300K. How long must I wait before the very top of the cylinder is depleted of xenon by a factor of 2? --HappyCamper 08:10, 13 December 2011 (UTC)[reply]

That's likely to involve complex math because the gases will work like liquids at some point and you'll get condensation at the walls of the container and fluid dynamic effects. (Think lava lamp. - Experts: I know not an exact analogy, but just to convey the idea.) OTOH there are probably standard formula that will get you a close enough approximation. 196.202.26.177 (talk) 09:59, 13 December 2011 (UTC)[reply]

Forever. At 300 K, the kinetic energy per particle is much larger than the gravitational potential energy difference associated with each species traveling the length of a gas cylinder. As a result, the amount of differentiation you can expect due to gravitational separation is very low, and will never reach anything close to a factor of 2 enrichment. Diffusive mixing due to the thermal energy in the gases will keep them well mixed. Dragons flight (talk) 10:24, 13 December 2011 (UTC)[reply]

I don't know about that. At 40 bar the mean free path would be quite tiny so you're really talking about a kind of fractional distillation process rather than them zipping from one end to the other. On the other hand the difference in pressure between the top and bottom would be quite small, if either gas was on its own it would fill the cylinder nearly uniformly. There probably is I feel some elegant and simple way of estimating the answer rather than going through all the maths but it eludes me at the moment. Also I would guess you'd have to wait a very long time to avoid convection effects. Dmcq (talk) 12:58, 13 December 2011 (UTC)[reply]

An important question would be: How tall must the cylinder be? Assuming ideal gases, the answer to this question is independent of pressure. It depends on the Boltzmann factors $e^{-{\frac {mgh}{kT}}}$ where m is the atomic (or molecular) mass, g is the gravitational acceleration, h is the height, k is Boltzmann's constant, and T is the temperature. The density of each gas is proportional to this Boltzmann factor (with the specific atomic mass), so you can calculate the height at which the ratio of the Boltzmann factors $e^{\frac {(m_{Xe}-m_{He})gh}{kT}}$ becomes 2 (about 1385 m). Icek (talk) 15:13, 13 December 2011 (UTC)[reply]

That sounds a bit high to me. I typed 'heavy gas' into googleand got this video cool trick with heavy gas and it certainly didn't look like it was set to become fairly evenly dispersed with the air. Not helium and xenon of course. Dmcq (talk) 16:57, 13 December 2011 (UTC)[reply]

It will disperse. Just give it some time. Dauto (talk) 18:22, 13 December 2011 (UTC)[reply]

Let's say the cylinder is 2m in height. Is a century long enough? Somehow I'm sure that there is a simple way to estimate the time scale, but I just can't think of it right now. It's not clear what's the most appropriate model for this system. --HappyCamper 04:07, 14 December 2011 (UTC)[reply]

Read Dragons flight's and Icek's comments above. The two gasses will never separate at room temperature. Dauto (talk) 05:13, 14 December 2011 (UTC)[reply]

The systems you can see in those videos aren't at equilibrium; if you wait long enough, the xenon (or sulfur hexafluoride) will mix with the room air. TenOfAllTrades(talk) 05:39, 14 December 2011 (UTC)[reply]

First-line, second-line, third-line defense against cancer... What's the point of this?

I've been reading about cancer treatment recently, and have noticed a pattern. Oncologists tend to use a certain treatment (first-line treatment), and when the cancer develops resistance to it, they bring in a second treatment (second-line treatment), and when the cancer develops resistance to it, they bring in a third treatment (third-line treatment), and so on.

In many cases, there's no obvious reason why these different treatments can't be used at the same time. Sometimes they're not even in the same treatment class (chemotherapy, radiotherapy, immunotherapy, tumor-treating fields, vaccines, etc.), so there isn't the problem of drug interaction or overtoxicity!

So, to me, using staged treatment instead of simultaneous treatment seems just as stupid as using one antibiotic at a time on tuberculosis, and allowing it to gain resistance to all antibiotics one step at a time.

Can someone please explain why we have different lines of treatment, instead of using them all at the same time to minimize the chances of the cancer becoming resistant? Thanks!--109.14.86.149 (talk) 08:53, 13 December 2011 (UTC)[reply]

If the other treatment options have the same kind of side effects that chemotherapy does, then using them all at once is pretty bad news for the patient... 80.122.178.68 (talk) 09:18, 13 December 2011 (UTC)[reply]

I'm far from a medical expert but have had some close relatives with "personal experience." First of all with most cancer treatments we're still at the stage of "Kill the cancer cells without killing the patient." Far from humans being all the same, every body is different. We have different tissue, immune systems, health history, endocrine reactions, etc., etc. Yet treatment is mostly based on the best available statistical data, what worked for a large number of patients in the past. By its very nature, that data is not for the most recent advances in treatment. Five year survival data is for people who lived that long after the treatment they received 5 years ago. So, we have some idea how well that worked for a statistical sample. That still means it may not work for you. There's also a "standard of care" which is based on those studies. But not just that. Increasingly "quality of life" is being considered. That is, keeping the patient alive but in constant pain, misery and barely functional may be considered not in the patient's best interest. A first line therapy may also be applied to make tumors shrink to be able to evaluate a patient's chances for surgical removing those later on. Some patients get so sick from that "first line" that they are no longer suitable candidates for surgery. People's stamina, pain threshold, general mental state and attitude towards risk, are also taken into consideration. You punch some person in the face and they go "Is that all you got?" and the next person doubles over and huddles in a corner for a good cry. BTW: Diet (the stuff you eat, not the weight loss kind) is still not being studied sufficiently, although it's becoming increasingly apparent that it's very important to the development and treatment of cancer. Our standard medical establishment is just not set up to cover that area. Hope this helps. 196.202.26.177 (talk) 09:52, 13 December 2011 (UTC)[reply]

Some forms of therapy are contradictory: immunotherapy can either strengthen the immune system or disable it. Chemotherapy typically depresses the immune system by killing fast-dividing cells, so it won't work well with therapies that strengthen the immune system. Something like boron neutron capture therapy which involves filling the tumor with boron may interfere with conventional radiotherapy. --Colapeninsula (talk) 11:18, 13 December 2011 (UTC)[reply]

US scientific association membership for teenagers

Where I live, in England, we have the Royal Institution, which has a category of membership for teenagers, to get them interested in science. I have a friend in the US, with a scientifically-minded daughter in her early teens, and I am looking for something roughly equivalent over there. Does anyone know of such a thing? It doesn't necessarily have to be general science; it can be more physicsy, for example. Thanks, everyone. The Wednesday Island (talk) 13:27, 13 December 2011 (UTC)in the USin the US[reply]

I don't know of any organizations like that, hopefully someone else does. I would suggest some of the science (related) competitions like FLL (middle school) or FRC (highschool) (robotics competions) if it doesn't have to be any particular subject, and if your friend wants something that's likely to be useful in the future. Heck froze over (talk) 14:56, 13 December 2011 (UTC)[reply]

Yes, see our article on FIRST, or their web site http://usfirst.org. As a former judge at a FIRST Robotics competition 3 years ago, I can heartily recommend this. I never saw such a large group of kids enthusiastic about science. A couple of the teams even had their own cheerleaders. For younger kids, FIRST also has a "Lego league" which is a competition using Lego robotics.

The Society of Physics Students also has some outreach programs for teenagers. ~Amatulić (talk) 18:04, 13 December 2011 (UTC)[reply]

You can become a member of the AAAS, the publishers of the prestigious journal Science. Their membership includes a subscription to the magazine and they have specific kids and student categories. Here's their website. Vespine (talk) 22:10, 13 December 2011 (UTC)[reply]

radiation

Is there any formula for the radiation of matterials other than black bodies, if not, can I at least look at a diagram or something? also, a black body emits every possible frequency at any temperature, what about others? are there "gaps" in their intensity-frequency diagrams?--Irrational number (talk) 14:27, 13 December 2011 (UTC)[reply]

The simplest modification to the black body formulae is just to introduce a constant emissivity and absorbance. That gives you a grey body (which just redirects to black body, but it's mentioned near the end of the "explanation" section). You can get more complicated things once you introduce chemistry, though. Different atoms and molecules have different absorption spectra. If you look at the spectrum of the Sun, there are lots of black lines corresponding to wavelengths that are absorbed by different atoms in it (that is, in fact, how helium was first discovered, hence the name - helios is Greek for sun). See Fraunhofer lines for more detail. --Tango (talk) 15:08, 13 December 2011 (UTC)[reply]

Born's Proof of uncertainty principle

can anyone explain the mathematical proof of the uncertainty principle in laymen's terms to someone who only understands basic Matrix algebra.Thanks.109.148.92.178 (talk) 16:59, 13 December 2011 (UTC)[reply]

If the extent of a wave is small compared to its wavelength (Small uncertainty in the position) than the wavelength itself becomes uncertain (Large uncertainty in the momentum). Dauto (talk) 19:15, 13 December 2011 (UTC)[reply]

If you don't understand the math (and want to learn it), Fourier Transform is the place to go. Dauto (talk) 19:17, 13 December 2011 (UTC)[reply]

as i said, in laymens terms, and wading through that article is a long way to find the solution; pretend you were writing a book for the people.Thanks — Preceding unsigned comment added by 109.148.122.30 (talk) 20:03, 13 December 2011 (UTC)[reply]

Alright. In lay terms. If you have a wild animal in a corner, the tighter you constrain his wiggle space (small uncertainty in position), the wilder it becomes jumping up and down and back and forth (large uncertainty in momentum). Dauto (talk) 02:01, 14 December 2011 (UTC)[reply]

Radio around the world

During the Second World War, WLW in Cincinnati, Ohio, USA broadcast at 500,000 watts, and there were reports of American military personnel in Alaska and England being able to hear it in optimal conditions. Given ideal conditions (including no interference), how many megawatts would a radio broadcast need to be able to be heard at the antipode? 140.182.232.225 (talk) 17:25, 13 December 2011 (UTC)[reply]

It's not just the power, it's the bounce. See DXing. Rmhermen (talk) 17:54, 13 December 2011 (UTC)[reply]

It is also stated as if this were a homework problem. That isn't what this page is for. ~Amatulić (talk) 17:56, 13 December 2011 (UTC)[reply]

And how can it be that the radio waves bounce around the planet? How is this effect called? could we all transmit a laser around the planet bouncing around? — Preceding unsigned comment added by 80.58.205.105 (talk) 18:07, 13 December 2011 (UTC)[reply]

See Radio propagation and Ionosphere Nanonic (talk) 18:47, 13 December 2011 (UTC)[reply]

.. and Tropospheric propagation Nanonic (talk) 18:48, 13 December 2011 (UTC)[reply]

My father explained it to me by telling me that the Van Allen belts reflected the radio waves back to earth. They are opaque to radio waves but don't absorb them, so the waves bounce. He was an ex-RAF radar operator so I thought he knew what he was talking about. The query about a laser, I'm not sure but as laser is a form of light, and as we can all see on a starry night the Van Allen belt is transparent to light, it wouldn't happen. --TammyMoet (talk) 18:57, 13 December 2011 (UTC)[reply]

The explanation I had heard growing up in Cincy was that WLW got an exemption becaue they broadcast farm reports acoss the midwest, not anything to do with WWII, but I had heard the same thing about them being occasionally picked up absurdly far away from Ohio. Their tower is quite impressive, a standard high mast radio tower turned upside down with another one right side up on top of it. Beeblebrox (talk) 19:43, 13 December 2011 (UTC)[reply]

The first paragraph of Radio#Audio says that stations that powerful had their transmitters "commandeered for military use by the US Government during World War II", and the WLW article says that the authorization for the station to broadcast at 500 kw was withdrawn in 1939, though "because of the impending war and the possible need for national broadcasting in an emergency, the W8XO experimental license for 500 kilowatts remained in effect until December 29, 1942." There seems to be a bit of a contradiction there. Deor (talk) 20:29, 13 December 2011 (UTC)[reply]

This isn't a homework question; my homework right now is related to matters of abstruse historical theory. It's simply that I know almost nothing about radio propagation. I'm guessing that you mean that it wouldn't really be possible, regardless of the wattage; am I right? 140.182.232.225 (talk) 21:06, 13 December 2011 (UTC)[reply]

At night time the radiation from the sun does not ionise the D layer of the ionosphere which cause absorption due to high density. They reflect off the lower side of the F layer. The main problem with long distance AM band is interference. So the far station cannot be received due to interference from closer stations. If there were no other stations reception may be possible at the antipodes at sunset/sunrise over the night path, and higher power would not be required. The signal becomes more concentrated at the antipodes due to focusing by the curved inner ionosphere. Graeme Bartlett (talk) 21:27, 13 December 2011 (UTC)[reply]

The receiving antenna is important. A Beverage antenna is suitable for this purpose. Count Iblis (talk) 00:09, 14 December 2011 (UTC)[reply]

The Kon-Tiki expedition carried 3 transmitters with 10 W power, and they managed to reach Oslo from their raft in the middle of the pacific, about 10000 miles away, sending only slightly delayed birthday greetings to King Haakon. See Kon-Tiki#Communications. --Stephan Schulz (talk) 13:32, 15 December 2011 (UTC)[reply]

Yes, but they used the 20 meters band and then it's a lot easier. Count Iblis (talk) 14:44, 15 December 2011 (UTC)[reply]

Sending a radio signal to the antipode is trivially easy, provided the correct frequency for the conditions is selected. In fact going completely around the world is routinely achieved by amateur radio operators. In good conditions a few milliwatts is enough power to do it. Roger (talk) 15:24, 15 December 2011 (UTC)[reply]

pleeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeease help me

two thin rods that have some angle with each other, and rotate along an axis perpendicular to the plane of the two rods, how do I calculate moment of inertia?I know that I mustn't simply add them up, because if this was the case, the MOI of a thin rod that rotates around an axis perpendicular to it's center could be the sum of MOIs of two thin rods with half the length that rotate around their end, which is not. so, how do I do it?--Irrational number (talk) 17:46, 13 December 2011 (UTC)[reply]

Is this a homework problem? ~Amatulić (talk) 17:54, 13 December 2011 (UTC)[reply]

Well, technically, I am solving the problems of my university's last year exams (my own is on this Thursday). but I have thought about it(the problem), and have no clue what to do!--Irrational number (talk) 18:21, 13 December 2011 (UTC)[reply]

can anyone please give me a hint?--Irrational number (talk) 18:56, 13 December 2011 (UTC)[reply]

I am not an expert in this field, but I'll try to help a bit. According to Moment_of_inertia#Properties,

"The moment of inertia of the body is additive. That is, if a body can be decomposed (either physically or conceptually) into several constituent parts, then the moment of inertia of the whole body about a given axis is equal to the sum of moments of inertia of each part around the same axis."

--This seems to me to contradict your claim on additivity, so you may wish to research further along these lines. Also you could also try to solve the problem as a system of point masses, and then take a limit as the point masses approach a continuous uniform mass distribution. If done correctly, you should get an integral similar to the one at the end of Moment_of_inertia#Scalar_moment_of_inertia_for_many_bodies. You'll have to be careful about how to set up the integral, but this might help you get started. SemanticMantis (talk) 19:15, 13 December 2011 (UTC)[reply]

Your first instinct was correct. Simply add the two MoIs. The MOI of a thin rod that rotates around an axis perpendicular to it's center IS the sum of MOIs of two thin rods with half the length (and half the mass) that rotate around their end. Dauto (talk) 19:10, 13 December 2011 (UTC)[reply]

That's right. What trips up a lot of people when comparing the two rod equations at list of moments of inertia is forgetting to adjust the mass when adjusting the length.

You didn't say if the axis of rotation was at the vertex of the angle, or through any arbitrary point perpendicular to the plane of the rods. The solution is trivial if it's the former, more complex if it's the latter. ~Amatulić (talk) 19:29, 13 December 2011 (UTC)[reply]

Ha, I get it, I found the MOI of the two rods, just forgot to add them up!sorry for disturbing you...--Irrational number (talk) 19:30, 13 December 2011 (UTC)[reply]

Kennedy's Disease

The current diagram shows only carrier mother. What is the diagram if the father is an affected father? — Preceding unsigned comment added by 99.243.178.111 (talk) 18:48, 13 December 2011 (UTC)[reply]

Here is a link to the article, for anyone who doesn't know what you're talking about: Kennedy's disease. StuRat (talk) 19:16, 13 December 2011 (UTC)[reply]

That's actually the most interesting case. In the case of daughters, they will be obligate carriers, since the farther only has the one, the affected, X chromosome to contribute the the daughter's XX genotype. Sons of affected fathers can never be carriers of the disease, since the father will necessarily have contributed the Y chromosome, so the X comes from the non-affected mother. Fgf10 (talk) 19:53, 13 December 2011 (UTC)[reply]

Forgot to add, this is not just the case in Kennedy's, but in all recessive X-linked diseases. (And for completeness, the inheritance is the same in a father affected by a dominant X-linked disease, with the difference that the daughters will be affected by the disease, rather than be just carriers) Fgf10 (talk) 19:56, 13 December 2011 (UTC)[reply]

focal length and depth of field and wavelength-dependence

I have a very narrow depth of field in a particular imaging experiment. I notice that if I switch from visible (red) light to near-infrared light, the image becomes out of focus. Does depth of field increase or decrease with increasing wavelength, and how much does the focal length for a given aperture change with wavelength? elle _{vécut heureuse} ^{à jamais} (be free) 19:44, 13 December 2011 (UTC)[reply]

First of all, what determines your focal length? (In other words, describe your optical system). If you have a lens made of optical glass - even high-grade scientific-quality optical glass, its index of refraction is wavelength-dependent (and therefore, its focal length is also wavelength-dependent - to say nothing about derived properties like its depth of field). This behavior is called chromatic aberration, and because you are working with infrared, it's very probable that even your high-quality glass optics, calibrated for minimal chromatic aberration in the visible spectrum, suffer from severe aberrations at infrared. The focal length of a single lens element can have a huge variance with wavelength - this is sometimes called the Vd or dispersion value of the index of refraction. Multi-element lens? All bets are off! You should try to get your hands on the excellent (but pricy) book, Applied Photographic Optics; you can read much of it online at Google Books.

In any event, single-element optical glass or multi-element optical system, the vd is the value that parameterizes wavelength-dependent behavior of your focusing system.

In the purely theoretical sense, all other items being equal, an aperture is measured relative to the wavelength you are imaging. So, shrinking the wavelength is, physically, similar to growing the aperture. Therefore, I would expect a shorter wavelength to have a tighter depth of field ("fewer things are in focus"). In other words, infrared light should have a larger region in focus than red light; and, the center of the focused region will occur at a different wavelength-dependent focal length. However, practical complications can severely limit the validity of this general rule. Nimur (talk) 20:34, 13 December 2011 (UTC)[reply]

Chapter 22, Depth of Field and Depth of Focus. Equations, diagrams, charts, tables of common values; pure physics, and on-the-ground realities for the common lenses that you can actually buy in stores. There's even a discussion of lensing for electron microscopy. This book is truly excellent. Nimur (talk) 19:06, 14 December 2011 (UTC)[reply]