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January 16

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What kills Sudden Oak Death?

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A preserve where I love to hike is currently being devastated by sudden oak death. I don't want to spread the pathogen, Phytophthora ramorum, around to other places. I can wash my boots but I don't know whether that's enough. Does anyone have any information on what kills it? Something simple that won't hurt either me or my boots (much), maybe rubbing alcohol? --Trovatore (talk) 02:51, 16 January 2011 (UTC)[reply]

If you find out, the information would be a valuable addition to the article. Dismas|(talk) 05:51, 16 January 2011 (UTC)[reply]
This intrigued me, so I poked around a bit. There is some information on Late blight of potato, caused by a very similar fungus-like-thingy. As a preventative measure, it suggests any fungicide, of which an interesting option is ordinary milk (see the sources), as well as various natural oils. The other low-damage approach may be special-order UV light. Just don't tell any customs agents where you've been - does someone know what happens if you re-enter the country and say "yes" to having been on farms? SamuelRiv (talk) 07:26, 16 January 2011 (UTC)[reply]
I've been in that situation twice, having arrived in the U.S. after having hiked across farmland in Scotland and Peru. All that happens is the customs agent uses some kind of disinfectant on the bottoms of your hiking boots, and then gives your boots back to you. The disinfectant doesn't seem to have any adverse affect on the boots that I've noticed, and it doesn't delay the entry procedure by more than about a minute. There's no good reason to lie to the customs agent about having hiked across farmland, and you help avoid the possibility of spreading some pathogen to U.S. soil by telling the truth. Red Act (talk) 15:05, 16 January 2011 (UTC)[reply]
I had to disclose that I was on a farm and in contact with livestock when returning to the US after a trip to the Netherlands. They didn't seem to care, but they may have been distracted while asking if I had brought any "blown glass tobacco-smoking paraphernalia". Always tell the truth - it's not a hassle and lying to government officials gets you nowhere. Freedomlinux (talk) 08:40, 17 January 2011 (UTC)[reply]
Ooh, for soil molds: "The way to sterilize soil is to cook it. I have always been told 175 degrees F for 2 hours."[1] Can your boots withstand it? SamuelRiv (talk) 07:30, 16 January 2011 (UTC)[reply]
Here are some comprehensive guidelines for professionals working in P. ramorum-infested sites. Sanitation measures to minimize pathogen spread. So it looks like it boils down to:
Brushing anything off clothing. Cleaning dirt off boots and any equipment and wiping over with ordinary dilute bleach. Flaming everything with napalm doesn't seem necessary. --Aspro (talk) 15:25, 16 January 2011 (UTC)[reply]

Mass of the Earth or any Planet?

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Here is the link which shows earth's mass calculation. http://www.wonderquest.com/calculation-mass-of-earth.htm

Question One

As Mass of Earth = Me = g r² / G. But Me varies when on-center distance between the two masses varies e.g moon on earth - is this right?

Question Two

Shouldn't the force that the object of mass m resists the accelaration due to gravity be = (Me)g1= weight of earth on the body of mass m, where g1= G (m) / r², accelaration due to gravity of a body of mass m OR accelaration of earth towards centre of a body of mass m. Where r is on-center distance between the two masses= radius of earth and shouldn't be confused with radius of a body of mass m.

Thus equating both forces if I'm not wrong

F1=F2; mg=Meg1, Me=m(g/g1)

Mass of the earth still varies in this way too.68.147.41.231 (talk) 05:36, 16 January 2011 (UTC)khattak#1-420[reply]

EDIT

Putting respective values of gs in both cases we get Me=Me w/o putting numerical values of g, G, r

Me = g r² / G= (GMe/r²)r²/G=Me

Similarly

Me = mg/g1=m(GMe/r²)/(Gm/r²)=Me

Mass of earth Me itself unknown in its equation of gravitational accelaration g = GMe/r² then how come we can get Me from equation Me = g r² / G.

68.147.41.231 (talk) 15:53, 16 January 2011 (UTC)khattak#1-420[reply]

For your first question: No. The in the formula must be measured at distance from the earth's center. If you move up into space, increases, but decreases such that your computed stays the same.
I'm not sure I understand your second question. –Henning Makholm (talk) 05:51, 16 January 2011 (UTC)[reply]
Note that the acceleration of gravity is something you can measure with simple mechanical experiments. Measuring the universal constant of gravity G is quite a bit harder; see torsion balance and gravitational constant for more details. –Henning Makholm (talk) 15:04, 17 January 2011 (UTC)[reply]
Does the 'center' in "on-center distance" between the Earth and Moon refer to their barycenter? The Moon's gravity does not change the Earth's mass, but there are changes in tidal forces (proportional to the inverse cube of distance) while gravity is proportional to the inverse square. ~AH1(TCU) 02:08, 18 January 2011 (UTC)[reply]

on-center OR center to center means the distance between the center of earth to the center of resting mass (may be of moon size) on earth.

As w = mg = m (MG/R^2) where R is radius of earth but actually the distance between center of earth to center of a resting mass on it. Here is how theoritically the weight of body decreases for the same mass due to increase in center to center distance between them.

1- Let a solid steel sphere of radius 100 m of mass m on earth

2- Let the same solid steel sphere remoulded to either hollow big shpere of thin shell or long vertical tube bar on earth.

The weight of a resting body of mass m on earth shown previously becomes zero when its size reaches to the size of earth exactly (if i'm not wrong)

The measurement of g through experiment is only useful for earth but not for the other celestial bodies 68.147.41.231 (talk) 07:18, 18 January 2011 (UTC)khattak#1-420[reply]

For planets that have moons, astronomical observations of how the moons move allows one to measure the acceleration of the moon directly. More careful observations of the orbits, including how the moons perturb each other away from pure Keplerian orbits, allow astronomers to calculate the masses of the moons themselves. The same technique must be used for planets that have no convenient moon to observe (Mercury and Venus). Later, of course, the orbits of artificial space probes that came close to those planets allowed more precise measurements.
Mass estimates for lone asteroids and trans-Neptunian objects tend to be very uncertain because there is nothing near them whose acceleration can be measured. –Henning Makholm (talk) 10:10, 18 January 2011 (UTC)[reply]

How are improved measurements of the speed of light reported?

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Since the definition of the meter fixes the value of the speed of light at exactly 299,792,458 m/s, how are improved measurements of the speed of light reported? —Bkell (talk) 06:06, 16 January 2011 (UTC)[reply]

They become advances in metrology: more precise ways to measure length (in practice, that means more precise calibration of existing length measurement methods). –Henning Makholm (talk) 06:55, 16 January 2011 (UTC)[reply]
You just contradicted yourself - you said the speed of light is fixed, yet you ask how its accuracy is improved. A fixed value can't be improved, then it wouldn't be fixed. --Plasmic Physics (talk) 08:26, 16 January 2011 (UTC)[reply]
The definition defines the length of the metre in terms of the fixed, constant, speed of light, so a "measurement" of the speed of light is used to refine the measured length of the metre, to define the metre more accurately, as Henning said above. Dbfirs 09:30, 16 January 2011 (UTC)[reply]

Yes, I understand all of these things. But presumably there are still efforts to reduce the measurement uncertainty of the speed of light (or, to be technical, of the length of the meter). Simply defining the meter in the way we do doesn't mean we are suddenly disinterested in determining more and more precisely what the speed of light is—it just means we can no longer express the results of our experiments as more significant digits in a quantity expressed in meters per second. Surely there continue to be ever more precise apparatus built to determine the speed of light length of the meter. My question is, how are the results of those experiments reported in the literature? It's a literal question—what kinds of words and phrases are used? It used to be possible (in 1972) to say, "We have determined that the speed of light is 299,792,456.2±1.1 m/s," but that is meaningless now. And surely it is just as meaningless to say something like, "We have determined that the length of the meter is 1.00000000000046 m." So how is it done? —Bkell (talk) 10:42, 16 January 2011 (UTC)[reply]

Hmm, after thinking about Henning's answer a little bit, perhaps I see. Is it the case that, since the redefinition of the meter, attempts to more accurately determine the speed of light per se have ceased, and instead other things are just measured more precisely against the speed of light? So the results of these experiments, which would previously have been interpreted and reported as more precise measurements of the speed of light, are now interpreted and reported as more precise measurements of some other quantity? —Bkell (talk) 10:48, 16 January 2011 (UTC)[reply]
Yes, the redefinition enabled a more accurate determination of the length of the metre, currently realised using the wavelength of a particular design of helium-neon laser, accurate to almost eleven significant figures, whereas the previous "krypton" definition (in terms of which the speed of light was measured) was accurate to less than nine s.f. Each redefinition of the metre throughout its history didn't change its length, but increased the accuracy with which we can measure the original length (though each was actually based on the previous definition, not on the original seconds pendulum or the ten-millionth of the half-meridian). Dbfirs 11:13, 16 January 2011 (UTC)[reply]
The definition of the meter is a ratio that defines the meter not the speed of light. If the uncertainty in the speed of light is reduced in the future, it will create options to keep the present meter definition or to redefine it by a ratio with more significant figures. The article Speed of light mentions the CGPM standardisation conference that ratified the present standard and a future revision might proceed in a similar way. Cuddlyable3 (talk) 11:29, 16 January 2011 (UTC)[reply]
Any subsequent redefinitions will be better measurements of the length of the metre. There is absolutely no intention to ever change the current value of the speed of light from 299,792,458 m/s or to change the definition of the metre. I assume that this is what Cuddlyable3 was saying, but the reply could possibly be mis-read. Any improvements will just give a more accurate realisation of the metre. There is now zero uncertainty in the speed of light, but there is uncertainty in super-accurate measurements of the metre to eleven significant figures. The announcement will be something along the lines of "The new recommended realisation of the metre is now [xxxxxxxxxxxx ±yy] wavelengths of [a particularly stable and easily measurable laser light] (or other radiation in the E-M spectrum) under certain conditions" (e.g. in zero gravity). This will not change the length of the metre because it will be within the previous limits for realisation, but it will give accuracy of perhaps twelve significant figures. Dbfirs 13:18, 16 January 2011 (UTC)[reply]
Bkell, that's exactly correct. The experiments are the same either way; what changes is the language used to describe them. -- BenRG (talk) 21:27, 16 January 2011 (UTC)[reply]
Asking how the accuracy of the speed of light is improved, is like asking how the accuracy of the number zero is improved. The speed of light was made to be exactly the current number. The number zero will never be made more accurate to change from 0.0000 to 0.0001. --Plasmic Physics (talk) 04:17, 17 January 2011 (UTC)[reply]
No, you're missing the point. An improved accuracy of the knowledge of the speed of light means improving the accuracy of the knowledge of the size of the unit m/s. It's like the electronvolt: the energy required to move an electron through a voltage difference of 1 volt is 1 eV by definition, but if we measure this energy more precisely, it doesn't mean we know a more precise value for the constant 1; it means we know the size of the eV more precisely. In this case we would express this knowledge by stating the value of the eV in joules, say. The original poster asks how we would express an improved precision of the speed of light, since there isn't an alternative unit suitable to convert m/s into for this purpose. I don't know the answer. --Anonymous, 07:47 UTC, January 18, 2011.
I gave an example above of how the "improvement" might be expressed as an improved realisation (not definition) of the length of the metre. I don't see what the problem is. The second is known to a high degree of accuracy. The speed of light is fixed, so the only variable is the accuracy with which the metre is known. Dbfirs 10:47, 17 January 2011 (UTC)[reply]
Measuring the length of a meter may sound absurd, but that's what happens when length is a derived unit of time. But for example, if you happen to have an old platinum-iridium meter bar lying around in your attic, with improved interferometry equipment you might take it out for grins and see if it's a little longer or shorter than a meter. You might then of course use the same improved device on a bridge, space elevator, Moon probe, meridian from North Pole to the Equator through Paris etc. - improving your measurement of the length of the distance, whatever it might be, in each case. (Well, except the North Pole - is there a North Pole at the nanometer scale? I'd think every time the ocean currents shift or a billion Chinese commuters try to get to work or (at least) when the currents in the outer core shift with the magnetic pole, that it must move too much to measure precisely, but I may be wrong. Wnt (talk) 05:48, 19 January 2011 (UTC)[reply]
Yes, the problem with the original metre as determined by the seconds pendulum was that it varied according to where the measurement was done. The fraction of the meridian was actually mis-calculated, so there is no point in re-measuring. The prototype bars were improvements, especially when temperature was specified, but even these changed over time. I think the current definition is based on the "best" historic measurements of the latest bar, but it would be interesting to re-measure. I'm sure someone must have done this. Dbfirs 18:55, 19 January 2011 (UTC)[reply]

Hexafluoroethane and fluorine

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Will hexafluoroethane react with fluorine? --84.61.131.41 (talk) 08:17, 16 January 2011 (UTC)[reply]

Possibly, under ultrviolet light, the hexafluoroethane may dissociate into trifluoromethyl radicals which would react with fluorine to produce tetrafluoromethane. --Plasmic Physics (talk) 08:26, 16 January 2011 (UTC)[reply]

Diesel passenger cars without diesel particulate filters

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Why aren't there any diesel passenger cars without diesel particulate filters anymore available in the European Union? --84.61.131.41 (talk) 08:55, 16 January 2011 (UTC)[reply]

Have you read the obvious article Diesel particulate filter which gives the IMHO fairly obvious answer:
While particulate emissions from diesel engines were first regulated in the United States, similar regulations have also been adopted by the European Union, most Asian countries, and the rest of North and South America World List of Standards.
While no jurisdiction has made filters mandatory, the increasingly stringent emissions regulations that engine manufactures must meet mean that eventually all on-road diesel engines will be fitted with them. In the European Union, filters are expected to be necessary to meet Euro.VI heavy truck engine emissions regulations currently under discussion and planned for the 2012-2013 time frame. PSA Peugeot Citroën was the first company to make them standard fit on passenger cars in 2000, in anticipation of the future Euro V regulations.
Nil Einne (talk) 13:12, 16 January 2011 (UTC)[reply]

What is density?

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Density is a function of mass and volume. Therefore, you can only measure density indirectly, having measured mass and volume beforehand.

Therefore, can we say that density is not a fundamental property of the universe? Can we say that it is a derived property?--Leptictidium (mt) 10:05, 16 January 2011 (UTC)[reply]

How do you define your term "fundamental property" ? Density can be measured directly - for example, a hydrometer measures the density of lqiuids. You could equally well argue that mass is derived from density and volume, or that volume is derived from density and mass. Gandalf61 (talk) 11:37, 16 January 2011 (UTC)[reply]
"Fundamental property of the Universe" is a very problematic term. Somewhat simpler is "fundamental property of a theory" - here it depends on what sort of system you look at (particle or fluid) and at what level you look at it (e.g. continuum approximation of a fluid vs. something like kinetic theory). If you talk about continuous systems, then density is the fundamental property and mass is derived by integrating over finite volumes. If you talk about particles, (particle) mass is fundamental and density is derived by averaging over finite volumes. --Wrongfilter (talk) 12:01, 16 January 2011 (UTC)[reply]

Is pH applicable to non-water solutions?

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e.g., can pure ethanol have a pH? 75.4.194.121 (talk) 10:16, 16 January 2011 (UTC)[reply]

Not really, no. The way that pH is defined will only really work in dilute aqueous solutions. However you can (and people do) define an acidity function for non-aqueous solvents: these are practical the same as pH if the solvent is water, but are easier to extend to other solvent systems. Physchim62 (talk) 13:13, 16 January 2011 (UTC)[reply]
Would it be proper to edit the lead of the pH article to say, "pH is a measure of the acidity or basicity of an aqueous solution? I would do it myself but I don't trust my knowledge of the matter. 75.4.194.121 (talk) 22:17, 16 January 2011 (UTC)[reply]
scratch that -- I will be bold and make the change, and if you disagree you can revert it :) 75.4.194.121 (talk) 22:18, 16 January 2011 (UTC)[reply]

Crap, I got reverted in less than ten minutes. OK, if the change is worth making, I leave the work to one of you. 75.4.194.121 (talk) 22:26, 16 January 2011 (UTC)[reply]

I think pH should apply to any solution that can produce hydrogen ions, such as sulfuric acid. After all, if you have a very concentrated solution of sulfuric acid (clouds of Venus, say), with maybe a little trace of water left, when do you say "well that tears it, no more pH measurements!" Wnt (talk) 05:53, 19 January 2011 (UTC)[reply]
(note, however, that p[H] is not precisely the same as pH, and the distinction is one of the more curious and obscure of chemistry's dark arts. Wnt (talk) 05:55, 19 January 2011 (UTC)[reply]

German dioxine contamination: how long does food take to pass through retail system?

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According to http://www.bbc.co.uk/news/world-europe-12200619 both German chickens and pigs have consumed dioxin contaminated animal feed.

How long does it take chicken, eggs, and pork to pass through the retail food system. If it is slaughtered or laid on Day 1, how long is it before it is all sold or disposed of? I imagine the figure is different for fresh, frozen, tinned, or otherwise processed food.

I know the dioxin is said to be very diluted by now, but the thought of food being possibly contaminated takes all the pleasure away from eating it. My local supermarket has large quantities of cakes being sold at reduced prices, so other people feel the same. Thanks 92.29.122.203 (talk) 10:45, 16 January 2011 (UTC)[reply]

For fresh egg this might be less than 10 days, but for deep frozen pork products or sausages this might be weeks or even months.--Stone (talk) 13:43, 16 January 2011 (UTC)[reply]

Image identification

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Can someone identify these? Doesn't have to be species name, anything will help :)--Abhishek Jacob (talk) 12:58, 16 January 2011 (UTC)[reply]

As I'm pretty good with herpetology I felt I should identify the second picture at least :)
Looked vaguely like a Wood Frog at first, but I checked the file description and it was found in India. Given our coverage of Indian frogs is all stubs, the exact species probably won't be clear, but I think I can narrow down the classification to a member of the Ranidae. Crimsonraptor | (Contact me) Dumpster dive if you must 13:24, 16 January 2011 (UTC)[reply]
You might have more luck asking at Mushroom Observer and whatsthatbug, I don't think there is an equivalent where you'll find herpetologists though unfortunately. SmartSE (talk) 14:09, 16 January 2011 (UTC)[reply]
FYI, they were all taken from my home in Palakkad, Kerala, India. Coordinates: 10°45'39"N 76°38'25"E --Abhishek Jacob (talk) 14:17, 16 January 2011 (UTC)[reply]
Ask a Biologist (the UK one, not the US one) might have some answers, as they have a number of reptile experts. Or, if there's a local zoo/wildlife place you might be able to ask them. Crimsonraptor | (Contact me) Dumpster dive if you must 17:47, 16 January 2011 (UTC)[reply]

Bug Guide is very good resource but unfortunately is strictly North American arthropods. The 5-part antennae suggest hemipteran to me. The lack of wings suggests either a nymph or a ground beetle. Do you have a ventral image that displays the mouthparts? That would clear things up significantly. -Craig Pemberton 06:38, 17 January 2011 (UTC)[reply]

Also, the time of year and a cross section of the mushrooms or at least an image of the gill attachment would be very helpful. -Craig Pemberton 06:44, 17 January 2011 (UTC)[reply]

can human beings destroy life on earth?

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I'm asking this because I hear such sentences a lot.but I think it is somehow impossible, because:

1.There are many extremophile forms of life in places that are somehow less affected by human activities, such as some caves,black smokers in deep oceans,etc.

2. we rely on other living things to live on, so before destroying other living things, we will die out. ( and therefore we won't be able to kill other things.)

3.Life (although very slowly) is adapting itself to some human made environment(or materials).the nylon eating bacteria, for example.

4. there is probably no one who intends to destroy life,I mean some people may not care about the environment,but they don't realy mean to destroy it.

I'm not neglecting the importance of taking care of the environment.but can human beingd destroy life in a way it can't recover itself again?--Sina-chemo (talk) 16:49, 16 January 2011 (UTC)[reply]

A cobalt bomb could theoretically move cockroaches to the top of the food chain. More info about these devices on Doomsday device. Do you ever suffer from happy thoughts?--Aspro (talk) 17:07, 16 January 2011 (UTC)[reply]
Right now? No. We can't. We can't even kill all the humans, much less all life. A Cobalt bomb would not work either - we don't have enough nukes to kill all humans using it. Ariel. (talk) 21:12, 16 January 2011 (UTC)[reply]
Though one should not neglect the fact that we could potentially make the world poisonous enough with such weapons as to make life pretty dang miserable, and to kill off a good percentage of people. Even a nuclear exchange between, say, just India and Pakistan would have global climate consequences, which would have severe global health consequences as well. That being said, it would be very hard to reliably kill off all humans with nuclear weapons. If I were a mad scientist, I would invest in infectious diseases, instead. You'd miss people who lived out of contact with other people, but you'd be able to kill off a massive number of people very quickly if you had the right pathogen. --Mr.98 (talk) 21:30, 16 January 2011 (UTC)[reply]
The key word, of course, is reliably. If you're a Vogon captain with a contract to destroy all human life on Earth, you can't be sure of fulfilling your obligations by setting off enough weapons to induce nuclear winter, partly because we don't know whether the models that predict it are correct. If you actually prefer not to extinguish humanity, it's probably an experiment you don't want to run. Even if, like me, you think the precautionary principle in general is a bunch of technophobe hogwash. --Trovatore (talk) 23:32, 16 January 2011 (UTC)[reply]
Although if you are a Vogon captain, you may have better models Nil Einne (talk) 00:03, 17 January 2011 (UTC)[reply]
or in any case better poetry --Trovatore (talk) 00:04, 17 January 2011 (UTC) [reply]
well, unless vaporized a la the Vogons, something is sure to survive no matter what we do to the planet, and whatever it is that is left will someday evolve to intelligent life, again. PЄTЄRS J VЄСRUМВАTALK 00:34, 17 January 2011 (UTC)[reply]
I expect nuclear winter would kill more people than a bioweapon. You can't find a cure for nuclear winter, you just have to try and endure it for a few years (which could only be done by reasonably small groups because of the need for food). It is possible someone would find a cure for a bioweapon. It would be very difficult to guarantee that everyone would die, though. It would be impossible to get even close to killing all life on Earth. --Tango (talk) 23:40, 16 January 2011 (UTC)[reply]

It shouldn't be too difficult to change the orbit of Ceres (dwarf planet) and let it collide with Earth. Count Iblis (talk) 23:19, 16 January 2011 (UTC)[reply]

It would be very difficult. The energy involved would be enormous. Even moving small asteroids enough to move them out of a collision course with Earth is at the limit of our abilities (I expect we could do it if we really needed to and spotted the asteroid long enough in advance). --Tango (talk) 23:40, 16 January 2011 (UTC)[reply]
I've read that an economical way to do this is by changing the course of smaller asteroids and let them move past or collide with the larger asteroid we're interested in. An extreme example was given in this paper (section 4.4). Here it is suggested that we should use asteroids to change the Earth's orbit to prevent it from being destroyed by the Sun. Count Iblis (talk) 23:55, 16 January 2011 (UTC)[reply]
Alternatively, we could use the method suggested in that paper to move the Earth closer to the Sun... Count Iblis (talk) 00:09, 17 January 2011 (UTC)[reply]
Runaway climate change offers a plausible option, though most authorities do not believe that it could reach Venus-like proportions. The key threshold of boiling the oceans and create water vapor as a greenhouse gas seems out of reach. Still, the Sun gets hotter all the time, and if we manage to put enough methane into the atmosphere abruptly (clathrate gun hypothesis, arctic methane release) we might find out for sure. Wnt (talk) 01:51, 17 January 2011 (UTC)[reply]

but we don't realy want to destroy the planet or life.Can i conclude that if we continue polluting the planet as we are now, Life won't disapear?I'm not saying that we are allowed to pollute the environment. but if we do, we will die out first, so there would be no one to continue polluting it. — Preceding unsigned comment added by Sina-chemo (talkcontribs) 05:54, 17 January 2011 (UTC)[reply]

No: Even if we pollute far more than we are now, life won't disappear, the more fragile creatures will die out, but the rest will adapt. And even if we pollute to the point that everything dies, we would be the last to die, since unlike them we can do something about it. And finally it won't happen, the peak for pollution on earth was years ago, it's only getting better from here. China and India have lots of work to do, but I expect an environmental revolution within the next generation. Ariel. (talk) 06:15, 17 January 2011 (UTC)[reply]
We would certainly not be the last to die - plenty of bacteria will survive longer. If nowhere else, then at least inside the corpse of the last human, but likely much longer in places like the deep hot biosphere. The claim that "pollution" is somewhat lower now than in its heyday is not implausible, but that depends very much on what you count as pollution. CO2 emissions are at (or near, due to minor fluctuations) an all-time high, and CO2 accumulates in the atmosphere, so CO2 levels are rising fast. --Stephan Schulz (talk) 08:10, 17 January 2011 (UTC)[reply]
We probably couldn't destroy all life but we could reduce it to extremophiles. If we put all our efforts into it, we could set up a sort of curtain between the Earth and Sun, blocking enough light to rather quickly produce a transition to a Snowball Earth state, in which the oceans would freeze over and the entire surface would resemble Antarctica. There would still probably be some life at deep-ocean vents, but that would probably be all. Looie496 (talk) 18:42, 17 January 2011 (UTC)[reply]
A true Venus-style runaway greenhouse would be a thorough steam-cleaning all the way down to the core. There would be no liquid water left, though it would take a while. Wnt (talk) 22:10, 17 January 2011 (UTC)[reply]
I am reminded of this website discussing ways to really and truly destroy the Earth, which would presumably destroy all life on it as well. Most of their methods are ridiculously impractical and completely unachievable, but it is an interesting idea to think about. For example, if you can find or make a black hole, that would potentially do a very good job of exterminating the Earth (and everything on it). Dragons flight (talk) 20:51, 17 January 2011 (UTC)[reply]
Even the worst mass extinctions that wiped out 90% of all species failed to destroy all life on Earth, especially prokaryotes which have more rapid microbe evolution capabilities, and life always recovers due to filling of niches by new species. However, even without "intending" to cause a mass extinction, the law of unintended consequences means that we may accomplish just that through environmental degradation (including business as usual scenarios). Even a Snowball Earth could have larger organisms that survive in warmer ocean pockets, likewise at Europa (purely hypothetical). The food web is an "interconnected system", meaning that the loss of one species could affect many others. A related fact is that mycorrhizae are threatened by environmental change, for example through plant biodiversity effects. Rapid global warming from anthropogenic causes could create novel biomes, leaving non-migratory species in new climates and vegetation patterns that in turn are slow to respond to rapid directional selection. Although the aforementioned mechanisms would not destroy all life, positive feedbacks could continue the trend well past the end of the initial "forcing", and ozone depletion would irridiate most complex life. Even an impact event that causes the crust to break and bring magma to the surface, as the Chicxulub crater impactor likely did would still not be enough to cause the extinction of all lifeforms. Some technological mishap such as grey goo might do it, however. ~AH1(TCU) 01:58, 18 January 2011 (UTC)[reply]
Genesplicing has the potential to wipe out any particular plant or animal by presenting some lethal pest. Mad scientists are at the early stages of splicing bits of DNA into the genomes of mice, goats, insects and bacteria. The intended consequences have included making goats produce spider silk precursors in their milk, making mice glow in the dark or have human brain cells, and encoding computerized information in the genomes of E coli. One day, such an experiment might make a germ vastly more lethal than its natural form, or might cause a pest such as venomous insects to become a lethal scourge. An agricultural pest might devastate staple food crops and cause widespread starvation. Military weaponization of disease organisms has horrible potential to wipe out humanity. Just as a billion monkeys might type out "Hamlet," a thousand biolabs tinkering with genomes might unleash selfperpetuationg and unstoppable havoc, with greater damage potential than one superbomb going off. Its easy to envision a bioexperiment going wrong in a way which might leave some lifeforms on Earth, but without vertebrates. I have seen something as simple as mosquitoes with naturally occurring West Nile destroy all birds in a large area. Fortunately the mosquitoes had limits to their range. Edison (talk) 17:41, 18 January 2011 (UTC)[reply]
This would seem to be true, not just for human experimenters but also the febrile imagination of Nature, but yet... so many of the great mass extinctions are explained not by disease but by some crude physical factor, that it makes me doubt the power of disease to finish such a task. Most of the genome is known to take a day off in someone or another - even those which appear "lethal" in mutants have only been proven to be lethal in one particular strain (or two, or three) under certain conditions. You can have a very nasty war indeed with bioweapons, but someone will probably live to write the history. Wnt (talk) 05:59, 19 January 2011 (UTC)[reply]

Massive spring

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If a spring with mass is pulled on both ends with different forces, by how much will the spring extend? 74.15.137.130 (talk) 16:50, 16 January 2011 (UTC)[reply]

It doesn't matter if the force is applied at both ends, or just the one end. Simply add (strictly it's subtract, due to opposite directions) the forces, and apply Hooke's law, if the spring is purely elastic. CS Miller (talk) 17:35, 16 January 2011 (UTC)[reply]
The force has to be the same size in both directions; otherwise, due to Newton's second law, the entire spring will start moving in the direction of the greatest force. One certainly shouldn't add or subtract the measurements from the two ends; at equilibrium it's the same force just measured at different places. And it only counts once in Hooke's law. –Henning Makholm (talk) 17:58, 16 January 2011 (UTC)[reply]
I know it would accelerate, but it would also stretch. How much would it stretch by? Suppose the spring was 1kg, with an outward force of 10N on one end and an outward force of 5N on the other. It would accelerate by 5m/s^2. If I understood Csmiller, it would stretch by 15N/k, right? If so, why? 74.15.137.130 (talk) 19:29, 16 January 2011 (UTC)[reply]
No, the situation is slightly more complex than that. Initially, one end of the spring will accelerate faster than the other until "accelerating equilibrium" (not a standard term) is reached. To determine both acceleration and extension, you need to be given the mass and the spring constant (often denoted λ in Hooke's law). Are you wanting to find both the extension and the acceleration? If we assume that the spring is uniform, then, once the two ends are accelerating at the same rate and oscillations have been damped out, the tension and thus the proportional extension (stretch) will increase linearly from the 5N end to the 10N end, but you can just use the average tension (7.5N) in Hookes law to give the total extension. Dbfirs 20:48, 16 January 2011 (UTC)[reply]
Thanks, but I have a few more questions (sorry!). First, is there a way to derive the average tension rule? Secondly, my textbook said that if a massive spring attached to a wall is stretched by a force F, the extension x will equal F/k. If I've understood things, this would only be true if the spring were static, right? Finally, if the same spring (attached to the wall) were connected to a mass, what would the total potential energy be? My textbook said that it's still (1/2)kx^2, but why would that be? I would think it would be (1/2)k(x/2)^2, because x/2 is the change in distance between the center of masses of the spring and the mass. It also assumed that the mass of the spring << the mass of the mass...would a more general case change the answer? Thanks! 74.15.137.130 (talk) 21:15, 16 January 2011 (UTC)[reply]
It should be easy to derive the "average tension rule" by considering forces on a small element δy of the spring and integrating, but I haven't thought through the exact method. Yes, the spring would be static if it is attached to a static wall and the free end is not moving. To answer your last question, you need to distinguish between potential energy stored in the spring and gravitational potential energy. It is not clear from your question exactly how the mass is attached, but the energy stored in the spring depends only on the extension, not on any attached masses. Dbfirs 21:56, 16 January 2011 (UTC)[reply]
By my second question, I meant that the extension was instantaneously x...would this only equal F/k if the spring were static (ie the end were not moving)? As for the last question, I'm sorry for the ambiguity but I had meant a spring-mass system that was horizontal (so no gravitational PE). To clarify my question...if a mass is subjected to a Hooke's Law force, its potential energy would be -∫-kxdx = (1/2)kx^2. But that only takes into account the energy of the moving mass. For a two-body system, the total potential energy would be -∫F(r)dr, where dr is the distance between the center of masses of the two bodies. So, shouldn't the potential energy be (1/2)k(x/2)^2? (x/2 being the distance between the COMs in terms of the extension x)74.15.137.130 (talk) 22:24, 16 January 2011 (UTC)[reply]
To solve that problem imagine that your spring is made of N small springs attached in series each one with a mass m_i = m/N and a spring constant k_i = kN. If the forces applied at either end of the whole spring are F_0 and F_N, and we define F_i as the force between the spring i and the spring (i+1) than we have F_i = F_0 + (i/N)(F_N-F_0) because that way all little springs will have the same acceleration. For a large N (-> infinity) The forces F_(i-1) and F_i acting on either and of the spring i will be almost the same and then it makes sense to use their average as an expression for the tension T_i on that spring: T_i = (1/2)(F_(i-1)+F_i) = F_0 + ((i-1) + i)/2N) (F_N - F_0) = F_i = F_0 + ((i-1/2)/N)(F_N-F_0). The spring i will stretch by an amount x_i given by x_i = T_i/k_i = 1/(kN) [F_0 + ((i-1/2)/N)(F_N-F_0)] = F_0(N-i+1/2)/(kN^2) + F_N(i-1/2)/(kN^2) and the total stretch x is given by x = SUM_(i=1)^N x_i = F_0/kN^2 SUM_(i=1)^N (N-i+1/2) + F_N/kN^2 SUM_(i=1)^N (i-1/2) = F_0/kN^2 (N^2 - N/2(N+1) + N/2) + F_N/kN^2 (N/2(N+1) - N/2) = k/2(F_0 + F_N). Conclusion: Use the average force 1/2(F_0 + F_N). Dauto (talk) 23:06, 16 January 2011 (UTC)[reply]
Nice! Can you also illuminate the potential energy issue (or let me know if I'm not explaining myself clearly)? 74.15.137.130 (talk) 00:26, 17 January 2011 (UTC)[reply]
I'm not clear what type of potential energy you are considering. Moving masses have kinetic energy, not potential (other than gravitational), and the spring potential energy is stored in the spring, not in the "mass". If a mass is attached to the spring, it is usually better to consider two separate springs. (My thanks to Dauto for setting out clearly what I was half-thinking.) You should be aware that the formula for energy stored in a spring is derived assuming that the tension and extension are uniform throughout the length. It will need to be modified (along the lines of Dauto's method) for an accelerating spring when the extension per unit length is not uniform. Dbfirs 10:30, 17 January 2011 (UTC)[reply]
I get V = [3F_0*F_N + (F_N-F_0)^2]/(6k). Dauto (talk) 20:29, 17 January 2011 (UTC)[reply]
I'm sorry for being persistent, but how did you get this expression? And why, in your previous derivation, does "it make sense to use the average force as the tension" for a small spring? 74.15.137.130 (talk) 05:41, 18 January 2011 (UTC)[reply]
When N -> infinity the difference between the two forces acting on either end of the microscopic string element becomes negligible F_i - F_(i-1) = (F_N - F_0)/N -> zero. So you can use either one of them or their average and it will give you the same results. My choice makes the algebra slightly easier and that's why I chose it. I don't have time right now to post my solution for the potential energy. I will try to come back to it later. Dauto (talk) 16:05, 18 January 2011 (UTC)[reply]

Makara Jyothi star

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Hi. Using Yoursky, I tried to identify the Makara Jyothi star that would have been seen on January 14 at Sabarimala. Since the "star", if it is a star, sets just after sunset, the actual star should set around that time (prior to 8 pm local time, or 14:30 UTC). The only two candidate stars I found that were bright enough and set in the general direction of west were Altair and Deneb. Since Altair set very close to sunset, and could only be seen in bright twillight, the star would probably be Deneb. This star sets around the same time that the stampede occurred, meaning it would likely disappear from the local horizon just before the disastrous incident, since the area is quite mountanous, and the star would appear to graze the horizon. However, it would require a program such as Stellarium to verify the exact setting time from the exact location. Is Deneb a likely candidate for this star, often called a "celestial event" by pilgrims? Thanks. ~AH1(TCU) 16:59, 16 January 2011 (UTC)[reply]

Sirius is known as Makarajyothi. But that's not what the pilgrims want to see. A light appears three times on top of the nearby mountain, known as Makaravilakku, which is actually lit by government officials but believed to be divine by the pilgrims.--Abhishek Jacob (talk) 19:12, 16 January 2011 (UTC)[reply]

Cardiac pressure-volume loop

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Can someone explain to me how the cardiac pressure-pressure volume loop can work without violating Boyle's law? I'm just not sure to understand how, with all valves closed, the ventricle can increase/decrease pressure without changing its volume. This would be the 2 vertical sections of the chart.

Thanks in advance

--Senorpurple (talk) 19:35, 16 January 2011 (UTC)[reply]

Boyle's law applies to gases and other compressible fluids. Blood is incompressible so you can increase the pressure by squeezing without changing the volume. (Pedantic note: Nothing is actually incompressible, so the volume does actually change, but by such a small amount you can ignore it.) Ariel. (talk) 21:18, 16 January 2011 (UTC)[reply]

Evolution of poison dart frogs

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I was watching this video http://www.youtube.com/watch?v=vxjc8m6s2p0 where they talked about the evolution of poison dart frogs. My question is how did they evolve the brightly colored patterns? A few points come to mind.

1) The poison secreted depends on their diet, when their diet is taken away they do not produce poison.
2) The coloration is independent of their ability to secrete poison. In other words, being brightly colored doesn't mean it's able to secrete poison.
3) Presumably they evolved from normal colored frogs, green, brown, grey, etc. Early in its evolution there were probably normal colored frogs that also secreted poison. Why would there be a selective advantage to being brightly colored over being normal colored if they both secrete poison? Wouldn't the predators be dissuaded from eating frogs in general if they became sick and died from eating frogs? If there were both dull and bright colored frogs that secreted poison, the bright colored frogs would stand out more. My only hypothesis is that the dull colored frogs would be harder to identify as a danger if it blends in more to its surroundings, but wouldn't that also be an advantage? Camouflage is a selective trait since it's harder to spot and eat.

So my question is what selected for brightly colored frogs, while dull colored frogs which ate the same diet and should secrete the same poison, were not selected for? ScienceApe (talk) 20:21, 16 January 2011 (UTC)[reply]

(edit conflict)My only guess would be that if all the frogs were the same dull green and whatnot, a poisonous frog could easily be mistaken for a non-poisonous frog. A predator, not knowing the difference, would perhaps try to eat the dart frog (possibly killing both in the process.) Simply, the colorations help the frogs' enemies distinguish it from other "safer" frogs to eat.[original research?] Like I said, though, merely a guess. :) Avicennasis @ 21:22, 11 Shevat 5771 / 16 January 2011 (UTC)
This is a textbook case of warning coloration, which can arise independently of the source of toxicity. Note that mimicry can result in similar warning colors on non-poisonous species. SemanticMantis (talk) 22:36, 16 January 2011 (UTC)[reply]
It must be all about the other frogs, the ones which are good to eat. If toxic frogs made up a large proportion of frogkind, I think you would be right - their coloration would serve no purpose; looking like a frog (an exaggerated frog?) would be sufficient. When their toxicity was just starting to evolve, though, the toxic ones were few. (Seems likely that they still are, among South American frogs in general.) Being toxic on its own wouldn't do them much good, since for predators the tiny risk of being poisoned might not outweigh the great benefit of eating mostly harmless frogs. Being toxic and slightly distinctive, though, to push the predators onto the other frogs, is an enormously successful strategy, which quickly leads to being brightly colored. (This might be accelerated by coevolution with the predator's instinct to dislike the brightly colored ones, if the predator has such an instinct.) 213.122.1.222 (talk) 00:02, 17 January 2011 (UTC)[reply]
Bear in mind that a frog which is non-poisonous due to its diet can still have a winning strategy as a Batesian mimic of its poisonous comrades. Since a predator eats many frogs and is not allowed any mistakes, I imagine that even a few poisonous members of the species cast a very long shadow. Wnt (talk) 01:36, 17 January 2011 (UTC)[reply]
Based on natural selection and predators that preyed on ancestral species of poison dart frogs, the species that did have bright colouration AND poison likely had a survival advantage over those that did not and were subsequently eaten, while camoflage and warning colours often co-compete to improve survival. Mimicry of existing poisonous species such as by the Viceroy butterfly also provides a similar advantage. ~AH1(TCU) 01:34, 18 January 2011 (UTC)[reply]

Medical question

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Is my hair more likely to fall of if I wear a baseball hat? A lot of people say so, but to me it doesnt make a lot of sense... --92.244.158.89 (talk) 21:21, 16 January 2011 (UTC)[reply]

I think some hairs will come off due to the rubbing of the hat on your scalp, dislodging some hairs. However, I doubt it would really be anything significant. --T H F S W (T · C · E) 22:21, 16 January 2011 (UTC)[reply]
From WP's article Baldness:
"Tight hats cause baldness."
  • While this may be a myth, hats do cause hair breakage and, to a lesser degree, split ends. Since hats are not washed as frequently as other clothing, they can also lead to scalp uncleanliness and possible Pityrosporum ovale contamination in men with naturally oily scalps. Some scalp infections, if left untreated, can cause hair loss. Bielle (talk) 22:28, 16 January 2011 (UTC)[reply]
I quite often wear a baseball cap when I'm out in the sun, but that's because most of my hair has already fallen out. HiLo48 (talk) 23:26, 16 January 2011 (UTC)[reply]
The quote from the article sounds suspiciously like a bit of original research at the Wikipedia end. It seems to imply, for example, that people with dandruff are more likely to develop baldness.
Now as for how hats could cause baldness, I should note PMID 8628793, which finds "a relative microvascular insufficiency to regions of the scalp that lose hair in male pattern baldness". Since this concerns blood flow in the scalp, which might be interfered with by a tight hat, I think the old wives may win the science award on this one. Wnt (talk) 01:33, 17 January 2011 (UTC)[reply]
If it's this hat, you might be tearing your hair out. Clarityfiend (talk) 01:54, 17 January 2011 (UTC)[reply]

Feynman diagram art help

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I had an idea in a dream Friday to make a Feynman diagram illustration of Keats's Ode to a Grecian Urn's most famous couplet. The preliminary result (which turned out amazingly better than I expected) is a weak interaction of bottom-down and top-down-up[2]. I'm wondering if someone might be willing to double-check the physics (my particle knowledge is terrible), and also maybe have a suggestion for a strong-mediated interaction that I can use as well, just because the coily gluon line looks neat. The only conditions are: a bottom quark becomes a top quark (Beauty is Truth), and the end products include a top-bottom meson (Truth Beauty). SamuelRiv (talk) 23:18, 16 January 2011 (UTC)[reply]

As far as the physics goes, unfortunately the baryon dut doesn't really exist because the quark t is too unstable and decays before it has enough time to form any bound states, and off course the bottom quark wouldn't back-decay into the top quark. But clearly those are minor points since you are not trying to describe a true physical process. Dauto (talk) 03:54, 17 January 2011 (UTC)[reply]