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April 8

Bread slice missing middle

I was nearing the end of a bag of sliced bread (branded, factory-made, bought at a major supermarket), and found that the second last piece (the piece before the end) consisted only of three edges edges (i.e. it was missing the middle and one edge) - more than just the crust, but most of the middle was gone. It looked torn, definitely not cut by a blade. The bag was closed with the original plastic clip, and there were no holes in the bag. I can't guarantee that the clip 100% securely closed the bag, but as far as I remember it was closed. The previous day I had taken two slices that sat above the odd slice in question, and they were fine. I think I would have noticed if the next slice was odd, but can't be sure. The last piece in the bag (the end) was intact.

What could have caused this? Some possibilities I thought of are: (i) at some point previously I had left the bag open and a mouse had gotten in and out, but then it would be odd for it to dig down to the second piece before last, eat it, and not disturb any of the others; (ii) there was some manufacturing error and that piece came like this; (iii) the bag was not securely closed the previous day, and a very nimble mouse or something smaller had gotten in during the night and eaten the bread; (iv) bread eating mould of some kind? The bread was a few days past its "best before" date. --PalaceGuard008 (Talk) 08:55, 8 April 2016 (UTC)[reply]

Possibly a damaged blade in the slicing machine. 217.44.50.87 (talk) 08:57, 8 April 2016 (UTC)[reply]

(ii) is definitely your answer. The bread could have been torn while slicing for some reason, or it could have been baked that way I suppose (an air bubble in the dough). shoy (reactions) 13:51, 8 April 2016 (UTC)[reply]

Perhaps the original Loaf had an oversize void due to uneven Leavening caused by inadequate mixing of the Bread dough. Mouse infiltration leaves tooth marks, crumbs or stools. AllBestFaith (talk) 13:54, 8 April 2016 (UTC)[reply]

I had a squirrel come down my chimney and attack a loaf of bread. It was fascinating what he did with it. He formed it into mouth sized balls, presumably planning to transport it back out by mouth. Then I showed up and ruined his plans by evicting him from my house and getting a chimney cap. StuRat (talk) 19:18, 8 April 2016 (UTC)[reply]

Thanks all, that must be it (faulty manufacturing, not squirrel - though that's fascinating as well)! --PalaceGuard008 (Talk) 09:11, 11 April 2016 (UTC)[reply]

Have we completely ruled out the possibility that Squirrel Girl was involved? :) --Guy Macon (talk) 12:24, 11 April 2016 (UTC)[reply]

Resolved

"Engagement length"

What does the "engagement length" indicate in ISO 965? There are many standard screws and nuts that go above and below this range, so what happens when you use such a combination? Let's say I try to seat a M2 screw 4 mm into a threaded hole, exceeding the "engagement length" range of 1 to 3 mm， what happens then? Will the thread strip? Or it is likely to get stuck or something? Johnson&Johnson&Son (talk) 10:25, 8 April 2016 (UTC)[reply]

Google "engagement length of bolt" and take your pick of the explanations on offer e.g. [1] "One of the main guidelines in designing fastener joints is that the length of thread engagement should be long enough so that the bolt will break rather than strip the threads in the tapped hole". Richerman (talk) 10:47, 8 April 2016 (UTC)[reply]

ISO 965 is used for pass/fail Acceptance testing of ISO threaded fasteners so it is desirable to set both minimum and maximum values for the engagement length that can be difficult to establish exactly but is critical for testing breaking strength. However the quoted maximum engagement length does not bear on usage and nothing bad happens if it is exceeded, as routinely happens with the long nut illustrated. AllBestFaith (talk) 13:37, 8 April 2016 (UTC)[reply]

Far as i know the minimum engagement length is defined as factor (usually 1.25x) of the screw diameter. Im irritated by this too, given that the list seems wrong in its "from"-row for minimum Engagement length. --Kharon (talk) 05:39, 9 April 2016 (UTC)[reply]

Psychology study that involves perception of pain of another person

I remember that there was this psychology study that involved a person tapping on something. If the person heard another person in the room answering the question wrong, then the other person seemed to be shocked literally. What was this study? 140.254.77.187 (talk) 17:36, 8 April 2016 (UTC)[reply]

Perhaps the Milgram experiment. Sean.hoyland - talk 17:39, 8 April 2016 (UTC)[reply]

The Milgram experiment.--TMCk (talk) 17:41, 8 April 2016 (UTC)[reply]

Damn, you both got it right, now I don't get to shock you. Aw, what the heck, BZZZZT ! BZZZZT ! StuRat (talk) 19:22, 8 April 2016 (UTC) [reply]

Arrgh, now I am dying! --69.159.61.172 (talk) 05:23, 9 April 2016 (UTC)[reply]

Engineering Physics

Light from a broad source source is obliquely incident on a thin transparent plate. Find the expression for the effective path difference between part of a ray reflected from upper surface and part that suffers reflection over surface internally . — Preceding unsigned comment added by 1.39.15.220 (talk) 19:28, 8 April 2016 (UTC)[reply]

Okay, I did it. Now you do it...seems like it's your homework not mine. DMacks (talk) 20:27, 8 April 2016 (UTC)[reply]

Depends on the thickness of the plate and the angle of incidence I would say. Any help?--178.101.224.162 (talk) 01:57, 9 April 2016 (UTC)[reply]

Just draw the path of the internally reflected ray. It forms two symmetrical right triangles. You can calculate the angle and hence the length of the hypoteneuses (don't forget to use Snell's law to take refraction into account - within the glass, the angle of the ray is different from outside the glass). You know the length of one side. The rest is sine and cosine applied repeatedly, I think. Wnt (talk) 16:57, 9 April 2016 (UTC)[reply]

On second consideration I realize I don't know... it's a matter of philosophy. :) The thing is, not only does the refracted light bend differently; it also moves more slowly than the light in vacuum. For most interesting measures of the path difference, like the spreading of a pulse laser when it bounces off the thing, you're measuring the time of transit, not the distance in glass + distance outside of glass. But that would be wrong since it's not truly the length... or is it? As I understand (the problem being, I don't), there are grand unified theories, M theory etc. that present electric fields as being a curvature of space, and the light is not really slowed, it just appears slowed as it wends and weaves through the fields of atoms. And so its squiggly path is exactly what the time suggests, and so the "length" of the path in glass should be multiplied by this time factor. But is curvature in space actually a part of distance? Wnt (talk) 17:10, 9 April 2016 (UTC)[reply]

Do hens ever reach a point where they are passed their egg-laying days?

Or do they lay eggs as long as they live? 140.254.70.33 (talk) 21:35, 8 April 2016 (UTC)[reply]

Are you talking about commercial egg laying birds or non-commercial birds? Commercial egg layers in the UK are usually kept for only 62 weeks as their egg production then starts to decrease. In some countries, including the US, the birds are force moulted (deprived of food and water for many days) which stimulates the hens to increase their egg laying again. This may happen several times. Let's not forget, these commercial birds lay approx 300 eggs per year whereas a non-commercial hen might lay only one clutch (e.g. 6) of eggs each year. DrChrissy ^(talk) 21:44, 8 April 2016 (UTC)[reply]

And in Britain, where there is a Royal Society for the Prevention of Cruelty to Animals, but merely a National Society for the Prevention of Cruelty to Children, there is the British Hen Welfare Trust, which enables members of the public to re-house hens past their laying days. "They are approximately 17 months old when we collect them from farms, and are off to slaughter because they are deemed no longer commercially viable as they may be laying fewer eggs." Carbon Caryatid (talk) 22:00, 8 April 2016 (UTC)[reply]

Yes, I should have mentioned that. When the commercial hens start to decrease their egg laying, the producer makes a financial decision to get rid of them. These are often called "spent hens". If people get spent hens, these often initially stop laying (probably because of the stress of the change of habitat) but very, very often start laying again at the usual pattern (they lay an egg each day for 7 days then miss a day). DrChrissy ^(talk) 22:12, 8 April 2016 (UTC)[reply]

"Spent hen"! Should I ever feel the need to insult a mother of eight, I am now armed. :) Wnt (talk) 17:14, 9 April 2016 (UTC)[reply]

Your mention of the National Society for the Prevention of Cruelty to Children baffles me. How is that relevant ? StuRat (talk) 15:28, 9 April 2016 (UTC) [reply]

I meant that, in some ways, British sensibilities afford more value to animals than to children, and therefore it isn't surprising that there is a specific organisation to enable well-wishers to be kind to chickens. Carbon Caryatid (talk) 16:13, 9 April 2016 (UTC)[reply]

So it being a "National" society versus a "Royal" one indicates it is a lower priority to the British public ? I can see why that might mean it's a lower priority to the monarchy, but that's not the same thing. StuRat (talk) 16:26, 9 April 2016 (UTC) [reply]

And of course if British sensibilities really did afford more value to animals than to children, we'd be slitting the throats of kiddiwinks who we decide are no longer useful to us. Henry ^Flower 16:35, 9 April 2016 (UTC)[reply]

See A Modest Proposal (yes, I do realize it's satire). StuRat (talk) 16:55, 9 April 2016 (UTC) [reply]

The RSPCA got its "Royal" title in 1840, thanks to Queen Victoria. The NSPCC also has a royal charter, even if it doesn't use the word in its name, and the Queen is in fact the patron of both charities. 217.44.50.87 (talk) 18:12, 9 April 2016 (UTC)[reply]

I'm not sure if anyone noticed but the Royal Society for the Prevention of Cruelty to Animals. It includes a very long list of controversies many of which seem to be mostly sourced to the Daily Mail. Nil Einne (talk) 19:30, 10 April 2016 (UTC)[reply]

Hens slow down as they age and they may stop egg production temporarily during the winter cold. They might also stop if they get sick, but otherwise no. My friend had a bird that lived and laid eggs for at least 9 years. This website mentions a hen that did so for 17 years. Modocc (talk) 20:19, 9 April 2016 (UTC)[reply]

April 9

Source of ammonia in life science lab?

What is a source of ammonia in a life science lab that might be used, for example, to clean an oven? --78.148.105.117 (talk) 13:13, 9 April 2016 (UTC)[reply]

I don't know why you need a life science lab. I can buy ammonia for cleaning at my local grocery store.--Jayron 32 13:23, 9 April 2016 (UTC)[reply]

It is no longer sold in stores in the UK. 192.41.128.212 (talk) 18:54, 9 April 2016 (UTC)[reply]

Apparently true, but you can still get it from Amazon. Tevildo (talk) 19:42, 9 April 2016 (UTC)[reply]

From what I read at [2], the UK actually loosened restrictions on ammonia in 2014, even while putting up roadblocks to other chemicals. (further info if you're curious) Of course, any kind of optimism seems out of place with the modern UK, but I think a citation needed is still in order. Wnt (talk) 23:43, 9 April 2016 (UTC)[reply]

Well, I had a quick look at the websites of various major UK retailers. Tesco, Sainsbury's, Homebase, Wickes and The Range don't list it at all - B&Q lists it, but show it as being out of stock in all their stores. It may be possible to obtain it in a UK store, but it's certainly not easy. Tevildo (talk) 09:14, 10 April 2016 (UTC)[reply]

Are you asking what lab activities might incidentally create ammonia fumes in that lab ? If so, the obvious answer is cleaning, specifically with glass cleaners, which often contain ammonia. It's also possible that some experiment might create ammonia. StuRat (talk) 15:25, 9 April 2016 (UTC)[reply]

Chemicals in laboratories are purchased from any number of sources, such as Sigma-Aldrich. --OuroborosCobra (talk) 15:42, 9 April 2016 (UTC)[reply]

This question is too unclear to answer. Ammonium hydroxide for most lab uses is highly concentrated and its fumes should NOT be inhaled. Using it to clean an oven would require substantial dilution in a well-ventilated area. Confusing the two could lead to severe injury, and I think it might also damage the oven (I don't know that). this source says 14-15 M for lab solution and 5% (NH3) for household; doing the calculation they suggest, I get 17 g/l is 1 M and so 50 g/l = about 3 M, or 5x diluted, but PLEASE DO NOT RELY ON THAT because neither source nor calculation is guaranteed here. Just buy your cleaner at the store, and do not fool with lab ammonia for anything but real chemistry. Wnt (talk) 16:42, 9 April 2016 (UTC)[reply]

What about solid ammonium bicarbonate? Ah, here we have solid ammonium chloride! Perfect! I heard I can just leave this to evaporate in the oven and do all the cleaning by magic 192.41.128.212 (talk) 18:54, 9 April 2016 (UTC)[reply]

- Lab ammonia? Chemicals do not care what their destination is, they just are. Any lab is going to either order the proper concentration and purity for their specific use, or be able to dilute down from a concentrated source. If I do not need HPLC grade purity for whatever work I am doing, I do not spend the money on HPLC grade. I don't know of anything called "lab <insert chemical name here>." Nor would I order already diluted HCl for almost anything (unless I really really need 1.000N HCl), I have a supply of glacial that I dilute down to whatever I need for either experimental or cleaning usage. --OuroborosCobra (talk) 17:51, 9 April 2016 (UTC)[reply]

Handedness and cheek concavity

I am interested in finding reports of evidence either for or against the hypothesis that left-handedness is correlated with concavity of the right cheek (more than the left cheek), because of the right cheek being supported by the right hand while the left hand is writing; and vice versa. Where can I find such reports?
—Wavelength (talk) 18:46, 9 April 2016 (UTC)[reply]

In the Eurasian Journal of Anthropology Analysis of facial directional asymmetry in extreme handed young males and females concluded that handedness may affect the level of directional facial asymmetry but only that this may be caused by the asymmetrical development of the cerebral hemispheres, with no speculations about writing habit itself as the cause. AllBestFaith (talk) 20:29, 10 April 2016 (UTC)[reply]

Thank you for your answer, although I am still interested in information about facial asymmetry caused by writing habits. (The EJA document introduced me to skewness and kurtosis.)

—Wavelength (talk) 18:46, 11 April 2016 (UTC)[reply]

US college admission through standardized tests alone

Does any US college admit students through standardized tests alone, either for undergraduate or graduate studies? The list of standardized tests in the United States confirms that there are plenty of options to choose from. Doesn't any college trust those test enough? --Scicurious (talk) 18:55, 9 April 2016 (UTC)[reply]

unless this has changed recently, Harvard requires neither high school diploma nor standardized tests for admissions...and there are instances of Harvard admitting unusual but gifted people who have neither...68.48.241.158 (talk) 19:39, 9 April 2016 (UTC)[reply]

You got the question the other way round. I don't want colleges that won't require any standardized test, but colleges that are happy with standardized test alone.--Scicurious (talk) 20:54, 9 April 2016 (UTC)[reply]

You would turn down Harvard? ←Baseball Bugs ^{What's up, Doc?} carrots→ 21:30, 9 April 2016 (UTC)[reply]

That's not the point. Where anyone applies or should apply, or turns down or whatever is a private issue.

The point is how much trust do colleges put on the reliability of those test. --Scicurious (talk) 22:15, 9 April 2016 (UTC)[reply]

You said, "I don't want colleges that won't require any standardized test." Therefore, you would reject Harvard. Unless you would make an exception. ←Baseball Bugs ^{What's up, Doc?} carrots→ 23:13, 9 April 2016 (UTC)[reply]

Come on, Bugs. He's not saying he would "reject" Harvard as a place to go, just that it's not an answer to the question. --Trovatore (talk) 23:16, 9 April 2016 (UTC)[reply]

Again, Bugs. That's not the point. Where anyone applies or should apply, or turns down or whatever is a private issue.

The point is how much trust do colleges put on the reliability of those test.Scicurious (talk) 23:39, 9 April 2016 (UTC)[reply]

College admissions is not just about academic abilities. They also want people who are a good fit, have social skills, are members of a minority (for being used as a token proving how liberal is the college), are children of alumni, have multiple interests, are famous, have stinking rich parents, or are good athletes.Llaanngg (talk) 22:42, 9 April 2016 (UTC)[reply]

Back when I was applying to schools (this was a long time ago), if memory serves, it was possible to qualify for admission to the University of California by test scores alone. That would be the SAT plus the "achievement tests"; I don't know what they call them now so I don't know where to link you to, but the SAT was considered an aptitude test whereas the "achievement tests" were about material you had mastered.

Oh, I think you still had to meet the content requirements — so many years of this, a different number of years of the other thing — but I believe the GPA requirement was waived.

There was an extra bonus. ~~At the time, you didn't apply to individual UC campuses. You applied to the UC as a whole~~, and listed the campuses you wanted to go to, but they were allowed to offer you admission but specify a campus different from the one you wanted. There were eight campuses at the time, not counting UC San Francisco which was more specialized, and I think two or three of them were competitive (probably at least Berkeley and UCLA) in the sense that there were students who preferred those schools, and who would be offered admission to UC, but not to the campus they wanted.

However, if you qualified by test scores alone, you were guaranteed admission to Berkeley if it was your choice. At least that's what the representative from UC told me. I never saw it written down anywhere. So Berkeley wound up being my "safety school"; I didn't get a good feel from the campus and didn't really want to go there, but it was good to have a backup. It has gotten much more competitive in intervening years. --Trovatore (talk) 23:05, 9 April 2016 (UTC)[reply]

Actually, it's filtering back to me that maybe you did apply to your first-choice campus directly, maybe listing your other choices in order of preference. The catch was that you were only allowed to apply to one campus. If you met the admission requirements, but your first-choice campus had more applicants meeting the requirements than they could accommodate, they would offer you admission to a different campus. So kind of the same in effect as what I said, but different in terms of where your application was physically sent and processed. --Trovatore (talk) 23:28, 9 April 2016 (UTC)[reply]

(This is really more of an RDH question but...) "Automatic admission" seems to be a big deal in Texas. Florida International University list GPA and test scores that can automatically admit an applicant, though it's unclear if you need one or both. Ditto for Louisiana State University. Google around some more for "automatic admissions" and you'll likely find a few other places. Do keep in mind that your state of residence is very, very important for any question pertaining to being admitted to or paying for a public university in the United States. Texas, at least, seems to only "automatically admit" students who are domiciled in Texas.

There are also likely to be some schools outside your state to which you could be admitted based on test scores alone, but for which you would wind up paying a lot more than you would going to an in-state public university (which, at least in most places, you have a better chance of being admitted to than as an out-of-state applicant elsewhere, even if your record is average-to-poor). This was actually quite similar to my situation when applying for undergrad admission: I had what I thought was a mediocre (homeschool) high school education, with notable highlights taking actual college courses, and wound up being admitted to two Public Ivys and the majority of the smaller liberal arts colleges I applied to. Like an idiot, I enrolled at the out-of-state Public Ivy and wound up paying triple what I could have paid in-state. If you're asking because you or someone you know is applying, you really owe it to yourself to give it a shot, think about finances, and not limit yourself based on your expectations. Every admissions department weighs the various criteria differently. Evan ^{(talk|contribs)} 03:41, 10 April 2016 (UTC)[reply]

The University of California used to have a legal mandate (maybe it still does, I'm not sure) to admit a certain top percentage of all California students. The practical result was that if you had a certain minimum SAT score, you would be guaranteed admission to at least one of the schools in the UC system. — Preceding unsigned comment added by 108.16.82.233 (talk) 11:34, 10 April 2016 (UTC)[reply]

The State of North Carolina will admit, to one of the schools of the University of North Carolina system, every North Carolina resident who has graduated an accredited North Carolina High School with a "college preperatory" diploma. Now, that doesn't mean all 17 of them would accept the applicant, merely that one is required to take them. --Jayron 32 22:56, 10 April 2016 (UTC)[reply]

That's an odd system, especially since the admissions departments of the 17 constituent universities are, in theory, completely separate entities. I wonder how "college preparatory diploma" is defined, also. Evan ^{(talk|contribs)} 23:40, 11 April 2016 (UTC)[reply]

April 10

Special relativity. Simultaneity 2.

https://en.wikipedia.org/w/index.php?title=Wikipedia:Reference_desk/Science&action=edit&oldid=714473425

Question

Remark

You said the hind clock, when it shows the numerical value

\theta _{0}^{[}

on its face, is at

t'=\theta _{0}^{[}

. It was not statement , but assumption as I'm not sure it's true (even for previous thread). I've taken it from previous thread (see Remark).

So, we shall heve next situation:

ε' frame			ε frame
time coordinate	clock reading		time coordinate	clock reading
$t'=0$	$\theta _{0,self}^{[}=-{\tfrac {L'}{2c}}$	$\theta _{0,self}^{]}=-{\tfrac {L'}{2c}}$	$t=0$	$\theta _{0}^{[}=\theta _{1}^{[}-\tau _{1}{\sqrt {...}}$ $=-\tau _{1}{\sqrt {1-u^{2}/c^{2}}}$	$\theta _{0}^{]}=\theta _{2}^{]}-\tau _{2}{\sqrt {...}}$ $=-\tau _{2}{\sqrt {1-u^{2}/c^{2}}}$
$t'=+{\tfrac {L'}{2c}}$	$\theta _{1,self}^{[}=\theta _{2,self}^{[}=0$	$\theta _{2,self}^{]}=\theta _{1,self}^{]}=0$	$t=\tau _{1}={\tfrac {L/2}{c+u}}$	$\theta _{1}^{[}=0$
			$t=\tau _{2}={\tfrac {L/2}{c-u}}$		$\theta _{2}^{]}=0$
$t'=\theta _{0}^{[}+{\tfrac {L'}{2c}}$	$\theta _{3,self}^{[}=\theta _{0}^{[}$	$\theta _{3,self}^{]}=\theta _{0}^{[}$
...
$t'=A$	$\theta _{(t'=A),self}^{[}=A-{\tfrac {L'}{2c}}$	$\theta _{(t'=A),self}^{]}=A-{\tfrac {L'}{2c}}$	$t=A$	$\theta _{(t=A)}^{[}=(A-\tau _{1}){\sqrt {..}}$	$\theta _{(t=A)}^{]}=(A-\tau _{2}){\sqrt {..}}$

Correct? Let's check.

Hind clock at start is situated in ε in point $(x,t)=(-L/2,0)$ and in ε' in point $(x',t')=(-L'/2,0)$ .

$(x',t')=({\tfrac {x-ut}{\sqrt {1-u^{2}/c^{2}}}},{\tfrac {t-xu/c^{2}}{\sqrt {1-u^{2}/c^{2}}}})$ ;

$(x,t)=({\tfrac {x'+ut'}{\sqrt {1-u^{2}/c^{2}}}},{\tfrac {t'+x'u/c^{2}}{\sqrt {1-u^{2}/c^{2}}}})$ .

What's next? What to substitute in and what should be compared with result?

https://en.wikipedia.org/w/index.php?title=Wikipedia:Reference_desk/Science&action=edit&oldid=713271421 Special relativity. Simultaneity.

In your problem, say the light is emitted at (0,0) (in either coordinate system). It reaches the walls at $(x',t')=(\pm L/2,L/2c)$ , or equivalently $(x,t)=({\tfrac {\pm Lc}{2(c\pm u)}},{\tfrac {L}{2(c\pm u)}})$ . Your first clock has the parametric equation (x,t)(τ) = (0,τ), which you can plug into the Lorentz transformation to find (x',t')(τ). To find the digital reading of that clock at a particular (x,t) or (x',t'), you just solve for τ. Your other three clocks have the parametric equations (x',t')(θ) = (0, θ) and (x',t')(θ^[]) = (±L/2, θ^[]), and you can plug those into the Lorentz transformation to find (x,t)(θ^whatever). Any question you have about what someone sees at a given (place,time) can be answered in this way. They are simple questions about coordinate geometry. -- BenRG (talk) 19:58, 29 March 2016 (UTC)[reply]

37.53.235.112 (talk) 12:02, 8 April 2016 (UTC)[reply]

OK. According BenRG's suggestions we have:

$(x';t')(A)=(-{\tfrac {L'}{2}};A-{\tfrac {L'}{2c}})$ ;

$(x,t)=({\tfrac {x'+ut'}{\sqrt {1-u^{2}/c^{2}}}},{\tfrac {t'+x'u/c^{2}}{\sqrt {1-u^{2}/c^{2}}}})$ ;

we will substitute x' and t' and should get next

$\color {red}{(x;t)(A)=(-{\tfrac {L}{2}}+uA;(A-\tau _{1}){\sqrt {1-u^{2}/c^{2}}})}$ .

After substitution:

$(x,t)(A)=({\tfrac {(-{\tfrac {L'}{2}})+u(A-{\tfrac {L'}{2c}})}{\sqrt {1-u^{2}/c^{2}}}},{\tfrac {(A-{\tfrac {L'}{2c}})+(-{\tfrac {L'}{2}})u/c^{2}}{\sqrt {1-u^{2}/c^{2}}}})$ ;

$(x,t)(A)=({\tfrac {uA}{\sqrt {1-u^{2}/c^{2}}}}-{\tfrac {L/2}{1-u/c}},{\tfrac {A}{\sqrt {1-u^{2}/c^{2}}}}-{\tfrac {L/2}{c+u}})$ .

Why result doesn't match with highlighted equation? 37.53.235.112 (talk) 18:02, 9 April 2016 (UTC)[reply]

There's a sign error in the first equation (below the box); it should be

(x';t')(A)=(-{\tfrac {L'}{2}};A+{\tfrac {L'}{2c}})

. With that change, the Lorentz transformation should get you

(x,t)(A)=(u\gamma A-c\tau _{1},\;\gamma A+\tau _{1})

, where I've introduced

\gamma =1/{\sqrt {1-u^{2}/c^{2}}}

and used your definition of

\tau _{1}

. This is close to the answer you got, and I think the difference is because of the sign error. You should be able to look at this answer and see that it makes sense. For example, if you plug in

A=0

, you get

(-c\tau _{1},\tau _{1})

, which is where the light reaches the clock. Also, the factor of

u

gives the clock the correct speed, and the factor of

\gamma

gives it the correct time dilation factor.

Your expected answer (in red) is wrong in several ways. Again, try plugging in

A=0

, think about whether the velocity of the clock (the slope of the line in spacetime) is

u

, and think about whether you should multiply or divide by the time dilation factor. -- BenRG (talk) 05:33, 10 April 2016 (UTC)[reply]

Is table ε' frame — ε frame correct? 37.53.235.112 (talk) 07:22, 10 April 2016 (UTC)[reply]

Yes, I think so. (Note that where you have

(A-\tau _{1}){\sqrt {..}}

in that table, it's a clock reading (correct), whereas in your prediction in red, it's a

t

coordinate (incorrect).) -- BenRG (talk) 07:30, 10 April 2016 (UTC)[reply]

How to check table ε' frame — ε frame through Lorentz transformations? Namely, is there correlation between left (ε') and rigth part (ε)? 37.53.235.112 (talk) 09:06, 10 April 2016 (UTC)[reply]

The Lorentz transformation will turn the

(x,t)

of a clock reading into the

(x',t')

of the same clock reading, or vice versa. In your table you don't have any

x

or

x'

, and most of the clock readings on the left aren't on the right and vice versa. You could add

x

and

x'

coordinates for the points where the clocks read

0

, then check that they match using the Lorentz transformation. But it makes more sense to derive a general equation for the

(x,t)

and

(x',t')

coordinates of the clocks as a function of their reading. You just did that for the hind clock (albeit with some errors). Then you can forget about the table, which just gives the values of those functions at certain points. -- BenRG (talk) 17:32, 10 April 2016 (UTC)[reply]

Help identify this plant, please

{ Moved from Miscellaneous Ref Desk }

A climber closely similar to Calycopteris floribunda but never seen to grow as big as that. The plant grows in Kerala, India and flowers during the heights of hard summer. Thrives in wild and don't seem to require watering at all for the flowering. Flowers are fragrant, though not intensely. What is it? — Preceding unsigned comment added by 117.253.195.109 (talk) 02:18, 10 April 2016 (UTC)[reply]

The hanging clusters of blossoms opening downward and the leaves resemble Hoya carnosa (familiar name: wax plant, porcelain flower). It's familiar to me as a shrub, possibly cultivated (i.e. not seen in the wild), in the arid subtropic climate zone. -- Deborahjay (talk) 07:34, 10 April 2016 (UTC)[reply]

Hoya carnosa is certainly is not by the photos I see on its page. The woody vine and leaves are very similar to Calycopteris floribunda but the flowers are not. --117.253.195.109 (talk) 08:20, 10 April 2016 (UTC)[reply]

To me this looks like Combretum indicum, or a closely related species of Quisqualis or Combretum. --Dr Dima (talk) 20:47, 11 April 2016 (UTC)[reply]

2016 Mercedes GLE 350d urea injection

My friend says that the 2016 Mercedes GLE 350d doesn't require urea fill-ups anymore unlike the previous version and most diesel cars on the market. Is this actual true?

Our article Mercedes-Benz_M-Class#Engines says "ML 350 BlueTEC 4MATIC", and BlueTEC is Mercedes' branding of urea injection, so just based on the model number I'm inclined to say he's wrong, but I can't find any sources to back me up. Johnson&Johnson&Son (talk) 10:42, 10 April 2016 (UTC)[reply]

Ok， so I found three sources[3][4][5] saying that there was a renaming of the model from "ML 350 BlueTEC 4MATIC" to "GLE 350d 4MATIC" back in 2015. Removing "BlueTEC" from the model name could count as circumstantial evidence.

I also found this Mercedes Australia page[6] which names 4 models that require urea injection and the GLE 350d 4MATIC isn't among the four. But this is circumstantial evidence again since it could simply be that the page is outdated or that they don't sell that model in Australia. Johnson&Johnson&Son (talk) 11:12, 10 April 2016 (UTC)[reply]

I added a redirect from urea injection to diesel exhaust fluid. StuRat (talk) 14:26, 10 April 2016 (UTC)[reply]

This which is about the 2016 Mercedes-Benz GLE 350d 4Matic Coupe [7] says "It involves automatically injecting a liquid called AdBlue into the exhaust stream at pre-determined intervals, breaking the harmful nitrogen oxides into harmless water and nitrogen." Nil Einne (talk) 16:04, 10 April 2016 (UTC)[reply]

This description of a very lightly used 2016 Diesel GLE 350d 4Matic AMG Line 5dr 9G-Tronic Auto [8] mentions "BlueTEC diesel emission control system including AdBlue reservoir". The current description doesn't mention that but has also been changed in other ways. This [9] says "The diesel GLE uses injections of AdBlue to reduce harmful emissions" although weirdly it doesn't talk about 350d despite being about the 2016 models. It does mention the GLE300d 4MATIC. This [10] isn't in English but despite not mentioning 2016, from the date and names appears to be about the 2016 models and you can see AdBlue is indicated even without machine translation. This [11] also isn't English but if you scroll right to the 4th of 6th photos at the top it shows an AdBlue tank and the headline machine translates to something like "AdBlue, the only solution for a less harmful diesel". This [12] also isn't in English but a machine translation says something like "Although already new signage does not know about it (déčko is replaced by the CDI, or with a BlueTEC) next to the conventional 93-litre fuel tank is also traditionally the urea tank (AdBlue liquid) that is used for injection into the exhaust system-most of the NOx into nitrogen and water changes". Nil Einne (talk) 16:47, 10 April 2016 (UTC)[reply]

The above used car which mentions the manuala got me thinking so I searched for Mercedes manuals and found [13] which is supposed to have 2016 manuals. Looking in the GLE section, there's a a link to [14] which has a filling capacity for the GLE 350 d 4MATIC Coupe for DEF (28.4L). It also links to [15] which has the same filling capacity for the GLE 250 d 4MATIC Sport Utility Vehicle and GLE 350 d 4MATICSport Utility Vehicle. Nil Einne (talk) 17:05, 10 April 2016 (UTC)[reply]

Many thanks!!!Johnson&Johnson&Son (talk) 02:41, 11 April 2016 (UTC)[reply]

Glad to help. Nil Einne (talk) 14:51, 11 April 2016 (UTC)[reply]

Resolved

Sunlight in June

The introduction to June claims that it is the month with greatest amount of daylight, in the Northern Hemisphere. Given the slow per-day changes in amount of daylight around the times of the solstices, how is June the correct answer? May and July both have an additional day, and July in particular is closer to the solstice than May, so how is it that July doesn't have the greatest amount of daylight? Of course, it makes sense for June to have the shortest amount of daylight in the Southern Hemisphere, since it has the winter solstice and one day fewer than either adjoining month. Nyttend (talk) 15:40, 10 April 2016 (UTC)[reply]

The quote from our article is "June is the month with the longest daylight hours of the year in the Northern Hemisphere and the shortest daylight hours of the year in the Southern Hemisphere". It's ambiguous if that means per day or the total for the entire month. And if if it does mean per month, there's only 3% more days in July, so even a minor difference per day could be more than that. Also note that how much the length of the day changes from June to July varies with the distance from the equator, so the answer might be different in different locations. StuRat (talk) 16:00, 10 April 2016 (UTC)[reply]

~~(ec)Looks to me like June would be correct. About three-fourths of the longest days of the season occur in June, and the fact of May and June having extra days does not make up for the difference...~~

The summer solstice typically occurs on June 21, but this year it's the 20th, this being a leap year. Checking my Old Farmer's Almanac for 2016 (which uses Boston, MA as its reference point, about 42 degrees north latitude):
May 1 has only 14 hours and 5 minutes of sunlight. Its first day with 15:00 or more sunlight is the 28th.

June 1 has 15:05 sunlight, peaking at 15:18 on the 20th and 21st, and is 15:14 on the 30th.

July 1 has 15:14, but its last day with 15:00 or more of sunlight is the 14th. The 31st is 14:28.

If the solstice occurred at midnight between June 30 and July 1, then July would have the most daylight, by 1 day. But because it's June 20 or 21, June wins. ←Baseball Bugs ^{What's up, Doc?} carrots→ 16:21, 10 April 2016 (UTC)[reply]

No, I've got it wrong. See calc's below. ←Baseball Bugs ^{What's up, Doc?} carrots→ 17:30, 10 April 2016 (UTC)[reply]

Let's do some math with those numbers, just using linear interpolation (hopefully close enough):

May average = {14 hrs 5 mins (14.083) + 15 hrs 3 mins (15.050)} / 2 = 14.567
May daylight hours = 31 × 14.567 = 451.577 hours

July average = {15 hrs 14 mins (15.233) + 14 hrs 28 mins (14.467)} / 2 = 14.850
July daylight hours = 31 × 14.85 = 460.35 hours

June average, up to solstice =  {15 hrs 5 mins (15.083) + 15 hrs 18 mins (15.300)} / 2 = 15.192
June daylight hours, up to solstice = 21 × 15.192 = 319.025 hours
June average, after to solstice =  {15 hrs 18 mins (15.300) + 15 hrs 14 mins (15.233)} / 2 = 15.267
June daylight hours, after to solstice = 9 × 15.267 = 137.4 hours
June daylight hours = 319.025 + 137.4 hours = 456.425 hours

So, if my math is correct, it does look like July has more total daylight hours in Boston, in July, than in June. StuRat (talk) 16:46, 10 April 2016 (UTC)[reply]

[ec] On the other hand, this page (which defaults to the latitude of Ghent) gives a comprehensive list of times. After some work with Excel, it appears that Nyttend's suspicions are correct - the exact figures are: May, 483 hours 37 mins; June, 494 hours 14 mins; July, 495 hours 31 mins. Presumably there's a critical latitude (which varies from year to year) at which June and July are equal, but the unqualified assertion that June has the greatest amount of daylight is apparently not true. Tevildo (talk) 16:40, 10 April 2016 (UTC)[reply]

Technically two critical latitudes, as there is also a line near the north pole, above which both June and July have 24 hours of sunlight every day. StuRat (talk) 16:55, 10 April 2016 (UTC)[reply]

Using this site,[16] note that at Barrow, Alaska, which is above the Arctic Circle, every day in both June and July is listed as 24 hours of sunlight. So at the North Pole, July beats June by a full 24 hours.[17] ←Baseball Bugs ^{What's up, Doc?} carrots→ 17:44, 10 April 2016 (UTC)[reply]

Note further that in Quito, which is very close to the equator,[18] daylight is 12:07 all year, except for two weeks before and after the December solstice, when it's 12:08. So at the equator, July beats June, by 12 hours. ←Baseball Bugs ^{What's up, Doc?} carrots→ 18:13, 10 April 2016 (UTC)[reply]

But the article actually says "longest daylight hours", not "most hours of daylight". Stu says it's ambiguous; I say the use of "longest" obviously implies that it refers to the average daylight per day. It's still badly phrased, though. As noted, there are latitudes where June and July have continuous daylight. --69.159.61.172 (talk) 19:46, 10 April 2016 (UTC)[reply]

I reworded the article to be unambiguous and correct: "June contains the summer solstice, the day with the most daylight hours in the Northern Hemisphere and the fewest daylight hours in the Southern Hemisphere (excluding polar regions in both cases)." StuRat (talk) 20:56, 10 April 2016 (UTC)[reply]

You created a further ambiguity by mentioning only the summer solstice, so I have reworded as follows: "June contains the summer solstice in the Northern Hemisphere, the day with the most daylight hours, and the winter solstice in the Southern Hemisphere, the day with the fewest daylight hours (excluding polar regions in both cases)." Akld guy (talk) 22:08, 10 April 2016 (UTC)[reply]

Thanks. Even the Jarada would be proud of that. :-) StuRat (talk) 22:16, 10 April 2016 (UTC)[reply]

And I have just changed December along the same lines with the same wording. Akld guy (talk) 22:22, 10 April 2016 (UTC)[reply]

Unfortunately, "the day with the most daylight hours" implies that it is the only such day. As soon as you get a little way beyond the Arctic Circle, there is more than one day each year with 24-hour daylight. --69.159.61.172 (talk) 23:11, 10 April 2016 (UTC)[reply]

But the wording specifically excludes polar regions. Nyttend (talk) 00:04, 11 April 2016 (UTC)[reply]

Yes, I thought it was necessary to exclude polar regions, since they would require more explanation than readers would want to read. StuRat (talk) 02:45, 11 April 2016 (UTC)[reply]

User:Nyttend, I'm late to this party, but do check out Insolation and refs therein. No need to resort to OR and back of the envelope, this stuff is seriously well-studied. Using that keyword, lat/long, etc can get you very good data, even stuff that includes cloud models and other advanced features if you like. Here's a slightly older but fairly comprehensive overview [19]. SemanticMantis (talk) 14:17, 11 April 2016 (UTC)[reply]

It's pretty obvious that the statistic refers to the average amount of daylight per day - and not the total for the month...the latter would be a really useless statistic. SteveBaker (talk) 17:36, 11 April 2016 (UTC)[reply]

If that's the approach, June wins: 456.425 / 30 = 15.2 while 460.35 / 31 = 14.8. ←Baseball Bugs ^{What's up, Doc?} carrots→ 21:39, 11 April 2016 (UTC)[reply]

Criticality and Neutron flux

If I had two pieces of Pu(238) (each weighing 5Kg), one in each hand and brought them togther quickly, how much netron flux would I get and how far away would someone have to be to survive such a criticality incident?--178.101.224.162 (talk) 21:02, 10 April 2016 (UTC)[reply]

I think your critical mass for plutonium is off, but in any case see Demon core (which is too mild, come to think of it, since it didn't blow itself apart), and fizzle yield (which is too harsh, since people are trying to jam that together with more effort). Wnt (talk) 21:34, 10 April 2016 (UTC)[reply]

Critical mass is between 9 and 10 kg according to our article. So whats wrong with my figures?--178.101.224.162 (talk) 21:39, 10 April 2016 (UTC)[reply]

I know you aren't talking about a "weapon" per se, however I think most of the same "issues" would apply as described in this article: Nuclear_weapon_yield#Calculating_yields_and_controversy . Vespine (talk) 23:52, 10 April 2016 (UTC)[reply]

No. Not a weapon. Im talking about a criticality 'accident' where the main effect is a large neutron flux and maybe some gamma rays. Definitely not a nuclear explosion as we know it (Jim).--178.101.224.162 (talk) 00:23, 11 April 2016 (UTC)[reply]

yes I know, I said I know you aren't talking about a weapon, but my point is that it's very hard to predict, even with accurate models, precisely what will "happen", whether something will reach criticality or not or how much reaction will take place. I recall seeing an article about software that was used for such calculations and a lot of details about it were confidential, except that it was extremely complex, extremely expensive and extremely classified. Vespine (talk) 02:23, 11 April 2016 (UTC)[reply]

I stand corrected! I had read some much smaller figures for bombs .... never realized that the critical mass could be changed that much by design considerations. I shoulda RTFA! Wnt (talk) 02:08, 11 April 2016 (UTC)[reply]

Yeah I'm not considering use of Neutron reflection, although the water content of my body may have a small effect.--178.101.224.162 (talk) 13:30, 11 April 2016 (UTC)[reply]

Why do Electric aircraft have poor thrust to weight ratio?

Is it because of battery limitations or electric motor limitations? Also if electric vehicles have such poor thrust to weight ratios, why is it that the Tesla Model S has such a good 0-60 acceleration time? ScienceApe (talk) 23:30, 10 April 2016 (UTC)[reply]

Electric motors are lovely. Small, powerful and efficient. Batteries carry both the fuel and the oxidizer, as rockets do, and don't work on chemicals with as much energy per weight. Power is splendid; thus you can accelerate quicly but batteries are heavy and poor in energy, hence can't go far. Jim.henderson (talk) 23:36, 10 April 2016 (UTC)[reply]

Plus, the motors they use have fairly stable "speed-torque" curves. In other words, they give you pretty much the same force whether they're turning slowly or quickly, and they don't generally need a complicated gear/transmission system (a few older electric trains had them). Fuelled vehicles by contrast get the best torque within a relatively narrow range of speeds, and you need a complicated gearbox to get the most out of them. Where acceleration is your biggest concern, electric vehicles are great (look at List of trolleybus systems in the United Kingdom - most electric bus systems were either in very big cities where you have lots of stopping and starting, or in mountainous areas like Yorkshire and the Welsh Valleys. Because it doesn't have the extra weight of batteries, a trolleybus will just fly up a hill almost as if it was flat road). Smurrayinchester 08:06, 11 April 2016 (UTC)[reply]

(edit conflict) What makes you think that electric aircraft have poor thrust to weight ratio? I could build you an electric aircraft with a fansastic thrust to weight ratio. It might run the capacitors dead in the first 30 seconds, but until then it would take off like a bat out of hell. Now if you want a good thrust to weight ratio and a decent range, then our article on Energy density has some interesting news about gasoline vs. batteries vs,. capacitors... --Guy Macon (talk) 23:45, 10 April 2016 (UTC)[reply]

That's roughly what small electric quadcopters do. Nobody minds the short flights caused by their low energy to weight ratio. Big airplanes, people like to fly far and long. So, they load up with energy storing or making things. This converts the problem to one of power to weight. 108.14.112.81 (talk) 02:24, 11 April 2016 (UTC)[reply]

As a bit of interesting trivia, the aircraft with the largest thrust to weight ratios are all blimps or dirigibles. :) --Guy Macon (talk) 04:19, 11 April 2016 (UTC)[reply]

Use mass instead. 1989 Goodyear blimp plus pilot, little fuel: 14.4 lb/hp. 1998 world's fastest accelerating production sedan, full tank, empty trunk, four 157 lb men, 4 ¼ lbs clothes (x4): 14.5 lb/hp. City block-sized balloons with 840 horsepower is where it's at. Sagittarian Milky Way (talk) 07:23, 11 April 2016 (UTC)[reply]

As for the Tesla, weight isn't as much of a problem in a land vehicle, as you don't need to provide enough lift to offset it. StuRat (talk) 04:28, 11 April 2016 (UTC)[reply]

I think to directly answer the question then: battery weight for an airplane is a much more significant issue than it is for a car...but even electric car batteries don't last that long and are very expensive...the limitation is not electric motors themselves (they can be plenty powerful to create great thrust) but powering a powerful electric motor...your flight time could be a matter of minutes instead of hours...68.48.241.158 (talk) 16:48, 11 April 2016 (UTC)[reply]

note, too, your tesla car can do what you describe...but will run down the battery in no time...68.48.241.158 (talk) 16:58, 11 April 2016 (UTC)[reply]

Besides the huge advantage in fuelefficiency, tru there is way more power stored in 1 kg kerosene then can be loaded into 1 kg of battery, electricity can only use a Propeller (aeronautics), which is limited to many conditions and not very effective in general, to generate reasonable thrust while you have multiple options with fuel, including very powerfull such as turbines (with afterburners), which have the highest thrust to weight ratio of all powermachines (some even excel rockets (which are not machines:)). --Kharon (talk) 17:45, 11 April 2016 (UTC)[reply]

I presume we're talking about propeller driven planes here. Jets are different for lots of reasons.

One of the reasons why electric cars have such good acceleration is that electric motors have very flat torque curves - they provide more or less the same torque for any given motor speed below the maximum. Since torque and acceleration are closely related, they have very uniform acceleration at all speeds. So the 0 to 60mph time for a Tesla is almost identical to it's 60 to 120mph time. With fast gasoline powered cars, the torque curve is anything but flat - there is a relatively narrow band of RPM's at which they produce peak torque - and if they have turbochargers, getting uniform torque is even harder.

The (partial) fix for that in a car is to use lots of gears - but there are losses involved in a complex gearbox - and the act of shifting gears (with the requirement to match the engine RPM while shifting) takes time and results in the engine RPM going in and out of the "sweet spot" where all the power is available. Most electric cars don't even have gearboxes.

So - how come gasoline engines for propeller-driven airplanes don't suffer the same problem? They don't even have shiftable gearboxes! Well, generally, they DO have variable pitch propellers - and that's a form of gearbox. It's actually a very efficient one because it has a continually adjustable "gear ratio" and no friction-creating gear teeth or clutch mechanism. This makes it much easier for the plane to keep it's motor running at the optimum RPM's.

Also, when acceleration is most important - on takeoff - the engine can be pushed to the optimum RPM, and the plane held on it's brakes until it needs to start moving. This means that the torque is already at the PERFECT amount when the plane starts rolling. With a car (and especially with a supercharger or a turbo) - the engine has to be at relatively low RPM's when you launch or you'll wreck the clutch or simply spin the wheels without much acceleration.

So many of the disadvantages of a gasoline engine aren't really a problem for airplanes - which makes the advantages of electric power very much less. For a car, the liability of a gasoline engine is a nightmare - and switching to electric motors solves a ton of issues.

Having said all of that - how much better is the acceleration of an electric car? Consider the 2013 BMW 1 series - which is the basis of the BMW ActiveE electric car. Let's consider the stats for a 2013 BMW 1 versus the Active E:

Weight: BMW1: 1500kg, ActiveE: 1800kg...a 20% overhead for the electric version. Of course a tankful of gasoline adds another 50kg to the BMW1 - so it's a bit less than that.
Torque: BMW1: 400Nm, ActiveE: 250Nm...oh...not good. But that's the PEAK torque for each car - and the ActiveE maintains that torque over a larger RPM range.
0-100kph time: BMW1: 6.3 seconds, ActiveE: 8.6 seconds - actually, better than you'd expect from the numbers given above. With 20% more weight, we'd expect 20% less acceleration for the same torque - but it also has vastly less peak torque. If BMW had put more powerful motors into the ActiveE, it could easily beat the BMW1 in a 0-100kph drag race - but with that much more weight, it's all-important recharge range would be much worse.

So read what you like into that.

SteveBaker (talk) 17:26, 11 April 2016 (UTC)[reply]

So the 0 to 60mph time for a Tesla is almost identical to it's 60 to 120mph time. -- I very much doubt that. With constant torque, you can bring constant force onto the road. But you need not double, but 4 times the kinetic energy when you go from 60 mph to 120 mph. --Stephan Schulz (talk) 17:53, 11 April 2016 (UTC)[reply]

might be true in a vaccum or something...but drag would indeed quadruple, I think...so the car would have less torque available for acceleration and more being used to maintain speed.....68.48.241.158 (talk) 18:14, 11 April 2016 (UTC)[reply]

Here's someone's graph of torque vs speed for a Model S. The motor is torque-limited (with power rising linearly) up to a certain speed, then power-limited (with torque dropping inverse-linearly). The crossover point is around 60 km/h ≈ 40 mph. So the Model S's 0–20 time should be similar to its 20–40 time (in a vacuum, at least), but 60–120 is a different matter. -- BenRG (talk) 21:14, 11 April 2016 (UTC)[reply]

April 11

Heat of deuteration and tritiation reactions

What are heat of the following reactions of deuteration and tritiation of water or heavy water by neutron capture?

H₂O + n → HOD

HOD + n → D₂O

H₂O + 2n → D₂O

D₂O + n → DOT

D₂O + 2n → T₂O

Are they exothermal or endothermal reactions? Is there some additional experimental complication in measuring the heat of these reactions compared to ordinary chemical reactions?--5.2.200.163 (talk) 13:24, 11 April 2016 (UTC)[reply]

This book discusses the mechanisms and energies of neutron capture by protium to form deuterium. --Jayron 32 13:42, 11 April 2016 (UTC)[reply]

What pages exactly? 391? Are they included in the free preview? Is the mentioned value of 2.23 MeV measured or calculated based on some (semiempirical) formula considering neutron binding energy?--5.2.200.163 (talk) 14:28, 11 April 2016 (UTC)[reply]

Is the formation of sodium-23 somehow included in the non-free preview?--5.2.200.163 (talk) 14:31, 11 April 2016 (UTC)[reply]

There is a thermodynamic isotope effect, though one hears much more about the rather modest kinetic isotope effect. [20] points me to a review by Galimov about it, only Google is not smiling on me today to tell me what that reference is. I should search on Sci-Hub, but generally prefer to make my interlibrary loans from the library wifi.

As written the reaction seems unlikely to be measured directly, as I assume the velocity of the neutron will blast the proton right out of the water molecule. There are two components: the thermodynamics of making D or T, and the thermodynamics of the exchange. The former is with very high energy and the latter with very low, so I assume that as written, your reactions can be replaced with bare H+, D+, and T+ transmutations for the same result as best as it can be measured. Wnt (talk) 14:38, 11 April 2016 (UTC)[reply]

Just to also expand on Wnt's solid answer, and clarify something: the thermodynamic relationships are state functions dependent only on the the starting and ending states, and thus independent of the path taken between the two states. The kinetic relationships DO depend on the path taken. In this case, given the fact that both deuterium and protium are stable, there is likely not a huge difference in potential energy between the two, so I would expect the thermodynamic changes (i.e. if they are exothermic or endothermic) to be relatively low. The kinetics are a different story; there may be a VERY large activation energy for the ¹H + n --> ²H reaction, but this is entirely independent of the thermodynamics. A reaction COULD be exothermic, and still have a very large activation energy. --Jayron 32 18:45, 11 April 2016 (UTC)[reply]

The activation energy of this reaction is close to zero as even thermal neutrons are easily captured with a cross-section roughly equal to the geometric cross-section of proton. Ruslik_Zero 20:35, 11 April 2016 (UTC)[reply]

Masturbation and Finasteride

Does masturbation decrease the efficiency of Finasteride? — Preceding unsigned comment added by 182.18.177.78 (talk) 16:29, 11 April 2016 (UTC)[reply]

I can't imagine there's any possible reason it would...I've never heard of such reducing the efficacy of any medication...did you read this somewhere and can cite? 68.48.241.158 (talk) 17:49, 11 April 2016 (UTC)[reply]

There seems to be a common myth that masturbation can contribute to hair loss [21] [22] [23] (see [24] [25] [26] for some examples of the myth in action). So the idea is perhaps not surprising. However as I said, the link seems to be a myth i.e. there's no scientific evidence for it as with many other rumours about masturbation. Given that there's no evidence masturbation affects hair loss and the mechanism of action of finasteride doesn't suggest any particular reason to think use of the drug will be affected by masturbation in some other way, there's no particular reason to think masturbation will affect efficiency of the drug. P.S. I don't think there's any reason to think testosterone levels will affect finasteride but in any case AFAIK the effect of masturbation on testosterone levels is also fairly complicated [27] [28] [29]. P.P.S. If you are taking finasteride you should speak to an appropriate medical professional if you have concerns about any possible interactions or side effects. P.P.P.S. I should perhaps also clarify I'm mostly thinking of something "resonable". If you're an adult masturbating an average of 20 times a day perhaps while watching a lot of porn and it's causing significant physical and psychological effects, then it's possible it could affect hair loss in some way. Nil Einne (talk) 19:56, 11 April 2016 (UTC)[reply]

(ec) Finasteride is a medication primarily for treating enlarged Prostate in men. Its reported possible side effect of diminishing libido is reported by the FDA who advise not to stop taking the medication without first consulting with one's health care provider. Masturbation has been purported to have many harmful effects, especially in the stigmatizing views of the Victorian era but today's medical consensus is that it is a medically healthy and psychologically normal habit. It is considered abnormal only when it inhibits partner-oriented behavior, is done in public, or is sufficiently compulsive to cause distress.(ref.). AllBestFaith (talk) 20:03, 11 April 2016 (UTC)[reply]

The question is not about the effect of Finasteride on the body or baldness. But of masturbation on Finasteride.--Scicurious (talk) 23:26, 11 April 2016 (UTC)[reply]

But if masturbation increases testosterone levels, and the goal of finasteride is to reduce testosterone level, wouldn't that count as an interference? Scicurious (talk) 21:42, 11 April 2016 (UTC)[reply]

Not unless there are citations to that effect. -- Jack of Oz ^{[pleasantries]} 23:06, 11 April 2016 (UTC)[reply]

Well there is a link between DHT and baldness: [[30]], and there are some endocrine effects linked to masturbation habits: [[31]]. Add to it that according to finasteride, it "prevents conversion of testosterone to dihydrotestosterone (DHT)." So, it's not too mad of a supposition that it could be a visible effect between masturbation on the effect of Finasteride. Scicurious (talk) 23:26, 11 April 2016 (UTC)[reply]

But if masturbation increases testosterone levels, and the goal of finasteride is to reduce testosterone level, wouldn't that count as an interference? Scicurious (talk) 21:42, 11 April 2016 (UTC)[reply]

Well, see, there's your first mistake — that's not the goal of finasteride. See the finasteride article. It's not meant to reduce testosterone levels at all. It's meant to inhibit the conversion of testosterone into dihydrotestosterone. --Trovatore (talk) 02:30, 12 April 2016 (UTC)[reply]

See response above. -- Jack of Oz ^{[pleasantries]} 02:22, 12 April 2016 (UTC)[reply]

momentum/weight

If an object (say on dry land, maybe a car) is traveling, the faster it moves it gains more momentum. What happens to the weight of the object (car?) Does the weight become displaced the more momentum the object generates? 199.19.248.20 (talk) 23:20, 11 April 2016 (UTC)[reply]

The weight of a car - the force exerted on it by gravity - depends only on its distance from the centre of the Earth, which won't vary significantly if it stays on the ground. The mass of the car depends only on what it's made of - for a car with an internal combustion engine, the mass will steadily decrease as the fuel is used up, but this isn't directly related to its speed. However, the vertical force exerted on the ground by the car - which might be thought of as "weight" - will vary with speed, depending on the aerodynamics of the car's body. Tevildo (talk) 00:09, 12 April 2016 (UTC)[reply]

In the strict interpretation of mass energy equivalence, I believe given two identical cars, where one is traveling faster than the other, the faster one will have more kinetic energy and will therefore have more mass and weigh more! Mass_in_special_relativity#Relativistic_mass. However for any speed which is not within an order of magnitude of the speed of light, the increased mass can pretty much be ignored, as it would be infinitesimal. Vespine (talk) 03:43, 12 April 2016 (UTC)[reply]

I could be totally wrong, but after a little playing with some online energy calculators, you would need 1,000,000,000 (one thousand million) 1 ton cars traveling at 100km/h to increase the relativistic mass of the whole system by 4 grams. Vespine (talk) 03:50, 12 April 2016 (UTC)[reply]

April 12

Methylene diphenyl diisocyanate (MDI)

I was excited yesterday to score three 55 gallon steel drums for five bucks each. I was a little bummed today to see that one, they were filled with MDI, and two, there is still a little left inside of each one. How bad is this stuff? There is probably less than 20 ounces in the bottom of each. It looks like I can treat it with water, ammonia, and detergent, but then how could I dispose of the results? The MSD sheets I've looked at show that this stuff is flammable but it does not appear to be very volatile that I can tell. After treating it I'm hoping it would be safe to use an angle grinder on it with no risk of explosion? Additional, how bad is it that a junk dealer in town has 50 or so of these drums behind his building that have not been cleaned or decontaminated and he is selling them to the public? Beach drifter (talk) 00:07, 12 April 2016 (UTC)[reply]

Well, the first thing to do is look up MSDS for it - see [32] for example. There are plenty of sour notes in there, read carefully. There is also an article Methylene diphenyl diisocyanate, though it isn't certified for staking your health on (see disclaimers!) it may help with curiosity. (The MSDS refers you to contact the company if you want to know more about safe disposal ... I dunno, that could be like calling the phone number on the barrel in Return of the Living Dead, in a good or a bad way) Wnt (talk) 00:46, 12 April 2016 (UTC)[reply]

Thanks, that MSDS was more informative than the one I had read. I'm satisfied I can clean these barrels well enough to use for some projects, but making a grill or smoker is out for these guys I think. Beach drifter (talk) 01:25, 12 April 2016 (UTC)[reply]

Sentient nerves

Do they exist in the large intestine?--178.101.224.162 (talk) 00:49, 12 April 2016 (UTC)[reply]

Sentience is not a quality that is usually held to be imbued in a given nerve. To be fair, not withstanding the great strides of modern neuroscience, we genuinely have almost no understanding of how the non-physical and nebulously-defined phenomena of consciousness arises out the physical mechanisms of the nervous system (see hard problem of consciousness), although we do have an increasingly detailed understanding of which regions of the nervous system (the brain in particular) are associated with particular mental functions, as a computational matter.

Now, there is a growing trend in neurology/cognitive science to not disassociate the brain from the rest of the body when looking at how the two give rise to a mind. But that being said, the innervation of the enteric nervous system are about as far removed from the mental processes we most associate with sentience as any part of the nervous system broadly. I think what might be causing this confusion is a phrase that has been making the rounds in recent years, that the enteric nervous system is "its own brain". This little piece of apocrypha originates from the fact that its been discovered that the enteric nervous system operates closed circuits which function all on their own, with very little in the way of feedback signals between it and the central nervous system by way of either the sympathetic or parasympathetic nervous sytems, which otherwise govern the body's autonomic functions. That is to say, the enteric nervous system knows how to keep your gut operating and generating peristalsis, even sometimes if there has been severe trauma to the pathways that link it to the rest of the nervous system. So it's a "brain" by way of it's closed functions and modularity, but the nerves within the system are not "thinking" in the traditional sense of cognitive function (though they may, in limited ways, deliver information which may influence the total human sense of self in subtle ways we don't yet appreciate, as with other quasi-somatosensory systems). Snow ^{let's rap} 02:10, 12 April 2016 (UTC)[reply]

Compression ratio vs displacement

I'm having a hard time understanding how this can be true. We are comparing 2 vehicles with different variables. Chevy Cobalt has 2.2 liter engine displacement. Mazda 3 has 2.3 liter engine displacement. So the mazda has 0.1 liters larger displacement than the chevy. However the chevy's compression ratio is 10:1 and the mazda's is 9:1. How can an engine have larger cylinder displacement yet a smaller compression ratio? (Both cars have 4 cylinders) 199.19.248.20 (talk) 04:10, 12 April 2016 (UTC)[reply]

Why do you think the amount of compression is related to the size of the engine ? There's often the reverse-correlation, because smaller engines need a higher compression ratio and/or turbochargers/superchargers, to produce comparable power to a larger engine. StuRat (talk) 04:29, 12 April 2016 (UTC)[reply]

Both cars do NOT have superchargers, tubros, etc. Both cars have dual overhead camshaft valve train configurations. Compression ratio is the ratio of the cylinder volume with the piston at bottom dead center, when compared to the remaining volume with the piston at top dead center. Cylinder displacement is the volume the cylinder holds between bottom dead center and top dead center. With both of these definitions I'm having a hard time understanding (what appears to me) as a contradiction.199.19.248.20 (talk) 05:21, 12 April 2016 (UTC)[reply]

I didn't say that turbo/superhchargers were present on that model. I only mentioned them as an alternative way to get more power out of smaller engines, besides a high compression ratio. StuRat (talk) 05:31, 12 April 2016 (UTC)[reply]

(ec) This comes down to two factors: bore/stroke, and cylinder head shape. The Chevy may have a smaller bore and longer stroke, thus compressing the combustion gases more, or it may have a less concave cylinder head shape. That is, a smaller combustion chamber into which the gases are more tightly compressed than they are in the Mazda. Or, it may be a combination of those two factors. One engine is not necessarily better than the other on compression ratio alone; there are sound reasons and trade-offs for choosing a particular compression ratio. Akld guy (talk) 05:28, 12 April 2016 (UTC)[reply]

The Chevrolet Cobalt has a four cylinder engine. Imagine that we made that a six cylinder engine, keeping the bore, stroke, compression ratio, etc. in each cylinder the same. Larger displacement yet same compression ratio, right? Now imagine that we made that a three cylinder engine, again keeping each cylinder the same. Smaller displacement yet same compression ratio, right? --Guy Macon (talk) 05:38, 12 April 2016 (UTC)[reply]

Makes so much sense now. Thanks!! 199.19.248.20 (talk) 06:18, 12 April 2016 (UTC)[reply]

Flamethrowers

In World War 2-era flamethrowers, could the fuel be transferred from one flamethrower to another? So, for example, if an Allied firebat ran out of fuel, could he pick up a discarded German flamethrower and transfer the fuel from that to his own, either by siphoning or by swapping cylinders? And vice versa, could a German soldier do the same with an Allied flamethrower? 2601:646:8E01:515D:8D43:54FA:2E2E:994A (talk) 05:35, 12 April 2016 (UTC)[reply]

Even if technically possible, I doubt if the situation would have come up, because flamethrowers were used by the side with near total ground superiority in an area, to "smoke out" any enemy troops hiding in caves or rubble. So, only one side would be likely to use flamethrowers at a time. StuRat (talk) 05:43, 12 April 2016 (UTC)[reply]