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March 12

pH value from dissociation constant

This is based on a homework question with specific values, but I am asking only for general formula.

If I know the acid dissociation constant (pKa) and the concentration of a weak acid, how would I find the pH? How do I find the pH after adding a certain volume of a strong base? − Twas Now ( talk • contribs • e-mail ) 00:58, 12 March 2007 (UTC)[reply]

Take a look at Henderson-Hasselbalch equation

{\textrm {pH}}={\textrm {pK}}_{a}+\log _{10}{\frac {[{\textrm {A}}^{-}]}{[{\textrm {HA}}]}}

From the pKa page the ionised acid and H+ concentration will be the same, so you end up with pH+pH=-log10(concentration of a weak acid)+pKa.

{\textrm {pH}}=({\textrm {pK}}_{a}-\log _{10}{[{\textrm {HA}}]})/2

GB 02:18, 12 March 2007 (UTC)[reply]

Thanks. I guess what confuses me is how do I determine [HA] and [A^-] for a given weak acid (and its conjugate base)? − Twas Now ( talk • contribs • e-mail ) 03:09, 12 March 2007 (UTC)[reply]

It's really very simple. All you need to do is set up an ICE table for the ionization equilibrium of the acid, i.e. HA ⇌ A^- + H⁺ :

	[HA]	[A^-]	[H⁺]
I	c₀	0	0
C	-x	+x	+x
E	c₀ - x	x	x

Because $pH=-\log {[H^{+}]}$ , and you can tell from the table that when the ionization is over [H⁺] = x, then $pH=-\log {x}$ . So, your job is to find x.

The equilibrium constant expression for the ionization is:

$K_{a}={\frac {[H^{+}][A^{-}]}{[HA]}}$

Substitute the concentrations with the values found in the last row of the ICE table.

$K_{a}={\frac {x.x}{c_{0}-x}}$

Now plug in the specfic values for c₀ and K_a ( $K_{a}=10^{-pK_{a}}$ ) provided in your question, then solve for x, and you're done!

If you do some math, you can quickly figure out that the relation GB provided is only a different (and handier) way of expressing K_a, especially when you're working with buffer solutions.

Now for the second part of your question: when you added a certain quantity of a strong base, the OH^- ions you just added will react with some of the H⁺ in solution. From the OH^- - H⁺ reaction stoichiometry, you can calculate the number of moles of H⁺ that have disappeared and how many moles are left. Then you can find the new concentration of H⁺ ions (keep in mind the change in volume) and the new pH. —LestatdeLioncour t 14:45, 12 March 2007 (UTC)[reply]

Tin can water battery sparking experiment thing :-)

I am trying to find if there is an article or other information about a physics set up of a water spark experiment.

Basically you have a tank of water with 2 holes in it. Water falls out each hole and falls through two cylinders and into two tin cans. Each tin can is wired to the opposite cylinder (not the one its own water falls through). As the water falls it appears a charge builds up and high kv spark can be generated across points that are connected to the tins. There are no other components or sources of power. I saw this in a movie of a lecture and wanted to read about it. As the charge builds up it causes the water falling to repel itself or something, such that as the charge builds up the trikles of water change from being smooth and straight to rain drop like and then when the energy is discharged it instantly goes back to smooth and straight.--155.144.251.120 02:42, 12 March 2007 (UTC)[reply]

Here it is. [1] --Zeizmic 13:27, 12 March 2007 (UTC)[reply]

And our article on it is here, though it's a tad stubbish at the moment. GeeJo ^(t)⁄_(c) • 07:32, 15 March 2007 (UTC)[reply]

Force of Impact?

Here's a question I've asked in the talk page of Impulse and I began to wonder if it would not be better of me to ask it here, so I quote it verbatim.

Often one hears the term "Force of Impact". Say you can calculate/estimate the impulse of an impact but you want to estimate (perhaps) the maximum force of collision, or possibly the average force of the collision, or maybe even the average force for the top 25% of the most forceful moments of the collision. Is any of that information relevant? I'm interested in the relationship between the fracture strength of materials and "how forceful" an impact an object of said material can take. It would seem that Impulse is a desired quantity to know, but what else should I know? IIRC to get average force, I just divide divide the impulse over the time it takes for the the impact to take place. But how would one estimate this time? How would I estimate the average force imparted by one billiard ball on another over the lifespan of the impact, which I cannot estimate with the unaided eye. How high could I drop a billiard ball in a frictionless environment from a building before cracks of a certain size develop on its surface (I.E. the force imparted exceeds locally the fracture strength of the material the billiard ball is made from but it does not fracture the ball). How do I estimate how much effect the ground chosen for the ball to impact has on this? If the fracture strength of the ground is quite low, energy from the impact will be dissipated by "flakes" of the ground being thrown outward, and it would seem that also needs to be taken into consideration.

It seems methods to estimate information is relevant and related to the article, but no description or references are included in the article. What quantities are these and how are they measured or estimated?

I'm not a physicist and I wouldn't even know where to begin to look (other than, say, research into stuff like bicycle helmets).

Might any of you be any help on this? I'm interested in any comment, even if you might consider it to be off-the-wall. Thanks.

Root⁴(one) 03:58, 12 March 2007 (UTC)[reply]

I think it is pretty rare to do such experiments or observations in the time-domain. The impact duration for a "mostly elastic" collision such as a billiard ball is extremely short. A few high-speed photography experiments have been done (this guy from MIT was famous for his photos of things colliding). My guess is that it is easier to measure the peak force than the impact time. You could then model the time-varying force with a Gaussian function force-vs.-time relationship based on the measured peak-force (or invent a different theoretical model). Then, you could integrate for the total impulse (momentum transferred).

As far as the material strength, you might look at bulk modulus or Young's modulus, which are useful for calculating solid object deformation as a function of force. Naturally, more sophisticated analysis can be performed per material, with loss of generality, to account for nonlinear, anisotropic effects, for example.

Here is a photograph of a golf-ball at the instant of impact. Clearly deformation occurs. If I recall, this image is taken in a dark room illuminated by a single strobe-flash, so the image exposure is on the order of 10 micro-seconds. Nimur 05:03, 12 March 2007 (UTC)[reply]

Thanks, your reply indeed does help, although it appears to mean I may need to learn/re-educate myself on a few maths I can't say I know at present. The Young's modulus article was quite insightful, as I had forgotten how important the cross sectional area is to what I'm trying to learn.

If you wanted to know what got me started on this road, its research into this "5 pounds to break a collarbone" myth. It is true that it does not take much force to break the clavicle, or rather, pressure solely exerted on the clavicle itself, but 5lb == 22 Newtons seems awfully low. however, if that 22 Newtons is spread over a small enough area on the clavicle, this maxim may be more true than I realize.

I also was involved a few months back in a minor accident on a bike that separated my shoulder, and I've often wondered how to demonstrate, in terms of physics, how and why it happened, which recommendations for prevention are better and why they are better in certain situations. I already know what I should have done, but knowing the physics behind why the technique works and in which situations the technique is better or worse, I think, would be helpful, and would help put my mind to rest as to what truly is good advice. I'm not sure how I could contribute this information to Wikipedia, as I'm sure this would qualify as "original research", but maybe at least I could use it to help defend against bad information and advice. At least maybe I could put in on a blog somewhere if I choose to set one up.

But given how complicated the mathematics appears be for simple collisions of "anisotropic" materials, I may need to research this for a while. I seriously doubt I'm the only one to have considered these questions, so hopefully with enough research I can find solutions to problems already answered relating to the physics of protecting the shoulder.

Root⁴(one) 11:58, 12 March 2007 (UTC)[reply]

In the special case of a ball or cylinder bouncing off a hard plate the answer is complex.

Not very surprisingly Timoshenko covers this problem in "Theory of Elasticity".

Unfortunately I didn't copy all the right equations down, so I'll just solve the simpler case of a steel ball bouncing off its twin. If you think about that case it is likely that this is also the same as the steel ball bouncing off an infinitely stiff flat plane.

The time of the contact is

t=2.94(1.25*sqrt(2)*pi*rho*(1-nu^2)/E)^0.4*R*v^-0.2 eqn 244

R=0.031 m

v=3.132 m/s

rho=7843 kg m-3

E=210*10^9 N m-2

nu=0.3

t=0.1493 ms

The average force, F during the contact is 2*m*v/t (ie the change of momentum divided by the time)

F=42 kN, ie a little over 4 tons force.

Timoshenko actually gives a direct solution for a ball on a flat plate, and the peak force rather than the average force, but it is spread over two pages. This solution assumes that the contact time is long compared with the period of the lowest modes of vibration. Greglocock 01:06, 14 March 2007 (UTC)[reply]

Artificial Sweeteners

Can chronic fatigue be from overuse of artificial sweeteners? I am 22 years old, in great health, sleep well, and take my vitamins, but I am tired all the time. Could it be from the 36 packets of splenda I have everyday?

Splenda#Safety does not list anemia or fatigue as a side effect. You might suffer from mild anemia, unrelated to your splenda consumption, that could be due to genetics, diet, or some chronic medical condition. This causes sleepiness and exhaustion. You might consider consulting a physician. Sometimes, the best solution is a change in diet (more iron-containing, green leafy vegetables and legumes, and more red-meat), and sometimes dietary supplements or medicine is necessary. See also Iron deficiency (medicine). P.S., this is not medical advice, consult a physician, etc. etc. Nimur 04:52, 12 March 2007 (UTC)[reply]

As Nimur said, you might want to consult a physician. However, fatique can come from many things, including the room being too hot / cold, or a non-fixed sleeping schedule (something a 22 year old is more likely to engage in). --Wirbelwindヴィルヴェルヴィント (talk) 04:58, 12 March 2007 (UTC)[reply]

doh. I think sugar substitutes may trigger the same insulin response as sugar. Jsut a guess and certainly not medical advice but maybe that much splenda is giving you the sugar crash without the prior rush. Use regular sugar for a week and see if it changes. --Tbeatty 05:19, 12 March 2007 (UTC)[reply]

I thought splenda is safe for diabetics because it doesn't trigger the insulin response? --Wirbelwindヴィルヴェルヴィント (talk) 05:28, 12 March 2007 (UTC)[reply]

Chronic fatigue can be a sign of a lot of things. That is, you could potentially have a real problem. Visit a doctor, and they'll be able to check if anything is wrong. Even if you don't have one of the really big problems, they will also be able to get a much better idea of what is causing your fatigue than we can, so they are more likely to be able to help you cure your fatigue, which I'm sure you see as a positive thing. Please, see a doctor. Skittle 16:56, 12 March 2007 (UTC)[reply]

You say you sleep well. Does this mean you fall asleep readily? It could be that you are not getting enough sleep and hence feel tired during the day. I suggest trying giving up all caffeine for several weeks. At first you will feel worse due to withdrawal, but after a few weeks you will all the time feel as alert as if you'd drunk strong coffee. I know, I've been caffeine free for years. I'd thoroughly recommend it. —The preceding unsigned comment was added by 62.253.53.39 (talk) 21:21, 12 March 2007 (UTC).[reply]

Feeling tired all the time could be a symptom of diabetes, so you must see a doctor about this. Untreated diabetes causes blindness, loss of limbs, etc. 62.253.44.18 19:54, 13 March 2007 (UTC)[reply]

If someone were to ask you, what do you think a potential side-effect of artificial sweetners would be, chronic fatigue sounds about right. Sugar is basically pure energy, and a substance that mimics the taste of sugar but isn't quite the same sounds like 'tainted fuel' to me. Vranak

My friend has Chronic fatigue and he has found the Gluten makes it much worse. Try going gluten free for 2 weeks, it really helped him. I know this doesn't really answer the question. CaptinJohn 12:30, 14 March 2007 (UTC)[reply]

Ionization of acetic acid

How much does acetic acid to ionize in water? For example, if you have one mole of acetic acid, would it ionize into only a small amount of CH₃COO^- and H₃O⁺ (e.g. 5% or 0.05 moles each), or would it be much more (e.g. 95% or 0.95 moles of each)?

I know acetic acid only ionizes partially, but both 5% and 95% are partial. So when chemists refer to "partial ionization", what is their threshold for "partiality"? − Twas Now ( talk • contribs • e-mail ) 05:43, 12 March 2007 (UTC)[reply]

I think the pH might be a starting place. The pH specifically gives the concentration of the ionized acid, (given in a logarithmic scale described in the article). You can then compare with the total molarity of the solution to determine the amount of "partiality." Actually, that article contains a pretty useful example here. Nimur 06:03, 12 March 2007 (UTC)[reply]

Also, directly from the acetic acid article, the constant is given: "A 1.0 M solution (about the concentration of domestic vinegar) has a pH of 2.4, indicating that merely 0.4% of the acetic acid molecules are dissociated." Nimur 06:09, 12 March 2007 (UTC)[reply]

Thank you. I understand pH, but the ionization percentage is not directly evident from a pH value. The second comment definitely helps though! Thanks. − Twas Now ( talk • contribs • e-mail ) 06:14, 12 March 2007 (UTC)[reply]

Yes, it's not directly evident, but it can be solved with the equations (as in the example above). Good luck! Nimur 06:36, 12 March 2007 (UTC)[reply]

How much should we trust in sunglasses?

As I know, it is dangerous to wear sunglasses without uv shielding, because the pupil will grow bigger to let more light in, so it would let more uv radiation in. So what is the most simple (or lest expensive) experiment to decide whether our sunglass does have uv protection or not? --V. Szabolcs 07:55, 12 March 2007 (UTC)[reply]

AFAIK glass is mostly opaque to UV anyways (that's why they need to use quartz to make Germicidal lamp), but most plastics are not, and should have a UV coating on it. --antilived^{T | C | G} 08:14, 12 March 2007 (UTC)[reply]

An easy way to determine the existance of UV shielding would be using a black light or a halogen light with its glass UV filter removed and some flourescent substance (glow in the dark toys, some especially bright t-shirts etc.), and compare the normal brightness to the brightness under the sunglasses. But any more than that I think you will have to visit an optometrist where I've seen a machine that does exactly what you said. --antilived^{T | C | G} 08:27, 12 March 2007 (UTC)[reply]

Oh man, you've got me all scared to ware my sunglasses now; and I need 'em man! Really I do - how else am I gona look cool? Think outside the box 12:45, 12 March 2007 (UTC)[reply]

By rights, you ought to repeat your experiment at several other ultraviolet wavelengths rather (UV-B and full-spectrum UV-A) rather than just the near-UV (360 nm light) emitted by blacklights.

Atlant 15:28, 12 March 2007 (UTC)[reply]

I heard recently that most modern mobile phone screens will turn black when seen through some sunglasses. Does it only prove that the glasses have some sort of polarization, or does it also have something with uv shielding? --V. Szabolcs 16:13, 12 March 2007 (UTC)[reply]

The light passing through liquid crystal displays definitely becomes polarized, but unfortunately, usually not at the same angle as that admitted by polarized sunglasses (which leads to people outdoors holding their heads tipped at a 45 degree angles to look at their laptop screens). The effect has nothing to do with UV light.

Atlant 16:58, 12 March 2007 (UTC)[reply]

If in Britain, I would look for the little sticker that told me they met EC or British Standards, and the one saying they blocked 100% of UVA and UVB [2][3][4]. I imagine the same applies elsewhere in the EU, and similar in the US. Skittle 16:52, 12 March 2007 (UTC)[reply]

How to add fat in raw milk

AS I WAS READING THE TEXT THE QUESTION WAS RAISED HOW TO REDUCE FAT, BUT I WANT TO KNOW HOW FAT CAN BE ADDED IN COW MILK THAT CAN BE INSTANTLY SOLUBLE OR MIXED. —The preceding unsigned comment was added by 203.101.172.199 (talk) 13:52, 12 March 2007 (UTC).[reply]

You can always add dairy creamer, thus putting back what was taken out. Root⁴(one) 14:15, 12 March 2007 (UTC)[reply]

The key question is how to keep the added fat from separating back out again. See homogenization (what a pathetic article!) and emulsification.

Atlant 15:25, 12 March 2007 (UTC)[reply]

The cream in whole (raw) milk naturally separates from the aqueous phase and rises to the top. Homogenisation only provides a temporary solution because the fat will separate out again given time. Hexane2000 16:09, 12 March 2007 (UTC)[reply]

Sure, but the milk will usually sour before commercial homogenization fails.

Atlant 16:54, 12 March 2007 (UTC)[reply]

Bizzare

I found this picture on the internet and I have absolutely no idea what it is. It looks like some sort of animal, having a body and arms, but I don't know. Maybe someone here knows.

File:BizzareThingy.jpg

I didn't take the picture myself, so i don't know where it is indeigenous to or anything like that. Thank you for your help. schyler 16:26, 12 March 2007 (UTC)[reply]

Jesus! After looking at it for a while trying to figure out what it is, I've decided that it is a bird. The photo is taken from behind (i.e. that pink thing at the top is the back of its head and those little things on the sides are wings). I'm no bird expert, so I don't know what type it is. GhostPirate 16:39, 12 March 2007 (UTC)[reply]

I agree, looks like a baby bird seen from behind. (thanks for switching my perception GhostPirate, that was disturbing) Skittle 16:42, 12 March 2007 (UTC)[reply]

Wow. Okay. What strange lookig bird! schyler 20:32, 12 March 2007 (UTC)[reply]

Maybe because my dad used to keep doves and pigeons, but I saw it as a bird right away. It looks like it could be a baby vulture from the back, before the feathers grew out. --Wirbelwindヴィルヴェルヴィント (talk) 23:53, 12 March 2007 (UTC)[reply]

You can see the very tip of the beak if you look closely --VectorPotential^Talk 00:08, 14 March 2007 (UTC)[reply]

OMG! (oh my gosh!) it looks like a person (possibly a baby) with no face!--Lerdthenerd 09:52, 14 March 2007 (UTC)[reply]

It is a baby bird, but difficult to tell which species. Might be chicken, turkey, any similar domesticated bird. --Nirajrm ^{talk ||| sign plz!} 22:40, 14 March 2007 (UTC)[reply]

Moden physics

if the probability density at certain points for a particle is zero, does this imply that the particle cannot move across such points—Preceding unsigned comment added by Hjchen1010 (talk • contribs)

No, it just means you can't ever find the particle at that point. To move past such a point or region is a phenomenon known as quantum tunneling. 131.215.159.161 17:35, 12 March 2007 (UTC)[reply]

Note that quantum tunnelling is a class of phenomena that involves particles passing through forbidden states under certain specific conditions (generally, allowed states on either side of a thin forbidden zone). -- mattb @ 2007-03-12T19:22Z

It depends on whether the probability density is zero before or after quantum effects are taken into account. While most problems that have a zero p.d. classically have a non-zero function for a quantum particle, it is technically possible to have a p.d. that is also zero in the quantum case, which the particle cannot enter. In nature it's practically impossible (very, very small, but not zero) and the above answers are correct; but it's quite a common thing to have in a homework or exam question (an infinite potential). Spiral Wave 21:28, 12 March 2007 (UTC)[reply]

Low pass filter

I gotta implement a Hanning window-shaped LPF. Have I to weighten the magnitude samples only or the phase too? tia —The preceding unsigned comment was added by Ulisse0 (talk • contribs) 17:17, 12 March 2007 (UTC).[reply]

If you are concerned about output phase, you must also include those in your specification of the filter design. If not, you don't. Either way, make sure your filter is stable (phase should be greater than -180°). Nimur 19:58, 12 March 2007 (UTC)[reply]

A window function is applied to the magnitude. A filter designed by windowing the impulse response will typically be FIR with real coefficients, unless you're working on complex-valued signals. Outside of a homework problem or laziness, there's better ways to design a low-pass filter. 24.91.135.162 15:51, 15 March 2007 (UTC)[reply]

Fireproofing MDF

For my friend's birthday, I'm going to drill holes in the form of "Happy Birthday Name" into a peice of wood and put matches in all these holes, the idea being that I can light one match and the fire will spread to present the message (in a darkened room). What can I cheaply coat the wood with so it wont burn if the matches burn all the way down and will MDF be suitable? --Seans Potato Business 17:32, 12 March 2007 (UTC)[reply]

Just soak the wood in water before you light. Of course, if it's a horrendously long name, you might start a firestorm. --Zeizmic 20:03, 12 March 2007 (UTC)[reply]

Lower pitch of voice

Anyone know of something that will lower the pitch of the voice like a thyroplasty but without surgery and at the same time won't cause side-effects on other parts of the body like taking steroids would? Perhaps a chemical applied just to the vocal chords? SakotGrimshine 19:35, 12 March 2007 (UTC)[reply]

<inappropriate suggestions removed to talk page>Edison 16:48, 13 March 2007 (UTC)[reply]

I was hoping for permanent. Basically it might be nice if I can get steroids just to go to my voice box and not elsewhere (like my hair folicles on my head). I doubt a rub-on cream would do it. SakotGrimshine 23:56, 12 March 2007 (UTC)[reply]

Going to a competent and qualified public speaking or singing teacher would help you find your natural speaking voice, which may well be pitched lower than you currently use it. It won't happen overnight, only with training and practice, but the permanent change would be well worth the effort. JackofOz 02:31, 13 March 2007 (UTC)[reply]

I don't think there is a quick/easy solution. I know that men who have undergone a sex-change have a lot of difficulty pushing their voices up in pitch (so they sound more feminine) - and the only thing they can do that really helps is professional voice training. I would guess (although I don't know for sure) that lowering the pitch of your voice would be a similar problem. SteveBaker 23:19, 13 March 2007 (UTC)[reply]

Well lowering it is posible if I take anabolic steroids, but those cause unwanted effects to the rest of my body when I only want their effects on my voicebox. SakotGrimshine 12:49, 14 March 2007 (UTC)[reply]

Some of you may recall the TV & Movie Star, Rock Hudson?? Among other things, he had a reputation for his deep, "swarthy" voice. I'm only paraphrasing, but his biography describes his initial entry into the entertainment business. Supposedly his agent thought he had the 'looks and the talent' to become a star, but his then "squeaky" voice would be his failing. His agent took him somewhere into the Hollywood hills...the middle of no-where...and made him scream at the top of his lungs, until he could scream no more. Subsequently he had NO voice for a few weeks, but upon getting it back, voila!, he had his legendary deep voice. (Perhaps an MD or, more specifically an ENT Dr. could explain whether his vocal cords were lengthened or shortened.) User: ashley604

Reccomendations for alarm clock

Strange question but does anyone know of an alarm clock available for purchase in the UK that is very loud and doesn't stop unless turned off? I am a very heavy sleeper and normal alarm clocks bleep or ring for a while and then stop, I want one I cannot sleep through! GaryReggae 19:46, 12 March 2007 (UTC)[reply]

Turn up the volume and put it far away from your bed so you have to walk to it. Nimur 19:55, 12 March 2007 (UTC)[reply]

Google 'very loud alarm clock' and it's hilarious! --Zeizmic 20:00, 12 March 2007 (UTC)[reply]

You can set your radio to function as an alarm clock and set it to a really loud volume. Also, set it to a talk radio station, rather than music, because we tend to orient towards voices.

If you have a good (or at least loud) set of speakers on your computer, you can use any of several different programs to act as an alarm clock. You can use whatever sound file is most likely to get you out of bed -- anything from a scream to a recorded threat to "Thus Spake Zarathustra." -Arch dude 21:35, 12 March 2007 (UTC)[reply]

If you regularly need an alarm clock then it suggests you are not getting enough sleep. You may be being kept awake in the evening by caffeine. I suggest giving up caffeine and going to bed earlier, you will feel much better after the first couple of weeks of caffeine withdrawal. Apart from that I second the suggestion of buying a clock/radio. I think ones from Argos only cost a few pounds. 62.253.49.0 21:59, 12 March 2007 (UTC)[reply]

Sounds like you need The sonic bomb with bed shaker alarm. :) Vespine 22:42, 12 March 2007 (UTC)[reply]

I use a $5 analog clock that I got from Goodwill. It takes a single N battery that lasta a couple years. Unfortunately, N batteries are hard to find in the US. Regardless, it is dirt cheap and the alarm is a high-pitched chirp that repeats forever until you turn it off. It wakes me up right away. Radio alarms never work for me unless I turn them up so loud that they wake the neighbors. Also, since most radio alarms plug into the wall, they reset to 12:00 every time power blinks. --Kainaw ^(talk) 00:14, 13 March 2007 (UTC)[reply]

My battery powered alarm reset to 00:00 occasionally for no reason :( You could get two alarms, on set a few minuites later than the other, and put it on the other side of the room, so you have to turn both off :] HS7 09:39, 13 March 2007 (UTC)[reply]

[5] looks entertaining, as does this, although it appears to be unavailable anywhere. Otherwise, try looking at alarms for the hearing impaired. Skittle 15:42, 13 March 2007 (UTC)[reply]

Green Beer

Does anyone know the specific ingredients in green St. Patrick's Day beer and why it results in horrible hangovers? I recall hearing somethign about the interaction of alcohol with dye. Any clarification from anyone? —The preceding unsigned comment was added by 74.12.102.224 (talk) 19:55, 12 March 2007 (UTC).[reply]

In general, green beer is just beer with added green food colouring. There shouldn't be any sort of interaction which results in a more severe hangover; any effects are due to excessive consumption – it's a special occasion, right? – and the nocebo effect. TenOfAllTrades(talk) 20:18, 12 March 2007 (UTC)[reply]

Exactly. I grew up in our family business, a tavern. We used McCormick Green Food Color. Of course, we would buy the dye in a pack of four colors and so had all sorts of blue, red, and yellow coloring left over that I as a child would divert to various projects... --BenBurch 00:07, 13 March 2007 (UTC)[reply]

OK: You can use two bottles of red food coloring to bake a Waldorf Red Cake. What can you do with blue and yellow? - Nunh-huh 00:10, 13 March 2007 (UTC)[reply]

Why, dye my sister's dolls hair. (this was not a popular experiment) And I used to dye 7-up odd colors too. --BenBurch 00:25, 13 March 2007 (UTC)[reply]

Of course, I should have thought of Barbie's blue "do". Five time science fair winner :) - Nunh-huh 00:57, 13 March 2007 (UTC)[reply]

Perhaps the focus on "green" instead of "quality" means that bars can charge premium price for a nasty beer (with green dye). I've found the extremely low quality beers make you feel far worse than higher quality ones. --Kainaw ^(talk) 00:10, 13 March 2007 (UTC)[reply]

Maybe it's because you just can't handle the Irish stuff? Nil Einne 20:40, 13 March 2007 (UTC)[reply]

Our article on Saint Patrick's Day shows an entire river dyed green--VectorPotential^Talk 00:14, 14 March 2007 (UTC)[reply]

Global Warming and Greenhouse Effect

What is the difference between global warming and the greehouse effect? —The preceding unsigned comment was added by 199.2.112.34 (talk) 21:02, 12 March 2007 (UTC).[reply]

Check out our articles on Global warming & Greenhouse effect. Simply, the greenhouse effect is one of the main mechanisms that may cause global warming, but not the only possible one. -- Scientizzle 21:08, 12 March 2007 (UTC)[reply]

Actually the oceans would be frozen over all or most of Earth's surface and Earth would not be inhabitable as we know it if we didn't have some greenhouse effect. Global warming refers only to the phenomenon of human-induced enhancement of the greenhouse effect. -User: Nightvid

Correction: Global warming refers not necessarily to a human-induced or anthropogenic global warming, but just the recent global warming activity of the Northern hemisphere. [Mαc Δαvιs] (How's my driving?) ❖ 18:13, 13 March 2007 (UTC)[reply]

Simply: The greenhouse effect is the cause. Global warming is the effect. An increase in the amount of greenhouse gasses in the upper atmosphere is causing more of the sun's heat to be trapped - which is a "warming" effect - and it's happening "globally" - hence Global Warming. SteveBaker 23:15, 13 March 2007 (UTC)[reply]

Tea: the cup that cheers

In Britain tea has sometimes been referred to as "the cup that cheers", meaning it cheers you up and improves your mood (it is nothing to do with cheering or shouting out loud). I know from personal experience that it often does this.

What is it in tea that improves mood? Drinking coffee or cocoa does not improve mood in my experience, so it is unlikely to be something like caffeine which is also in those other beverages. I have looked through the tea article and linked articles and this mood enhancing effect is not mentioned.

Second question about tea: how does the amount of catechins in leaves of the tea plant compare with the amount in vegetables? How much vegetables or fruit would I need to eat to get the same amount of catechin as found in say, one cup of black or green tea? Also not mentioned anywhere. 62.253.49.0 21:38, 12 March 2007 (UTC)[reply]

I've read some articles on wilderness survival that recommend brewing a cup of tea when you find yourself in a survival situation. The reason for this is that it keeps you from panicking unproductively, it gets you to do something productive (building a fire), and it's a familiar and soothing activity. Perhaps the latter is the reason tea cheers you up. Making tea is a very personal and calming activity, where coffee is usually made by machine. Shui9 02:13, 13 March 2007 (UTC)[reply]

Caffeine cheers me up. So that could be it as well. Both coffee and tea make me happier for that reason. They clear my head. 70.108.199.130 04:59, 13 March 2007 (UTC)[reply]

For your first question, have a look at our article on theanine. — Matt Eason ^{(Talk  Contribs)} 14:13, 13 March 2007 (UTC)[reply]