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April 22[edit]

Caffeine in Nesquik[edit]

How much caffeine is there in a 16 oz bottle of Nestle Nesquick fat free chocolate milk? The label simply says "99% caffeine free" without giving a mg amount. Magog the Ogre (talk) 00:32, 22 April 2012 (UTC)[reply]

I don't know the answer, but I'd like to take this moment to laugh out loud at the phrasing. Most of the drinks on this list have less than 1% caffeine by weight. One fluid ounce of water-based drink weighs approximately 30 grams, so anything listed there with less than 300 mg/fluid ounce will technically fit this description.-RunningOnBrains^(talk) 02:25, 22 April 2012 (UTC)[reply]

Ha! Good point! I wonder, does that phrase even have a legal meaning e.g. in the USA? Definitely <1% caffeine by weight is a low bar to cross :) SemanticMantis (talk) 03:25, 22 April 2012 (UTC)[reply]

Any amount higher than that should be fatal. Maybe the marketing department of Nestle can run with that and promote Nesquick with the slogan "It won't kill you." SkyMachine ⁽⁺⁺⁾ 08:17, 22 April 2012 (UTC)[reply]

"...at least not immediately." StuRat (talk) 13:47, 22 April 2012 (UTC) [reply]

This article appears from a google snippet to have the answer (it mentions chocolate Nesquik and has a table of caffeine content) but I don't have full access. By the way, the Nesquik company site makes the same claim for both the powder and the bottled drink. I am guessing that the "99% free" claim was merely transferred from the powdered product without any actual though being put into its meaning. One could deduce from that that the powder contains about 1% caffeine which will be diluted depending on how much milk it is mixed with. Spinning Spark 10:38, 22 April 2012 (UTC)[reply]

Flying car[edit]

Your proposed design has four fans each of radius 50 cm. Each fan sucks in air from above the car and blows it downward. The downward flow of air can be approximated as a column of radius equal to the radius of the fan traveling at uniform speed downwards.
A) How fast must the air be blown downwards if the car plus passangers weigh 1000 kg?
B) If the car is 100% efficient, how much power would it use while hovering?
The densidad of air is $1.2kgm^{-3}$

Is this the correct calculation for the upward force: $F={\frac {p}{t}}={\frac {mv}{t}}={\frac {n\pi r^{2}\rho hv}{t}}$ , and I think ${\frac {h}{t}}=v$ so $F=n\pi r^{2}\rho v^{2}$ . Overall I got $51ms^{-1}$ for part A and $2.5\times 10^{5}W$ for part B. Widener (talk) 08:25, 22 April 2012 (UTC)[reply]

Do your own homework. Whoop whoop pull up ^{Bitching Betty | Averted crashes} 17:50, 22 April 2012 (UTC)[reply]

The poster has done their homework. They have already shown an attempt at the question and are asking if they have made any errors. Spinning Spark 18:12, 22 April 2012 (UTC)[reply]

Your answers look correct. If this is for a calculus-based physics class, they're fairly certainly going to want you to work this out in terms of derivatives, i.e., start with F=dp/dt instead of F=p/t, but your equations work OK as long as you're careful about the definitions and assumptions being used. Red Act (talk) 01:19, 23 April 2012 (UTC)[reply]

For a general audience such as Wikipedia editors, please define each variable you use rather than just throwing out random Greek or Latin letters. Some readers took physics decades ago and the letters used might have been different. Few are likely enrolled in your particular class with access to your lectures, homework assignments and textbook.F=force?, t=time?, s=seconds?, mv=millivolts? What is nu?, What is rho?? Is h=Planck's constant? Meet us halfway. Edison (talk) 03:14, 23 April 2012 (UTC)[reply]

I wouldn't sweat this so much—the people who are likely to be able to assess this type of equation are probably fluent in the usual meanings of these symbols, and it's only important to define unusual or non-standard symbols. For basic physics formulas, F is always a force, p is always momentum, t is always time, m is always mass, v is always velocity, and ρ is always a density. h is generally either a height or Planck's constant; the correct choice should be obvious from context in nearly all circumstances (as it is here). There is no nu in the above equations; I'm guessing that you're (mis)reading the stylized v used in Wikipedia formulas.

There's no need to define standard SI units. ms^-1 is potentially ambiguous, as it could represent meters per second (m/s) or inverse milliseconds: (ms)^(-1); it may be best to write it as either m·s^-1 or m/s to emphasize that the m isn't a prefix. Not that it particularly matters in this situation, as the context clearly demands a velocity. W, finally, is always a watt.

If there are editors who aren't familiar with the notation used in these types of problems, I encourage them to ask—but there's no need to browbeat the OP. This usage would be familiar to any first-year undergraduate physics student. TenOfAllTrades(talk) 03:42, 23 April 2012 (UTC)[reply]

Perhaps to "any first year physics student" in recent years. "m is always mass" (except when it is "meters.") Right. There is probably a convenient listing somewhere in Wikipedia of today's common usages of Latin and Greek letters to represent things in science. A link to it would be appreciated. I have seen different textbooks use different conventions than those you cited, although I grant that there might be more standardization today than there was many years ago. Edison (talk) 17:53, 24 April 2012 (UTC)[reply]

At long last I found List of common physics notations, which is helpful in figuring out what is intended in such a question as this. Even that list has "n" only as as "refractive index" or "principal quantum number", which do not seem likely candidates. Amount of substance seems more likely. Would it be appropriate to add the latter meaning to the list of common physics notations, or would it have to go in nonexistent List of common chemistry notations? I Edison (talk) 17:20, 25 April 2012 (UTC)[reply]

Edison: italic m is mass, upright m is meters. Measurement units are written in upright letters. OP didn't get this right. – b_jonas 09:48, 26 April 2012 (UTC)[reply]

High buoyancy and low drag[edit]

If you build a very large capsule (hence with enormous buoyancy) but with a narrow pointed shap (hence very low drag), could you get this capsule to rise so fast it could actually rise up a waterfall?

According to my calculations, I see no reason why this could not be done. I calculate the terminal velocity as $v={\sqrt {\frac {2Vg(\rho _{w}-\rho _{c})}{C\rho _{w}A}}}$ . ( $\rho _{w}$ is the density of the water and $\rho _{c}$ is the density of the capsule). If you make ${\frac {V}{A}}$ sufficiently large then you can make $v$ as large as you want. If you can release it at any depth, then it can come as close to this terminal velocity as you want. Does anyone object to this? Widener (talk) 17:25, 22 April 2012 (UTC)[reply]

For a capsule in water, you cannot ignore parasitic drag, which increases linearly (IIRC) with total wet area. So your very aqua-dynamic capsule will still experience significant drag that is proportional to its volume. So in practical terms, I doubt it's possible. --Stephan Schulz (talk) 17:38, 22 April 2012 (UTC)[reply]

Oh I see, of course at high velocities the drag equation underestimates the true drag. Widener (talk) 17:47, 22 April 2012 (UTC)[reply]

I'm not much of a scientist, but I imagine that there aren't many waterfalls which are a column of solid water without any voids in them. As soon as your capsule finds some air, it's going to stop rising and fall out of the side of the waterfall. While I was typing that, I remembered that water with a lot of air bubbles in it (ie "whitewater") is a lot less bouyant than water without, so your caclations are going to go awry before you start. Good luck with your experiment. Alansplodge (talk) 01:14, 23 April 2012 (UTC)[reply]

You use hydrostatic pressure to calculate the buoyancy. It does not apply very well to falling water, whether it has air in it or not. Any vertical pressure gradient in the water would act against gravity and slow down the falling, but as far as I know waterfalls are close to free fall. --145.94.77.43 (talk) 05:06, 23 April 2012 (UTC) Though on a second reading I take you are planning to "jump" from a depth rather than starting at the base of the fall? --145.94.77.43 (talk) 05:12, 23 April 2012 (UTC)[reply]

There's a much more fundamental problem here than those discussed so far: bouyancy is driven entirely by the pressure gradient in a standing fluid. (It just happens that it can be stated simply in terms of density in that case.) In free fall, there is no such pressure variation (as the fluid is free to move around to suppress it) and bouyancy just doesn't happen. --Tardis (talk) 13:00, 23 April 2012 (UTC)[reply]

On rereading I see that this is fundamentally what 145.* already said. I should point out that an idealized cylinder of water falling through air will of course eventually reach a sort of terminal velocity where the shear from the air balances the acceleration due to gravity. There's no reason you can't (in principle) float up somewhat past this point before you reach too good an approximation to free fall to continue. --Tardis (talk) 13:11, 23 April 2012 (UTC)[reply]

Some Science questions[edit]

1)When the electric charge is quantized, how the quarks having fraction electric charge exists?

2)Why can't we use D.C. current everywhere? I mean, what if all the devices are also made to use dc. and a rectifier is added at each house.

3)Here is a practical problem.I have ammeter , 2 volt battery and a 2 volt dc rectifier. When i connect both voltage sources to ammeter why it shows different readings?V and R of ammeter is constant, so I must be constant according to Ohm's law.

For 1: Charge is still quantized, just maybe not in the size that was originally thought (i.e. quarks always have a charge that is a multiple of 1/3).

For 2: War of Currents. The problems are largely historical. There are some places with High-voltage direct current distribution. You could use DC current everywhere, but we don't, because it used to be unsuitable for long-distance distribution. Buddy431 (talk) 21:31, 22 April 2012 (UTC)[reply]

For number 3: How different? What are the actual readings for the battery and the rectifier? If we knew that, it would help answer the question. --Jayron32 22:01, 22 April 2012 (UTC)[reply]

This entirely depends on what the rectifier is being powered from. If it is the battery, a diode voltage drop would be expected at the output. Two volts is an unusual value for both batteries and rectifiers by the way. Also, you appear to be connecting the ammeter directly across the voltage sources. If this is so, there will be a significant volt drop at the source output and in general this will not be the same for both sources. Spinning Spark 22:10, 22 April 2012 (UTC)[reply]

There's something weird about the fractional electric charge. Quarks can't exist in combinations that have non-fractional charge, and can't transition in ways that would work out to an odd fraction. They go, say from 2/3 to -1/3, not 2/3 to 1/3, emitting W particles. This is sort of the way that electron spin can be +1/2 or -1/2, but the photons have to carry an angular momentum of just plain 1 (Planck's constant, that is). Not sure why things are that way at the bottom - is there a way to understand that intuitively? Wnt (talk) 00:50, 23 April 2012 (UTC)[reply]

For question 2: It is really simple to have an electric generation and distribution to serve a large area wherein the electricity is generated, then stepped up by simple transformers to a very high voltage AC at the generating station and sent in all directions, then stepped down to say 12 KV at substations every few miles, and distributed to a neighborhood, then stepped down to say 120 or 240 volts (depending on the country) by a transformer every block or so, to be sent into houses. With DC, the generating stations would have to be every 2 miles or so apart, producing 120 (or 240) volts, because of "copper losses." The same power could be provided by DC, but much, much more copper would be required. Copper currently costs close to 4$ (US) per pound. Edison (talk) 03:09, 23 April 2012 (UTC)[reply]
It makes me kind of uneasy that AC power is being advocated here by Edison, of all people. —Akrabbim^talk 12:40, 23 April 2012 (UTC)[reply]

Ha! DMacks (talk) 15:22, 23 April 2012 (UTC)[reply]

So far as I know the War of Currents stuff is sort of out of date. As I recall HVDC is now a serious contender for very long range distribution networks, like proposed transmission from solar or wind stations in the Sahara to Europe. (Of course, as you can tell from the sound of an idea like that, it may be a little starry eyed) In former times, changing DC voltage from one value to another was difficult, making it easier to use high-voltage lines (less loss) with AC, but as I understand it this is no longer such an obstacle. Wnt (talk) 14:09, 23 April 2012 (UTC)[reply]

What can be done now that could not be done decades ago is the use of switchmode power convertors to change the voltage. Until the development of high power Field Effect Transistors, efficient low cost switch mode convertors (aka DC to DC convertors) could not be made. Until 30 years ago all power operated consumer electronics used transformers to bring the 120V or 240V house power down to whatever voltage was required internally - now virtually all consumer electronics uses switchmode. However, don't expect switchmode to be used in electricity substations and street distribution any time soon, as the reliability, while good enough for consumer products, is no where near that of a transformer, and not good enough for power companies. Transformers are also very tolerant of short term (<1 second or so) overload, which tends to be frequent in power distribution. Switchmode on even small overload must shut down to protect itself, whereas transformers even on 300% or more overload just keep on delivering close to the normal voltage, allowing fuses & circuit breakers to operate where necessary, with discrimination. Discrimination is the feature provided by a hierachial arrangment of fuses or circuit breakers that result in only the minimum of consumers suffer blackouts due to faults. Keit120.145.185.73 (talk) 14:34, 23 April 2012 (UTC)[reply]

Thanks for a really informative response. I put the link above into standard Wikilink format (hope you don't mind). Wnt (talk) 23:29, 23 April 2012 (UTC)[reply]

For question 2, some devices, such as AC motors, require AC. Of the devices that use DC internally, such as radios, televisions, and computers, they need different values of DC, so it has been easier to have one standard AC voltage and build transformer-rectifier units to step down to the DC voltage needed by an individual device. The evolution of cheap and efficient DC to DC converters is much more recent than most of the electrical infrastructure. Jc3s5h (talk) 14:22, 23 April 2012 (UTC)[reply]

For question 1, it's hard to give a definite answer because we don't really know why charge is quantized in the first place! However, there are a few lines of reasoning that may shed some light on it. Quarks, electrons, etc. are all part of the standard model of particle physics, which is built on certain fundamental ideas like symmetry (specifically, gauge symmetry). In order to have a working gauge theory, there must be an exact balance between certain properties of the different particles. In particular, there's something called a gauge anomaly that will completely break the theory if it doesn't balance out. And one of the things that needs to balance is the total charge of all the particles in one "generation." That means that the charge of (electron + neutrino + red up quark + red down quark + blue up quark + blue down quark + green up quark + green down quark) = 0. Since the each quark comes in three colors, its charge contributes three times. So you can say in a sense that quarks have multiples of 1/3 times the electron charge because three quarks have to balance against one electron. Of course, this is a hand-wave-y argument, and there's a lot that we still don't understand about this topic. --Amble (talk) 02:42, 26 April 2012 (UTC)[reply]

Cholera rates in Kenya[edit]

Hi Reference Desk,

I'm look for the prevalence of cholera in Kenya. I can find [rates], but nothing on cholera. Please help me. Wiki or external links that directly inform me of this information would be great! Thank you. 208.22.79.249 (talk) 23:15, 22 April 2012 (UTC)[reply]

There's some information at WHO: [1]. It lists the number of cases of Cholera for many countries, including Kenya. RudolfRed (talk) 00:32, 23 April 2012 (UTC)[reply]

The WHO also has a short article here. Someguy1221 (talk) 00:36, 23 April 2012 (UTC)[reply]

Thank you. If anyone else has anything, please help. -original poster 208.22.79.249 (talk) 04:11, 23 April 2012 (UTC)[reply]