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January 29

Sweet crude oil

I was in class and my prof mentioned that there is a type of crude oil called Sweet crude oil, and he said that workers would actually taste the oil. Isn't oil toxic? Wouldn't tasting it be detrimental to their health? ScienceApe (talk) 02:34, 29 January 2011 (UTC)[reply]

According to this document from the Oil Spill Academic Task Force of the State of Florida, "In the 19th century, oil workers would taste and smell small amount [sic] of oil to determine its quality." Clarityfiend (talk) 02:56, 29 January 2011 (UTC)[reply]

Many geologists are rather partial to having the occasional lick of anything that looks (or smells) interesting. DuncanHill (talk) 03:07, 29 January 2011 (UTC)[reply]

It looks like geologists like to put dirt in their mouth for several purposes, also to evaluate a sample's texture. See, for example, siltstone or, regarding oil, old formation evaluation tools to detect oil and gas by grounding the well cuttings between their teeth. They "tasted to see if crude oil was present", according to that article, unreferenced. Other sources seem to suggest it was for detecting texture, not taste. ---Sluzzelin talk 05:06, 29 January 2011 (UTC)[reply]

And they're not the only ones. Ask an archaeologist some time about how to tell the difference between a bone fragment and a flake of stone in the field. (The bone sticks to your tongue, but stones don't). Matt Deres (talk) 14:42, 31 January 2011 (UTC)[reply]

Tasting this oil is not detrimental to the health? ScienceApe (talk) 16:25, 29 January 2011 (UTC)[reply]

No one is claiming that! This (commercial) site lists some figures and hazards, toxicity in case of ingestion or skin contact included. The data are referenced. They don't directly answer your question, but it's pretty clear this isn't something you want to try at home, and certainly not repeatedly. Some of the long-term hazards from doing this repeatedly, such as carcinogenicity, probably couldn't have been detected soon enough to make a connection, at the time. (I wanted to point you to an article on the history of occupational health and safety, but, judging from the few articles that link to Bernardino Ramazzini, we have no such article, nor is it well-covered in any of the history sections). ---Sluzzelin talk 16:53, 29 January 2011 (UTC)[reply]

The level of benzene in crude oil before cracking is not very large, and cancer is definitely not guaranteed - though no level of exposure is truly acceptable. But bear in mind that petroleum was taken as a "medicine" in Western countries at a time when its use for fuel was obscure if not forgotten. Wnt (talk) 21:39, 29 January 2011 (UTC)[reply]

giving m₂=0 for newton formula

what happens for field when we give m₂=0 for newton gravity formula.(G) a. mohammadzade —Preceding unsigned comment added by 78.38.28.3 (talk) 11:26, 29 January 2011 (UTC)[reply]

See the article about Newton's law of universal gravitation which is expressed by the equation

F=G{\frac {m_{1}m_{2}}{r^{2}}}\

Any real physical object has mass, but if a hypothetical object had zero mass it would have no gravitational attraction to any other object. Cuddlyable3 (talk) 15:23, 29 January 2011 (UTC)[reply]

Well, there would at least be zero gravitational force acting on the zero mass object. But from F=ma (Newton's second law), calculating what acceleration results when an object with zero mass is acted upon by a zero force results in the indeterminate form a=0/0, so you can't conclude that the zero-mass object experiences no acceleration. What works better in this case is to combine Newton's two equations to give

a_{2}=G{\frac {m_{1}}{r^{2}}}\

,

where a₂ is the acceleration of the object you're calculating the acceleration of, and m₁ is the mass of the other object. That combined equation avoids the indeterminate form when m₂=0, and helps lead toward the more appropriate understanding of gravity as involving a coordinate acceleration as is produced by a fictitious force, rather than being a real force, as per Einstein's equivalence principle. As per that combined equation, a theoretical zero-mass object will still experience the same acceleration due to gravity as an object with nonzero mass, even though there is no force acting on it. Red Act (talk) 18:06, 29 January 2011 (UTC)[reply]

Yes, except that equation arises by dividing by zero. The best we can say in the Newtonian theory is that following the ordinary acceleration of gravity is the limiting behavior as the mass of a body goes towards zero. However, if there were any force acting on the body except for gravity, that force would completely dominate its behavior when its mass gets small enough. It could be a force that is so tiny that its effect is not observable for anything but massless bodies, and so the Newtonian theory cannot actually predict that massless objects move with the acceleration you quote (that is, "predict" in the sense that it would automatically falsify the theory to find a massless object that did not move so).

And actually, the massless objects we know (that is, photons) do not experimentally behave in gravity fields like Newton says they ought to. They are deflected by more than the Newtonian limit, thanks to GR and space curvature. –Henning Makholm (talk) 01:30, 30 January 2011 (UTC)[reply]

Perhaps I should have given more emphasis to the requirement that the above equation would only apply in the case of there being absolutely no (nongravitational) forces acting on the object, either forces that are now known or that might be discovered in the future (or perhaps even some physical phenomenon discovered in the future that results in a proper acceleration but can't be expressed as a force). But I did end the post with the phrase "even though there is no force acting on it", so that limitation should have at least been rather clearly apparent at that point, even if it hadn't been obvious enough before then.

Newtonian gravity as an approximation to general relativity falls apart at high relative speeds, regardless of whether or not the object in question has a nonzero mass. And the approximation falls apart at high speed for even geometrical considerations alone, so the approximation's breakdown can't just be fixed by special relativistic corrections to the mass. So I was treating the question as pertaining to a hypothetical zero-mass object moving at low speed, since behavior at low speed is the only condition under which any discussion of Newtonian gravity has any validity at all. So the behavior of photons is irrelevant to the hypothetical situation that I was considering.

The equation above is the correct equation for the (coordinate) acceleration due to gravity of low speed objects in the absence of any (real) forces in the weak-field approximation as predicted by general relativity, and the equation above can be derived using general relativity without ever taking the object's mass into account (assuming the mass isn't large enough to itself cause nonnegligible curvature of spacetime). The geodesic followed by a object experiencing no real forces is completely governed by the geometry of the spacetime on which the object is traveling; the object's mass has nothing to do with it. So there isn't a division by zero involved if you derive the equation using general relativity.

Indeed, any other behavior for even an object of zero mass would be nonsensical for even geometrical considerations alone. In the absence of any real forces, the object must have a zero proper acceleration, because the local isotropy of space would prevent the object from having any way to even choose a direction in which to accelerate. And if the object has no proper acceleration, the only other "acceleration" the object can have is the coordinate "acceleration" of the object's local inertial frame with respect to the chosen noninertial coordinate system. And the coordinate "acceleration" is a purely mathematical device that has nothing to do with actual physics, so any physical properties of an object like whether or not it has a mass is irrelevant to the expression of the magnitude of the coordinate acceleration as expressed in the equation above.

As with any fictitious force, the apparent acceleration of objects under the "influence" of gravity is most appropriately viewed as more fundamentally being a coordinate acceleration, that only looks like a "force" if you artificially multiply that acceleration by an object's mass. That's the general relativistic perspective, and that's actually the entire reason why I pointed out in my earlier post how gravity can be described as being an acceleration, with any notion of a "gravitational force" being removed. I wasn't trying to fully explore the acceleration of a hypothetical low speed massless particle that might actually be discovered in the future. So bringing up tiny additional forces that might be discovered in the future and what-not is really tangential to my intended point. Red Act (talk) 08:06, 30 January 2011 (UTC)[reply]

Anal orgasm

Is it possible for woman to have orgasm from anal stimulation alone, and how is it distinct from vaginal orgasm? --78.150.235.125 (talk) 11:43, 29 January 2011 (UTC)[reply]

Orgasm#Anal_stimulation provides some referenced information for this and this Slate article may also be useful. SmartSE (talk) 12:34, 29 January 2011 (UTC)[reply]

Balloons

How much force upward does a balloon exert? I recently saw Up, and I know getting a house airborne would be impossible, but I am curious as to how much helium would be needed to lift a person or a basket. And how about hydrogen balloons? What is hydrogen and helium's lifting power per square inch? --T H F S W (T · C · E) 18:15, 29 January 2011 (UTC)[reply]

Per square inch it's zero ;-). The molecular weight of Hydrogen (H₂) is 2 g/mol, the molecular weight of Helium (He) is 4 g/mol, and the average molecular weight of air is 80%*28+20%*32 (for Nitrogen and Oxygen, both diatomic gases), or just shy of 29g/mol. So one mol of Hydrogen can lift 27g, one mol of helium can lift 25g. One mol of gas under standard condition occupies 22.4l, so the lifting power of either gas is about 1g/l or 1 kg/m³. Of course, that is the gross lift - for the net lifting capacity, you need to subtract the weight of the balloon. If you need more exact numbers, apply a calculator ;-) --Stephan Schulz (talk) 18:27, 29 January 2011 (UTC)[reply]

There was the interesting case of Lawnchair Larry, who dared in 1982 to try this experiement "at home". He attached 45 helium-filled weather balloons to a lawn chair in his back yard and flew to an altitude of over 15,000 ft! (Asked why he did it, he replied, "[Because] a man can't just sit around.";) WikiDao ☯ 18:28, 29 January 2011 (UTC)[reply]

OK, thanks, I'll start collecting weather balloons! Joke - for now.--T H F S W (T · C · E) 18:48, 29 January 2011 (UTC)[reply]

N.B. his "honorable mention" at the Darwin Awards...;) WikiDao ☯ 19:36, 29 January 2011 (UTC)[reply]

Anyone know the area of a normal party balloon? And @Stephen, yeah, I meant a cubic inch. Heh heh. --T H F S W (T · C · E) 19:58, 29 January 2011 (UTC)[reply]

You're a quick study, right? ;-). Do you mean the surface area or the projection onto the ground? Or do you actually mean the volume? A 30 cm ballon has a volume of around 14 l. It weights maybe 1.5 g, so it can lift around 12 g. --Stephan Schulz (talk) 21:16, 29 January 2011 (UTC)[reply]

In real units. 1000 cubic foot of hydrogen at STP will lift 74 lb. So a cubic foot balloon is err.. I think 1.184 ounces of lift. Helium has a lift of 72 lb per 1000 cu ft.--Aspro (talk) 21:29, 29 January 2011 (UTC)[reply]

Fully saturated moist air weighs only 5/8 of dry air and is very much cheaper than H or He. You can create some by burning (say) propane or butane gas in a large bag. That will also heat it and increase the lift still further. Thinks.... If one were to leave said bag open at the bottom you could keep it hot by continuously burning more gas. Hey! I wonder if Microsoft has patented this idea yet? --Aspro (talk) 21:47, 29 January 2011 (UTC)[reply]

While water vapor (gaseous H₂O) is about 5/8 the mass of an equivalent volume of dry air at the same temperature and pressure, it's highly incorrect to say that moist air is 5/8 that of dry air. Humidity notes that at 30°C, saturated air contains 30 g per cubic meter of water, which equates to at most 18 g of lift per cubic meter. This compares to an air density of 1.164 kg/m³, so humid air would be at most 1.5% lighter than dry air. This compares to the ca. 80g/m³ lift you get by heating 15°C air to 35°C (cf. Density of air). It's true that if you had an equivalent volume of water vapor at the same temperature and pressure, you'd get 3/8ths lift from it, but at temperatures below 100°C, that's not possible, as all the water would condense, and you'd have a deflated container [1], or a partial vacuum. -- 174.21.236.191 (talk) 00:25, 30 January 2011 (UTC)[reply]

Quite right -the lift is pitiful. Maybe instead, a sealed balloon of pure water vapour, fabricated from aluminium foil backed bubble-wrap, with argon filled bubbles to reduce conductive heat loss. Then a small radioisotope heater unit to maintain the vapour at ≥100 deg C temperature. --Aspro (talk) 18:32, 30 January 2011 (UTC)[reply]

Of course, the weight of the balloon itself also has to be lifted. My intuition says that a single large balloon should require less gas to lift the same weight than a large number of small ones, although this is not 100% obvious, because the large balloon might need to be thicker. Anyway, balloonists normally do use a single large balloon, so it makes sense that that's best. (The great airships like the Hindenburg did use separate gasbags within an outer cover, but I think that was mainly in order to contain the effects of any damage.) --Anonymous, 00:05 UTC, January 30, 2011.

While not an WP:RS, this forum thread reckons 60lbs per square foot of living space, or 120lb per square foot of footprint for a house is the approx weight. Exxolon (talk) 16:55, 30 January 2011 (UTC)[reply]

Buckminster Fuller went as far as to propose airborne towns.--Aspro (talk) 18:21, 30 January 2011 (UTC)[reply]

Graphite

In graphite, are the layers stacked so that the carbon atoms in one layer are directly on top of the carbon atoms in the lower layer, or are they over the centers of the other layer's hexagonal holes? --75.15.161.185 (talk) 19:39, 29 January 2011 (UTC)[reply]

The latter. See: graphite. Dragons flight (talk) 19:41, 29 January 2011 (UTC)[reply]

Converting ohm to siemens

I'm doing a science project on the electrical conductivity of different materials. I have an ohmmeter and know how to convert ohms to siemens. The problem is, I'm testing piece of rubber and the reading just shows up as 1 on the 2M setting. How would I convert this value? Writing down 1 [S/m] doesn't seem correct. —Preceding unsigned comment added by 68.230.183.227 (talk) 19:49, 29 January 2011 (UTC)[reply]

A multimeter reading of "1" typically means the value is unmeasurable because it is beyond the range of that meter. Since you put the multimeter on 2 MΩ, you know that the resistance of the rubber is greater than that (R > 2 MΩ), so the conductance is less than (1÷(2 MΩ) = 500 nS), so G < 500 nS. --Link ^(t•c•m) 21:04, 29 January 2011 (UTC)[reply]

I talked with my dad and he basically said the same thing. Thank you so much though. —Preceding unsigned comment added by 68.230.183.227 (talk) 21:51, 29 January 2011 (UTC)[reply]

Central American Bat Identification

I recently photographed this individual roosting on a tree trunk, approx 8-10feet above ground level. Accompanied by one other individual, roosting a few feet above. Osa Peninsula, Costa Rica. Hopefully those lines of white pale fur are destinctive, thanks in advance Benjamint 22:36, 29 January 2011 (UTC)[reply]

Looks very like the bats shown in this query, which have been identified by a respondant as either Lesser Sac-winged bats (Saccopteryx leptura) or Greater Sac-winged bats (Saccopteryx bilineata). 87.81.230.195 (talk) 02:45, 30 January 2011 (UTC)[reply]