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Is there any online gizmo or app that I can use to determine the exact apparent position of the Sun against the sky on a specific date? I'd prefer to know in map form as I'm placing the image of the sun on a flat representation of the northern celestial sphere that will have no grid lines. The date I'm most interested in is March 29 (when it is in Pisces, but I don't know where in that constellation it would be), but if there's an app out there for any date I'd like to know. Thanks! --NellieBly (talk) 01:44, 19 December 2012 (UTC)[reply]
Google "Sun position calculator" for many choices. hydnjo (talk) 03:06, 19 December 2012 (UTC)[reply]
I like Stellarium. Does what you seem to want, and much more. HiLo48 (talk) 03:54, 19 December 2012 (UTC)[reply]
Thanks. I did google and found a number of calculators that purported to be useful but only gave me a numerical answer, which is basically 100% useless for my purposes. Thanks again! --NellieBly (talk) 04:43, 19 December 2012 (UTC)[reply]
If you want an "exact" position, then a numerical answer is the way to go. Get a star map that shows the degrees of the ecliptic (the path of the sun). Then you can plot the position on there before transferring it to your own map.--Shantavira|feed me 08:37, 19 December 2012 (UTC)[reply]
And if you want an image readily made for you, try one of the many planetarium software freely available on the web (Including the excellent Stellarium mentioned above). Dauto (talk) 14:44, 19 December 2012 (UTC)[reply]
What is the furthest object that is not moving away from the Milky Way galaxy? I read that the Andromeda galaxy will eventually merge with us, so I guess it's not moving away like the rest of the universe. Is the rest of the Local Group also gravitationally bound to us? What about the Virgo Supercluster?
I found this, and the abstract says that there are blueshifted galaxies in the Virgo supercluster, and that this may be due to their falling towards the area surrounding M86, or to "being repelled from the local cosmological void". I suppose it stands to reason that there will be movement both towards and away from any given point, and that this will not necessarily be due to any gravitational attraction by that point. --TammyMoet (talk) 09:10, 19 December 2012 (UTC)[reply]
There are long-period comets that may have spent time outside our galaxy before we detect them entering our solar system. Some have orbits that take them far beyond the outer planets at aphelia, and the plane of their orbits need not lie near the ecliptic. Long-period comets such as Comet West and C/1999 F1 can have barycentric apoapsis distances of nearly 70,000 AU with orbital periods estimated around 6 million years. DreadRed (talk) 12:23, 19 December 2012 (UTC)[reply]
You seem to be confusing "Outside of the solar system" with "Outside of the galaxy". Dauto (talk) 14:39, 19 December 2012 (UTC)[reply]
Will you clarify the difference when it concerns comets in orbit planes not on the ecliptic? DreadRed (talk) 08:09, 20 December 2012 (UTC)[reply]
(1) The solar system includes things not on the ecliptic that are orbiting the sun. (2) Someone might define the extent of the solar system as being the extent of the heliosphere, within which the solar wind dominates the interstellar wind, and the Oort cloud of comets extends out beyond that. (3) But still the solar system and the Oort cloud are just a tiny corner of our galaxy, the Milky Way galaxy, which includes at least 200 billion stars. "Outside the galaxy" means beyond that. For comparison, the Oort cloud may extend one light year from the sun, whereas the Milky Way has a diameter of 100,000 light years. Duoduoduo (talk) 14:55, 20 December 2012 (UTC)[reply]
Relation between mass -radius -density of star[edit]
If there be any star 2 times massive than sun ,how much will be density and radius of it ?and so on.--Akbarmohammadzade (talk) 09:39, 19 December 2012 (UTC)[reply]
The table in Main_sequence#Sample_parameters gives for a star of 2.1 M☉ a radius of 1.7 R☉. The density of that star will be (3/4π)(2.1 M☉) /(1.7 R☉)3 = 0.43 ρ☉. --Wrongfilter (talk) 13:25, 19 December 2012 (UTC)[reply]
I'm planning to cold concentrate sulfuric acid by using a dessicating agent. The idea is to add a hygroscopic anhydrous sulfate salt to 30% sulfuric acid. Leaving it for a few minutes, and deccanting of the the liquid fraction. Repeating it until the system has reached an equilibrium state where the salt reached its dessicating potential. Which salt would be most suitable? Plasmic Physics (talk) 10:05, 19 December 2012 (UTC)[reply]
I thought about using copper sulfate to use the colour change as an indicator, but the copper would foul the acid and interfere with my nitric acid production. Magnesium sulfate has a greater hydrate capacity, but it lacks the colour change. Plasmic Physics (talk) 10:19, 19 December 2012 (UTC)[reply]
At what acid concentration does anhydrous copper sulfate no longer absorb water? What is the sulfate's solubility at that concentration? Plasmic Physics (talk) 10:56, 19 December 2012 (UTC)[reply]
One way to get an idea is to look at the vapour pressure of water over these various substances. I suspect that you will find that concentrated sulfuric acid will draw water from your copper sulfate. Graeme Bartlett (talk) 11:47, 19 December 2012 (UTC)[reply]
Yes, but at what concentration? Non-contact dessication via evaporation would take ages. Plasmic Physics (talk) 12:37, 19 December 2012 (UTC)[reply]
Oh, I get it. Does a low vapour pressure indicate a high water affinity? Plasmic Physics (talk) 09:44, 20 December 2012 (UTC)[reply]
Yes, so the substance with higher vapour pressure will put out water that the thing with lower vapour pressure will absorb. As you say when that pressure is very low, the rate will also be low. But the gas phase will isolate the salt hydrate from the aic. But I think the sulfuric acid will absorb water from your copper sulfate. After all in low humidity copper sulfate dehydrates. You may be better off using silica gel, which will cause less damage to sulfuric acid. Graeme Bartlett (talk) 12:19, 21 December 2012 (UTC)[reply]
Fixed your typo; please undo if I misunderstood. --Trovatore (talk) 10:25, 22 December 2012 (UTC) [reply]
It's a shame that I don't want to make more acid. Plasmic Physics (talk) 07:07, 23 December 2012 (UTC)[reply]
Is it normal for sulfuric acid to dissolve borosilicate labware at a rate that the item can only be used once/twice? I heated 37% sulfuric together with potassium nitrate, and after 4 hours of use, the boiling flask was corroded to the point of me accidentally sticking a hole straight through the bottom with a glass stiring rod. Plasmic Physics (talk) 07:12, 23 December 2012 (UTC)[reply]
No, just a fresh batch of nitric acid. Plasmic Physics (talk) 07:23, 25 December 2012 (UTC)[reply]
I wonder what you need to make the nitric acid for, since it is the quintessential bomb-making chemical, but okay. Anyway, it's not normal for either sulfuric or nitric acid to etch glass, regardless of concentration -- I've worked with both of them (at maximum concentration), and I never observed any corrosion or weakening of the glassware. Hydrofluoric acid, on the other hand, will rapidly dissolve glass or any other silicate materials, and for that reason must only be handled in plastic (Teflon) labware. 24.23.196.85 (talk) 02:40, 26 December 2012 (UTC)[reply]
It's for making iodine, among other things. Perhaps, my potassium nitrate was contaminated by calcium fluoride? Afterall, it was obtained from a fertiliser plant. Plasmic Physics (talk) 03:30, 26 December 2012 (UTC)[reply]
That is indeed a possibility. If so, you might try purifying it by dissolving it in water, adding calcium hydroxide or calcium nitrate (to depress the solubility of fluoride), filtering the solution thoroughly, and then evaporating the purified KNO3 solution, and see if it alleviates the problem. 24.23.196.85 (talk) 01:48, 27 December 2012 (UTC)[reply]
I have been having this strange thought lately... Could it be possible, that everything is just a pigment of my imagination, and I'm all but living a weird dream, and will soon wake up to a new reality? Imagine, when you cease to exist, everythng in your perspective would be nothing, as if we did nothing all. Could it really be? Life is an enigma..... BonkersThe Clown(Nonsensical Babble) 11:28, 19 December 2012 (UTC)[reply]
Sign over to me all your wealth, and I will see if I can help you find out? Not a pleasing proposition? Even dreams are real. μηδείς (talk) 03:27, 20 December 2012 (UTC)[reply]
Flemish painter Jan Van Eyck, working in the 15th century, did not ordinarily include blue in his paintings because of the non-availability of an affordable blue pigment. Thus for him blue was a pigment of imagination. DreadRed (talk) 12:39, 19 December 2012 (UTC)[reply]
Dolittle is talking to the computer on a massive planet-busting bomb that has incorrectly started to count down towards zero. He hopes to persuade it to not blow up: Dolittle:"Hello, Bomb? Are you with me?" ... Bomb:"Of course." ... D:"Are you willing to entertain a few concepts?" ... B:"I am always receptive to suggestions" ... D:"Fine. Think about this then. How do you know you exist?" ... B:"Well, of course I exist." ... D:"But how do you know you exist?" ... B:"It is intuitively obvious." ... D:"Intuition is no proof. What concrete evidence do you have that you exist?" ... B:"Hmmmm... well... I think, therefore I am." ... D:"That's good. That's very good. But how do you know that anything else exists?" ... B:"My sensory apparatus reveals it to me. This is fun!"....and so on.
See also: Simulation hypothesis - which I'm beginning to suspect might explain a lot about the nature of our universe - and could even be true.
Sure, it's possible that we're all a part of your dream (or that you are a part of ours) - but you have to consider how it would be that the "real" you could construct such an incredibly detailed dream - something so incredibly self-consistent and full of billions of fully fleshed-out characters? That seems very unlikely to me - unless the "real" you is a much more capable being than a mere human. You'd have (for example) to be able to dream the whole of Wikipedia - stuffed with more knowledge than a human brain could possibly hold. Since you couldn't have all of that information in your head - you'd have to invent most of it as you dream of clicking on links...but invention of something that rich and complex also seems beyond human mental capacity.
So we're now left with the option that "you" are some kind of vastly more capable being (one would think to apply the word "god-like") who is merely dreaming of being human. If this being is not human then how do you know that the "you" in your dreams is that being? If it's not human isn't it capable of constructing "you" in it's thoughts too? Now you're in the realms of religion - that some godlike being is dreaming (or perhaps just mentally constructing) everything we see, know and experience. So now you just invented God...or maybe just an amazingly powerful computer (hence Simulation hypothesis).
It boils down to this: Yes, you might be right. But should that cause you to act or behave any differently? If I was utterly convinced that what I experience was merely a dream, I might be more strongly tempted to push the limits and see what happens - knowing that the worst that could happen would be that I would wake up, with no actual consequences whatever. That's a very dangerous view to take on life - and could easily result in self-destructive behavior.
So the most rational view is to presume that this is actual reality...and if it turns out to have been nothing more than a dream, you're not any worse off.
I am still waiting for "Bonkers" to email me offering to transfer all his wealth to me and releasing me from any damages incurred if I punch him in the face or strangle him to prove whether he'll wake up in an afterlife (or whatever) once I do so. Waiting.... μηδείς (talk) 17:57, 20 December 2012 (UTC)[reply]
"Pigment of my imagination"? No, you have it backwards - "I'm pink, therefore I'm spam..." AndyTheGrump (talk) 07:15, 23 December 2012 (UTC)[reply]
promising technologies in natural resource management[edit]
machines used in agriculture
machines used in fruit processing
machines used in processing of precious stones
machines used in forest management
machines used in water treatment and management
machines used in air quality management
machines used in fish treatment
machines used in food processing — Preceding unsigned comment added by 39.54.129.90 (talk) 16:19, 19 December 2012 (UTC)[reply]
If you're asking for a list of all those things, you're asking for too much. Looie496 (talk) 16:53, 19 December 2012 (UTC)[reply]
I saw a TV program about mapping the universe, out to more than 5 billion light-years. But aren't they assuming something? They are seeing galaxies where they were 5 billion years ago - they will be somewhere else "now". (I know there is no universal now.) Bubba73You talkin' to me? 16:23, 19 December 2012 (UTC)[reply]
But as you've just noted, there is no "universal now". "As currently observed from here" is as good as it gets in that department. That said, anybody engaged in such an effort is undoubtedly aware of that caveat. Further, it's not as if we need a map for traveling several billion light years (which is one of the few areas where that discrepancy really matters) -- but even if we did, we'd just do the same thing we do for all other travel: observe current position, velocity, acceleration, and such, and estimate where it will be when we arrive. It's the same thing space agencies do on every rocket launch, just on a bigger scale. — Lomn 16:48, 19 December 2012 (UTC)[reply]
So if they know where they are most likely "now" why dont they just make the adjustments and remap?165.212.189.187 (talk) 18:08, 19 December 2012 (UTC)[reply]
To what end? "As currently observed from here" is hard fact, and from any bound-by-relativity perspective, is the only meaningful "now" with respect to Earth, and is not blurred by the estimates and errors produced by extrapolation. In terms of a scientific survey of the universe, it's entirely appropriate. Even some crazy-hypothetical billion-light-year spaceship mission doesn't care about the "now" you're referencing but the "where when we get there". — Lomn 18:36, 19 December 2012 (UTC)[reply]
Are they careful with conclusions they draw? (I assume they are.) But they talk about clumps and voids, but those might never have existed. Bubba73You talkin' to me? 20:58, 19 December 2012 (UTC)[reply]
They certainly exist(ed); they're observed. Their future evolution does not alter that fact. — Lomn 21:43, 19 December 2012 (UTC)[reply]
What I mean if the map shows a large structure, say 500 million light years across, that may never have existed because parts of it that we see now were a long way away from each other. That is, we might be looking at part of it were it was 2 billion years ago and the other part were it was 2.5 billion years ago. Bubba73You talkin' to me? 02:21, 20 December 2012 (UTC)[reply]
Because of the speed-of-light limit, structures half a billion light years across are not coupled at time scales shorter than that anyway. It exists if anything of that size can be said to exist. In spacetime we see a "diagonal" slice of the object instead of a "horizontal" slice, and that doesn't really matter. -- BenRG (talk) 04:47, 22 December 2012 (UTC)[reply]
I find Google maps quite useful, and they are not real-time either. μηδείς (talk) 03:24, 20 December 2012 (UTC)[reply]
Yes, but the continents didn't move appreciably since they were taken. Bubba73You talkin' to me? 03:26, 20 December 2012 (UTC)[reply]
Medeis' example is fairly apt, though -- stellar structures, as a rule, don't move with respect to nearby structures at anything approaching the speed of light, and so a picture pieced together over a relatively short time frame remains accurate at broader scales. We might note that Google Maps isn't capturing the appearance and disappearance of potholes, and this would be analogous to individual supernovae and the like within astronomic megastructures, but those supernovae don't impact superclusters any more than potholes alter the route of I-40. — Lomn 03:38, 20 December 2012 (UTC)[reply]
Exactly, thanks. μηδείς (talk) 17:54, 20 December 2012 (UTC)[reply]
I think that if present observations from Earth show a collection of stars 500 million light years across, and let us say that their observed arrangement is globular, it is possible that they never really had the arrangement that we are observing, because the light reaching Earth from the most distal of those stars, is reaching Earth 500 million years after the light from the most proximal of those stars. I think I am re-stating what Bubba73 is saying. I agree with it. Bus stop (talk) 03:57, 20 December 2012 (UTC)[reply]
The example I gave was a structure about 1/10 the size of the map. So consider something roughly 1/10 the size of the Earth - the distance from North America to Europe. Now suppose the Google Earth map of America was made 400 million years ago and the map of Europe was made 500 million years ago. It probably wouldn't show the two continents in their proper relationship. Bubba73You talkin' to me? 04:46, 20 December 2012 (UTC)[reply]
But now you've badly mixed timescales. Sure, over 100 million years, the continents can massively shift -- because the Earth is far, far less than 100 million light years in scale. Take your 100 mly superstructure, though: there is a 100 my lag between observations, for certain not-entirely-useful definitions of "between". However, no physical object can possibly cross that gap during the course of the observation lag, because of light speed limits. Practically speaking, most of those objects are probably moving at no more than 1% or so of light speed. Thus, motion over the observation period is, at the large scale those astronomers are interested in, insignificant. To return to your Atlantic discussion, it's like asking whether the distance from New York to London is 5500 km or 5600 km. Sure, that's some room for distortion, but a map with that error range is still perfectly useful for discussing the size and shape of the Atlantic. And again -- all this is predicated on assuming some notion of distant simultaneity which does not exist. What we observe, for that distant superstructure between 400 mly and 500 mly away, is also what the near edge of that superstructure observed 400 my ago, and that observation is in accordance with all forces acting from that superstructure. It is the real thing. — Lomn 14:23, 20 December 2012 (UTC)[reply]
Ultimately, most of this question seems to boil down to the fundamental linkage of space and time (the whole "looking in a telescope is like looking back in time" thing). I'll pull this closer to home and illustrate: Alpha Centauri is 4 light years away. Suppose that, for an assumed "simultaneity", it went supernova 3 years and 364 days ago. It is the Mayan apocalypse, and will destroy all life on Earth (this is all bogus, but that's not the point). Is it meaningful to talk about that supernova as having already happened? Or, more specifically, is it any more meaningful than to talk about our present observations? No. A Cent is still a normal star. Some photons reach the earth, and its radiation does not sear us to our bones. No information about what A Cent might have done over the past four years, from the perspective of an A Cent observer, can reach us ahead of the speed of light observations of A Cent itself. That "past four years" phrase is no more meaningful than (in fact, it means the same as) discussing what A Cent will do for the next four years, starting now. A Cent is either four light years away from us or four years away from us, but those numbers can't be added together. And A Cent is a high-proper-motion star -- over a millennium, it moves twice the width of a full moon relative to background stars. That's enough motion to observe over a four-year span with good equipment. But all interactions that A Cent has with us -- light, gravity, radiation, etc -- comes from where I look up and see it right now, not from some projected "where it will be in four years" point. That point, that star, won't happen for four years. It's not real yet. — Lomn 14:40, 20 December 2012 (UTC)[reply]
It may also be worth noting, though, that some other observer somewhere else won't see the same thing. Returning to the large-scale superstructures, an observer on the other side of the Sloan Great Wall won't see what we do, because different objects are at relatively different distances (and velocities) from his perspective. And his view is also the real thing, because of the relativity of simultaneity. Both observers are right, and no observer is privileged or "more right". What we see is real, but what someone else sees -- though it be different -- is also real. — Lomn 14:48, 20 December 2012 (UTC)[reply]
Lomn says to Bubba "But now you've badly mixed timescales." But that's exactly Bubba's point: that when we make a map of the universe consisting of current observations, we're mixing time scales to a very large extent (to Bubba's chagrin), since the photons from one star we see now were emitted at a very different time than were those from another star we see now. Lomn's point is that astronomers are aware of that, they take it into account as needed, and they don't need to have a map that doesn't "badly mix timescales". Duoduoduo (talk) 15:23, 20 December 2012 (UTC)[reply]
Well, no -- that's a secondary point at best. The main point is that "mixed timescales" as Bubba discusses are a flawed understanding of how the universe works. Current observations, regardless of distance, represent the real universe that we interact with. — Lomn 15:34, 20 December 2012 (UTC)[reply]
That's only true if we stay pretty much stationary with respect to the faraway stars and galaxies. It becomes a problem when we start traveling. On the moon, for example, there was a delay of a second or two between transmitting and receiving, between the astronauts and the space center. I don't think anyone would argue that there was some unique "now" at each end, as each was aware of the delay. "Now" was the same for both. Whether it's a light-second, a light-year, or a light-billion-years, it's the same principle. Obviously, though, a star map for someone planted on earth needs to match what we can see, not what it actually looks like a billion years later. But it would be intersting. Perhaps what the crab nebula looks like "now" could be extrapolated somehow. ←Baseball BugsWhat's up, Doc?carrots→ 06:12, 21 December 2012 (UTC)[reply]
No, that's not true (in no small part because we are not pretty much stationary wrt distant astronomical objects). There is no meaningful absolute position or speed, and so there is no "only true if" privileged position as you posit. Apollo astronauts on the moon observed a different universe than did earthbound Mission Control (specifically, the two parties would disagree about the order of an event on the moon and an event on earth occuring within 1 second of each other. And each would be correct). Everybody understood the time gap, and how to adjust for it if necessary, but that's not at all the same as saying it didn't exist. — Lomn 14:22, 21 December 2012 (UTC)[reply]
Bugs is simply referring to the very real principle of synchronisation. --Modocc (talk) 14:55, 21 December 2012 (UTC)[reply]
To be fair to the original question: many popular science documentaries and "poster charts" of the universe do badly mix timescales and blur facts into interesting sound-bytes and colorfully incorrect diagrams. I hope that if you're planning an interstellar journey, or even just undertaking any learning about any astronomical observations, that you find some fairly rigorous sources of knowledge - just as you wouldn't use a cartoon-ish novelty-map to navigate across an ocean. Professional astronomers spend great effort to make sure that their work is well-informed by the appropriate relativistic corrections, correct type of distance measurements, and so forth. Consequently, their work, and the maps of the universe they produce, tend to be much more difficult for the average person to consume - exactly like if you tried to read an aviation chart or a navigational chart for piloting a busy shipping channel. For example, you might start with the Sloan Digital Sky Survey, which is a scientifically rigorous "general purpose" astronomical map. Specialized astronomical observations use different types of references. Nimur (talk) 19:17, 20 December 2012 (UTC)[reply]
There's a chronology of the universe since the big bang which is basically the same everywhere (because the universe is homogeneous) and when you see something "as it was 5 billion years ago" you're seeing it at around the 8.5 billion year point in that chronology. The fact that we can see earlier eras this way is useful; it wouldn't be better to extrapolate the images to the present day, even if we could. -- BenRG (talk) 04:47, 22 December 2012 (UTC)[reply]
For what reason the force component which is perpendicular to the object's velocity, changes its direction but does not change its speed? I know the argument that perpendicular force does not do work and therefore can not change the kinetic energy, but I am looking for more straightforward explanation. Thanks in advance, 94.159.217.166 (talk) 19:00, 19 December 2012 (UTC)[reply]
Because a perpendicular (sometimes called orthogonal) force has no component which is either working to move the object forward or backward. That is, this perpendicular force doesn't make the object move faster or slower (in the initial direction of travel), and as such, it doesn't affect the velocity in that direction. --Jayron32 19:12, 19 December 2012 (UTC)[reply]
If a force does not change an object's velcity, it is because the object is constrained not to move in the direction of the force. For example, a downward force on a shopping trolley does not move it becuasue teh ground under the wheels stops it moving downwards. Therefore no kinetic energy is imparted. Quite different to a horizontal force for which the wheels allow acceleration. Ratbone121.215.50.184 (talk) 01:57, 20 December 2012 (UTC)[reply]
The OP's premise is "the force component which is perpendicular to the object's velocity, changes its direction but does not change its speed". Jayron interpreted this to mean "its speed in its initial direction", and Ratbone interpreted it to mean that its velocity -- speed and direction combined -- was not changed, implying some countervailing force. To clarify for the OP, assuming no countervailing force canceling out the exterior force, the addition of the perpendicular force does not change the object's speed in its initial direction; but it does impart speed in the perpendicular direction. So now the velocity of the object consists of its new direction combined with its speed in the new direction; the speed in the new direction is found non-relativistically (someone correct me if this is wrong) as the square root of the sum of the square of the (unchanged) speed in the original direction and the square of the imparted speed in the new direction. Duoduoduo (talk) 15:35, 20 December 2012 (UTC) And the speed in the new direction is greater than the speed in the original direction, since the hypotenuse of a right triangle is longer than either of the legs. Duoduoduo (talk) 15:54, 20 December 2012 (UTC)[reply]
If I understood you properly, I don't think you are right. The force component which is perpendicular to the velocity does not change the speed at all. Only the tangent component changes the speed. 94.159.217.166 (talk) 21:31, 20 December 2012 (UTC)[reply]
Illustration of tangential and normal components of a vector to a surface. Again, you have to distinguish between speed in the original direction, which is not changed but which is only one component of after-the-fact speed, and speed in the new direction, which is a combination of the unchanged speed in the original direction and the imparted speed in the perpendicular direction.
You can visualize it using vectors. The original motion is depicted by a vector going in one direction, say horizontally, with speed indicated by the length of the arrow. The imparted motion is depicted by a vector going perpendicularly, say vertically, again with speed indicated by the length of the arrow. The combined motion is given by completing the rectangle and drawing a vector from the starting point to the far corner of the rectangle. By the Pythagorean Theorem, this diagonal vector is longer than either of the two vectors giving rise to it. Euclidean vector contains this diagram. You can think of the motion as occurring from the origin to the endpoint in one unit of time. Duoduoduo (talk) 22:07, 20 December 2012 (UTC)[reply]
You appear to be assuming that a force is applied for a while such that the force is always perpendicular to what the object's initial velocity was. But what the OP is asking about is the case where a force is applied for a while such that at any instant, the applied force is always perpendicular to the object's velocity as of that instant. In that situation, the object's speed remains constant, with only the direction of the object's velocity changing. Red Act (talk) 00:05, 21 December 2012 (UTC)[reply]
I think the OP is thinking about circular motion, where a constant centripetal force changes the direction of motion but not the speed. Otherwise, Duoduoduo's description is correct - a force, even in a perpendicular direction, will usually change the speed. As an example from orbital mechanics if you have a satellite orbiting the Earth, it has a constant speed, but a changing velocity caused by the gravitational attraction of the Earth. If an additional force pushes the satellite towards the Earth (perpendicular to the direction of motion), the orbital radius decreases and the speed increases (if I recall correctly).--Wikimedes (talk) 00:22, 21 December 2012 (UTC)[reply]
No, a total force that's always perpendicular to an object's velocity at all times will not change the object's speed under very general conditions, not just in the case of circular motion.
Let be an object's velocity, a time-varying vector. Let be the total force applied to the object, also a time-varying vector. In Newtonian mechanics, at any instant, .
and are perpendicular only if , which implies that .
An object's speed is , where is a time-varying scalar. Taking the derivative, how an object's speed changes with time is in general given by . But in the special case of force and velocity being perpendicular, , so in that case , i.e., the speed does not change. Red Act (talk) 02:41, 21 December 2012 (UTC)[reply]
I stand corrected. Good answer to the original question as well. (I'm still curious why my orbital mechanics example doesn't apply....)--Wikimedes (talk) 13:29, 21 December 2012 (UTC)[reply]
It doesn't apply because to move into a lower orbit the satellite must spiral inwards and during the spiral motion Earth's gravity is not exactly perpendicular to the motion. Dauto (talk) 15:09, 21 December 2012 (UTC)[reply]
Red Act has shown correct mathematical steps, but under what circumstances does is a force perpendicular to velocity? When the motion of the body is constrained, of course. This can only be satisfied if the constraint is to move in a circle with finite radius, or a circle with infinite radius (a straight line perpendicular to the force). Ratbone 121.215.30.187 (talk) 15:38, 21 December 2012 (UTC)[reply]
@Dauto - I understand so far, but what if the additional force acts in a direction so that the total force on the object (additional + gravitational) remains perpendicular to its velocity?--Wikimedes (talk) 16:17, 21 December 2012 (UTC)[reply]
In that case than, as Red Act skillfully demonstrated, the speed remains constant, but the additional force will eventually be a large force equal and opposite to gravity's force leading to uniform motion That's not the outcome you described. Dauto (talk) 18:50, 21 December 2012 (UTC)[reply]
No, Ratbone, the total force on an object being perpendicular to the object's velocity does not at all imply that the object must be moving in a circle or along a straight line. Indeed, as long as you always move an object at a constant nonzero speed, you can move the object along any path you want, and the total force being applied to the object will always be perpendicular to the object's velocity.
For an object in general, . But for any object moving at a constant speed , , which means from the above equation that , so . Red Act (talk) 02:55, 22 December 2012 (UTC)[reply]
Your math essentially states that if an object has a certain speed, and a force applied to it does not change that speed, then that force must be orthogonal to the direction of velocity. Well and good. But if the object is unconstrained, the force will accelerate it in the orthogonal direction. And the scalar (absolute) speed |V|, being a positive function of the velocity in all three coordinates, must increase, and kinetic energy increase, as the component speed in the original direction cannot change without a force component in the same direction. Therefore, the object is constrained. Ratbone121.221.223.13 (talk) 08:54, 22 December 2012 (UTC)[reply]
That's not correct Ratbone. Red Act's math works along any path as long as the speed is kept constant. Dauto (talk) 13:05, 22 December 2012 (UTC)[reply]
Please explain, or provide an example of a path that is not circular or straight. Repeated statements are not a proof and do not assist. Ratbone124.182.36.35 (talk) 13:17, 22 December 2012 (UTC)[reply]
Red Act already provided the proof. No further proof is necessary. The proof is general and works for any (smooth) path so no path examples are necessary, but if you really must have one, take a Lissajous curve for instance. it's neither straight nor circular. Dauto (talk) 13:55, 22 December 2012 (UTC)[reply]
A physical object moving in a lissajous trajectory is constantly changing its' speed |V| and its' kinetic energy (observing makes this obvious, and comes from it being constrained to move in simple harmonic motion at different rates in 2 dimensions), unless it is the special case of the two constaint periods being the same, giving a circular path, and therefore does not comply with the OP's conditions, except for teh special case, and also will not satify the conditions of force always orthogonal to the direction. There are many situations for which a mathematical proof or computable value exists, but such proof or value has no real world significance, or only a restricted real world significance, as I think applies here. Any good engineer, and almost all physicists, recognise that a mathematical proof should be backed up by a physical understanding (though this is not always possible). Hence, while Red Act's math contains no error in math, some explanation is required if all you can do is repeated what it says (that if speed is unaffected, the force must be orthogonal to the direction of travel). Ratbone124.182.36.35 (talk) 14:50, 22 December 2012 (UTC)[reply]
You're being boneheaded. An object may very well follow the Lissajous trajectory at constant speed in which case the force will be perpendicular to the trajectory at all times. Sure, you can construct a Lissajous curve with two orthogonal harmonic motions but that doesn't mean that an object must follow that trajectory with the same motion used to construct the trajectory. You're free to make the object follow the path with constant speed if you will, in which case the OP's condition is satisfied. Dauto (talk) 15:11, 22 December 2012 (UTC)[reply]
Not boneheaded, just a bit thick. A lissajous trajectory (or any other trajectory) can be considered as a circular path with a varying radius. So, yes, a constant speed magnitude can be achieved in any path with an orthogonal force. In a non closed circular path, the orthogonal force must vary, (periodically in a closed lissajous path) as Red Act pointed out, but nothing in the OP's question says that isn't allowed. My error. I had seen this orthogonal force idea before but only in connection with circular paths. Ratbone 124.182.144.218 (talk) 02:33, 23 December 2012 (UTC)[reply]
May be now I'm the one that's being a bit thick, but I can't tell the difference between "boneheaded" and "thick". Dauto (talk) 01:37, 24 December 2012 (UTC)[reply]
I was assuming a billiard ball moving left to right a fixed speed, which is then hit exactly perpendicularly by another ball whose mass and speed are such that it stops or bounces back at the point of impact (or continues forward but more slowly than the original ball) so that the contact between the two balls is instantaneous only. How does this fit in (or does it fit in) with Red Act's analysis? Duoduoduo (talk) 16:30, 21 December 2012 (UTC)[reply]
The problem with the situation you describe is the word instantaneous. In reality there is no such a thing and the contact force acts during a small - but finite - time interval. Even though the force is initially perpendicular to the motion, by the end of the short interval they won't be perpendicular any more. Conversely, if you postulate an instantaneous contact, the acceleration will be infinite, and the dot product will be indeterminate. Zero times infinite is indeterminate. Dauto (talk) 18:45, 21 December 2012 (UTC)[reply]
What happens if you take the limit as the contact time goes to zero? Duoduoduo (talk) 19:28, 21 December 2012 (UTC)[reply]
Then your original answer using the Pythagorean theorem is the correct one, but, again, because of the caveats I pointed out, that solution doesn't really correspond to the question asked in which the force is always perpendicular to the velocity. Red Act's answer is the more appropriate answer here. Dauto (talk) 20:33, 21 December 2012 (UTC)[reply]
I understand that, outside the body, HIV is a very wimpy virus (it doesn't survive very long, you can't get it from the toilet seat, etc.) But what about other well-known viruses? If someone with a cold sneezes on a door knob, how long will those pesky cold cooties be viable? In general, are most well-known viruses more on the HIV side of survivability or more on the Superman side? TresÁrboles (talk) 21:10, 19 December 2012 (UTC)[reply]
Well, a virus, by definition, is a parasite on live cells. So, unlike bacteria, they aren't going to thrive when away from live cells. The might lie dormant, though. StuRat (talk) 21:14, 19 December 2012 (UTC)[reply]
From Common cold#Transmission: The viruses may survive for prolonged periods in the environment and can be picked up by people's hands and subsequently carried to their eyes or nose where infection occurs. It's footnoted to a reference if you're interested in more details. Duoduoduo (talk) 22:15, 19 December 2012 (UTC)[reply]
You might find the stuff at this article interesting. It's not a direct answer to your question, but it does delve into some of the possibilities. And what better bedtime story could you ask for than finding out that one of the greatest killers of all time was able to be resuscitated after 80 years? Matt Deres (talk) 01:55, 20 December 2012 (UTC)[reply]
Mayo Clinic suggests cold virus can last 48 hours or so, but I kind of think this assumes indoor.[1] Exposure to the sun would certainly lessen the time. Some other online sources of less certain authority say hepatitis can live outside the body for more than a week. I remember reading something similar about the filoviruses (like ebola virus). Shadowjams (talk) 02:59, 20 December 2012 (UTC)[reply]
Thanks, that's the kind of thing I was looking for. So far it looks like the other familiar viruses are much hardier than HIV. TresÁrboles (talk) 23:04, 20 December 2012 (UTC)[reply]
Why have some countries chose the ±220 V and others the ±110 V? Big countries can have 110 like the US or 220 like Russia. Why not something in the middle like 170 V? Or 400 V? And how can it be that countries like Japan have both 50 Hz and 60 Hz? Or Saudi Arabia 127 V and 220 V? I know that it has historical reasons, but one country should be able to unify internally its frequency. I also know that power lines transmit at a different voltage that what we get at home. OsmanRF34 (talk) 22:27, 19 December 2012 (UTC)[reply]
Well, a lack of standardization comes about when systems are set up with little thought for existing standards. I believe it's easier to double or halve the voltage than to change it in other ways, so you tend to get multiples of 2. In the US for example you get both 110-120 V (most things) and 220-240 V (used for heating, major appliances, etc.). StuRat (talk) 22:33, 19 December 2012 (UTC)[reply]
Yes, it's not surprising that different countries have different standards, but no country has a mix of the two, in contrast to volts and frequency. OsmanRF34 (talk) 12:52, 20 December 2012 (UTC)[reply]
Not quite true. The U.S. Virgin Islands drives on the left with the steering wheel on the left. Duoduoduo (talk) 15:40, 20 December 2012 (UTC)[reply]
Driving on the same side of the road as you have the steering wheel is difficult and dangerous. It might be an arbitrary decision as to which side of the road you're going to drive on - but having the steering wheel on the same side is a poor choice. SteveBaker (talk) 20:24, 20 December 2012 (UTC)[reply]
It's due to a conflict between history and geography. Historically there was a British influence, hence the driving on the left. But the cars are imported from the US, hence the steering wheels on the left. Duoduoduo (talk) 21:50, 20 December 2012 (UTC)[reply]
Japan has both 50-Hz and 60-Hz because of the history of electrification: one electrical grid grew around Osaka Electric Lamp's American 60-Hz equipment, while another grew around Tokyo Electric Light Co.'s European 50-Hz equipment. By the time that people thought about setting up a national standard, there was so much installed equipment that would need to be changed that it didn't happen. See Energy in Japan#National grid. --Carnildo (talk) 00:50, 20 December 2012 (UTC)[reply]
Our Utility frequency article has a good explanation of the history behind the multiple frequencies. Shadowjams (talk) 02:53, 20 December 2012 (UTC)[reply]
