Schrödinger's hat

Uhmm, depends on what you are trying to achieve I guess. The net result is that you will get 1 Helium 4 from every 4 hydrogens, as the extra 2 hydrogens are available for future reactions (as you correctly surmised). But that doesn't mean they aren't involved in the reaction (and in the extreme case of only having six hydrogen atoms, you would need them for the reaction). The full reaction goes like this (D in this case is Deuterium, or Hydrogen-2): [math] 2 _1H^1 \rightarrow 1 _1D^2[/math] [math]1 _1D^2 + 1_1H^1 \rightarrow 1 _2 He^3 [/math] This first part has to happen twice to give you two Helium-3 atoms (or just double all the numbers). Then: [math] 2 _2He^3 \rightarrow _2He^4 + 2_1H^1 [/math] Note that 3 hydrogens go into each Helium-3, and two are released when it becomes Helium-4. If you only had 4 atoms you would not be able to do this reaction.

Not at all. It's just that you have to be quite specific as to what you mean by order and disorder. This is why we use words like entropy and free energy which have very specific meanings (whereas, in a non physics context order and disorder mean subtly different things). Our solar system (when taken along with the gas and photons that left it over the past billions of years) is a higher entropy state than the gas cloud it evolved from, even if it is more orderly when you look at it on a large scale. Further thoughts: If you look at what remains in the sun and planets it might be more ordered than a hydrogen cloud of equal mass. I do not know one way or the other, but it would not surprise me. Systems can have localised pockets of low entropy (such as life, or the LCD monitor I am staring at right now), this is quite permissable (and even expected -- by the same logic that entropy increases, there will be more ways for the system to increase in entropy non-uniformly and uniformly so we expect any given system to have pockets of low entropy) as long as the whole system increases in entropy.

This is more of a data collection than an experiment. With an experiment you are usually testing a hypothesis. Something along the lines of 'is a correlated to b?'. You could ask: 'Is the amount of lead present correlated to the reading on my meter?' In that case your independent variable would be the amount of lead, and the dependent would be the reading on the meter. A control group would be some un-tampered with samples of whatever you're adding lead to. Presumable the device is already calibrated, so this probably won't be very useful. Or you could ask: 'Is the level of lead in beach a higher than the level of lead in beach b?' In this case you'd take a number samples from each beach and do some statistics to see if they are significantly different. Or maybe something like: 'Are beaches near x location higher in lead than other beaches?' In this case you'd need some beaches near location x, some distant. A beach is not really something a single person can alter that much, so a control would be quite difficult (you can't easily prepare a set of beaches which are identical other than the absence of location x). In effect the other beaches you measured to get your baseline would be the control. If you suspected that there was some difference between the beaches you measured other than proximity to location x you might try and find a similar group of beaches (to see if the other differences are contributing factors).

Position of an initially stationary object with respect to time for constant acceleration is: [math]x=\frac{1}{2}at^2[/math] a in this case is 1g, so you wind up with: [math]t=\sqrt{\frac{2}{g}\times 1m}[/math] Or about 0.456+/-0.004 seconds. (456 +/- 4 ms) If you wanted to be more precise for a changing acceleration you use a differential equation: Umm, it's not really exponential, it's inversely proportional to the square. [math] \frac{d^2x}{dt^2} = \frac{F}{m} = \frac{GM}{x^2} [/math] Where x=0 represents the centre of mass of earth. Solving this is a little bit complicated, but there are a number of reasons why we can consider something at or near the earth's surface to have a constant acceleration (and thus we don't have to). Firstly, earth is spinning, so when you are at the equator some of the gravity is 'used up' keeping you moving around in circles. On top of this earth isn't exactly spherical, and even where it is there are fluctuations in its density. These factors add up to 'exactly 1 g' not really being overly precise. Wiki says it's [math]9.80\pm0.02 ms^{-2}[/math] Or varying by about 0.2% Earth is about 7000km in radius, or 7000000m. If you fall 1 metre then you are only changing in height by 1/70000%. So the change due to a little change in altitude is going to be insignificant compared to doing your experiment where there are iron deposits, or moving north or south a bit.

(Apologies if this is not quite the correct category, I couldn't really find any single category in which to discuss this. The primary query is at least a matter of engineering more than anything else.) Does anyone know of any figures for the energy return on invested of the nascent organic PV technology? Or perhaps know of anywhere I could find figures with which to calculate this. I'm given to understand that the state of the art has rather fantastic payback time (measured in months rather than years), but rather short total lifetime. With this being the case I guess one would have to consider transportation/mounting/wiring/installation/etc costs in addition to the panel. Another important factor is the very small initial investment required for organic PV. This thought came up upon reading Tom Murphy's blog, specifically this. This got me thinking along the lines of solar potentially surpassing other infrastructure in terms of EROI. Subsidies and other government incentives aside (from my reading, there are arguments that both solar and the oil/coal industry have more subsidies than the other in various places), would we see a sudden shift towards solar if it were to achieve a better EROI than oil/coal plants+fuel? Are there other factors that would prevent this? Some that come to mind would be: Cost of land -- parking lots and roofs come to mind. Distribution -- The sunniest areas are not always where you want power. Especially in winter Everything else that goes with non base-load power

It's incoherent to the point where I'm not entirely sure what he's claiming. As far as I can see he's saying 'if we do calculations where we mix variables from different frames we get different answers', and somehow claiming that disproves relativity. He's also mixing up the concept of rest mass and relativistic mass. Relativistic mass is a bit of a tricky thing, because it does change significantly between different frames.

Here's some for you. Roughly in order of increasing difficulty. Depending on what level the course is/background you have, I wouldn't be surprised if you get stumped (possibly in part due to my poor wording). Archimedes famously said: "Give me a place to stand and I shall move the earth." Using torques, work out where you would have to stand to lift the Earth against the Sun's gravity (using an appropriate approximation for your strength) given that the fulcrum is 1 metre away from the contact point. How much torque (from the fulcrum) is there on the lever? (Yes, I know, it's moving; it's still a fun exercise even if it doesn't make sense. The force of gravity due to the sun at earth is [math]\frac{GM}{r^2}[/math] or an acceleration of g=0.006m/s^2, mass of earth is about m=6x10^24kg) What advantage does having a patella give to a human? Show using a diagram and equations. One way of measuring energy is force multiplied by (really integrated over, but not too important here) distance. Power is energy divided by time. With this in mind, what quantity in addition to torque would you need to measure to get the power of an engine? Look up some figures about boat propellers (speed of rotation and horsepower of the engines). Using what you worked out in the engine question, work out how strong the propeller shaft would need to be (how much torque it needs to withstand). Model a motorcycle as a rectangular box of uniform density 1 m high and 1.8m long weighing 250kg including its rider. Given that there is one force acting down on its centre of mass of mg, and a reaction force acting straight up at the 'rear wheel' (back of the rectangle) with equal magnitude: What is the force (straight forward, from the same point -- back of the recangle -- as the upwards force) required to do a wheel stand? Assume the bike is doing 80mph when this occurs, and that the engine is revving at 7000rpm. What torque must the engine be producing? (NB: this is an extremely simplistic model, so don't be surprised if the answer you get is at least a factor of two or so off) Build a similar model for a car (maybe move the centre of mass further forward) and try to work out what power a car would need to produce to do a wheel stand. Sail-boats Exhibit a phenomenon known as weather helm, wherein the tiller (stick on the rudder used to control it) will pull against the person steering the boat and cause a tendency to turn up-wind. The opposite of this (called lee helm) is considered undesirable. An un-piloted boat pointing up-wind will come to a near-stop, whereas an un-piloted boat pointing down-wind will move at close to top speed. Build a simple two dimensional (as in a childrens' drawing) model of a boat out of basic geometric shapes. Include a rudder and centreboard or keel. Find the centre of resistance (directly analagous to centre of mass for constant density plane of uniform thickness) of the keel, rudder, and portion of the hull that is under water. Find the centre of effort (same story) of the sails. If these are not in the same place, what happens? What is the mechanism for this being exhibited as a torque on the rudder? What might you do to the position of the sails/keel to ensure that the boat does not develop lee helm -- whilst keeping the effect of weather helm small?

You can derive them a number of other ways as well. I could take the odd part of [math]\frac{-i}{2}e^{ix}[/math] I could take the taylor expansion of that as the fundamental and ask 'why does nature use infinite series?' Or I could use a circle of different radius. If I considered a circle of radius [math]\frac{180}{pi}[/math] and measured the ratios of the size, I'd get sin tables with the circumference of the circle being the angle in degrees. I could even use the side lengths rather than angles. I'd have a bunch of random 360s in all my equations, but it'd still work. I could consider a mass-spring-damper system [math]\left(\frac{d^2}{dt^2}y = -y\right)[/math] as the fundamental definition. In this case it is something I can point to in nature (any kind of spring or r^2 potential such as gravity/electric force). Also we only get pure sine waves when we consider special cases, usually something involving circles or some kind of springy force. Everywhere else we tend to approximate things as sine waves because they're easy.

It does...sort of. You're conflating dimension and power of x. If x were in distance then [math]x^2[/math] would have units of [math]m^2[/math], which would represent some kind of area-like or two dimensional thing. But that doesn't mean everything with [math]x^2[/math] in it will be area-like. Take [math]F = \frac{GMm}{r^2}[/math] for example. There's an [math]r^2[/math], but when you take the whole thing into consideration it has units of force. [math]\sin{(x^2 + y^3)}[/math] isn't the kind of formula that would come up in physics. For one we can't really take the sin of something with units. Secondly the x^2 and the y^3 have different units. But if you are only considering x and y to be numbers (no units, so x^2 is just a number, it doesn't have dimension or units) then this is perfectly valid. Think of it as just a relationship between three numbers.

Neodymium magnets are a few dollars each on ebay. On top of that you'll need string, maybe some super glue (if you can't find a magnet with a hole in the middle) and I've seen electronic kitchen scales that go down to 0.1 grams -- maybe ask your neighbours/friends if you can borrow theirs?

Depends. If it's hot then your body is spending energy on cooling. If you eat something cold then that does the cooling for you and you actually use less energy. If it's cold then you will use a little bit more energy, but this is comparable to taking off your hat or socks for a few minutes. Your body is also quite good at maintaining heat, so you may just make it go into heat conservation mode and wind up with cold fingers and burn exactly as many calories as you would otherwise. No. Although concentrated heat can be partially converted into useful energy as it diffuses (this is how a car engine works, the petrol burning makes hot gas which does work as it comes into equilibrium with the surroundings), your body has no way of doing this. It could, however, result in you spending less energy heating yourself. If you are shivering and drink some hot cocoa you will spend fewer calories getting warm than if you drink iced chocolate.

My first thought if budget is limited would be to use string and gravity for something like this. Tie the magnets to strings in different places, then adjust the length so that they almost, but can't quite touch. With a bit of fiddling you should be able to get them to sit in equilibrium. Then measure the distance between and the angle. Use your knowledge of forces (or ask if you don't know much about forces) and a measurement of the weight to work out their force on each other. It may be easier if you use one fixed magnet and one string, then move the other end up as well as away. Or maybe one very short string (so the magnet always faces the correct direction) and one long string. Experiment a bit If you have trouble with strings, it can be easier to manage a stiff rod, but it's harder to work out the forces in this case (because the rod has significant weight). As to your exponentially increasing force hypothesis. Why did you pick exponential? Can you think of any other types of function that increase as r approaches 0? Maybe try more than one? If you can find other functions you could compare them to an exponential. See what fits best.

Unless of course that's the exclusive or, in which case the answer is "No".

Most scientific equations you'll see are in SI units. So energy is Joules. I don't quite follow what you're asking here.

The rate of hawking radiation is inversely proportional to the mass of the black hole. So for a small enough black hole, you would have an acceptable rate of energy emission. You would have to have some way of containing the matter while you excited it to a state in which it would turn into energy. The only containment we know of that is strong enough to do this is a black hole (the process baric described above). With regards to sustaining or creating a black hole, This would be pure science fiction. I'm not sure if we know of any principle around which a machine could be based that could do this.

Couple of readability tips to make people enjoy reading your questions more (and just be more likely to answer). 1. Brackets please.e^rt is somewhat ambiguous as to whether you mean e^(rt) or (e^r)t. 2. (Not as necessary) Read up on the latex tutorial in the maths section if you're posting a lot of maths. It makes it a bit easier to read. That equation would be something along the lines of: [math]A=xe^{(rt)}[/math] Which looks like: [math]A=xe^{(rt)}[/math] (You can also click on the images to pull up the latex that produced them) If you use latex then the brackets aren't necessary (as all the indeces will be superscript), but I left them in to demonstrate what I meant. Doing the rearrangement: [math]A=xe^{(rt)}[/math] [math]\implies \ln{(A)}=\ln{\left( x \times e^{(rt)}\right)}[/math] [math]\implies \ln{(A)}=\ln{(x)} + (rt)[/math]

The calculator just uses the same approximation methods but does them very quickly. If it gets the approximation accurate to the number of digits it can display (plus a few extra for rounding error), you don't notice that it's approximate. There are also things like computer algebra systems that work a little bit differently (more like you or I would treat a logarithm, only evaluating it if asked for a number).

<facetious and unhelpful> Here you go, just don't worry about any of that gravity nonsense. </facetious and unhelpful>

As well as the series expansion that bignose linked to (which will give you an arbitrarily precise value if you are patient enough). If you just want a very approximate value, it's useful to know that [math]e^3[/math] is very nearly 20. So if we have some random number: 7052443 We can compare it to powers of 20 20 400 8000 160000 3200000 64000000 And see that It's about double fifth power of 20 [math](2\times e^{15})[/math]. e is about 3, so we can say it'll be fairly close to [math]e^{16}[/math]. Or that the natural log of this number is about 16. Doing it on a calculator gives me 15.769, so I was reasonably close. Not really close enough for precise work, but if you just want an order of magnitude then it's good enough. With a bit of practise (and maybe memorising a few more numbers) you can get one or two sig figs doing it in your head.

Hmm. I'm not even sure if this is the case. The precursor to our discovery of relativity was knowledge of Maxwell's equations. This in turn came from knowing about magnetism, which we can't sense directly. Certainly the idea that light had a speed (which was proposed since classical times) would have helped, but discovering the concept of a speed as a law of physics is what lead to relativity. I think it could even be argued that not being used to light and thus used to the idea of knowing about remote things immediately (or indistinguishably close to immediately) could be helpful for understanding relativity. If we'd never developed a concept of the present, but instead only thought of 'things we are effected by', 'things we can effect', and 'other' then I think relativity would seem quite natural, and maybe even self-apparent upon discovering the equations governing radio waves. But this is all wild speculation, there's no way of knowing this unless we genetically engineer completely blind but super-intelligent bats and watch them for a few Millennia without interfering in their development.

If that happens one of two assumptions must be abandoned: Causality -- Effects happen after causes. The principle of relativity -- There is no preferred reference frame. The problems with the former are probably self-apparent, but there are also issues with the latter. Instead of a nice unified theory, we'd have a hodgepodge of different mathematical relationships that just happen to be lorentz invariant.

1) If you want to know if your answer is correct, try putting it in the original function 2) Not quite sure what you're doing with this line: [math] So -2x^2 + 1 = 0 [/math] Maybe try from here: [math] 4x^2 + \frac{1}{x^2} = 0 [/math] [math] 4x^4 =-1 [/math] Other things to think about that are good for checking your answers ahead of time: If you put a real number in [math]x[/math], what is the sign of [math]x^2[/math]. When is [math]4x^2[/math] less than 4? When is [math]\frac{1}{x^2}[/math] less than 1? Are either of these ever negative (for real [math]x[/math])? What does that mean if [math]4x^2 + \frac{1}{x^2}[/math] is less than 1? If you ask questions like these you can usually figure out what your answer should be like without doing the rearranging. Also a good idea is to plot it on a calculator or similar.

expandCapacity() is O(n), but it doubles the capacity of the stack when called. size() == stack.length will only return true on average 1/n times.

Possibly because everyone got tired of the constant bickering over minutae and reiteration of the same points. I'm not even sure what you were trying to say with that last post. If you were trying to say that we are saying, "The (3d) shape of an object depends on the reference frame you choose." Then yes, this is what we were saying all along. (3d) Shape is not frame invariant. This in no way implies that reality is frame dependant or subjective. Given the measurements of a system in any frame, the measurements in any other system can be calculated. There are certain quantities (interval, magnitude of any four-vectors such as energy-momentum, order of causally connected events) that do not change between frames. There are other quantities (distances, times, order of non-causally connected events) that do change. You can pick a certain set of events which are simultaneous in one frame and call it 'now' if you really want thus declaring any measurements taken outside of this frame as incorrect. It's just that there's never been any evidence that the 'now' you pick is the correct one (or any more likely to be correct than any other set of events that someone measures to be simultaneous). If you're trying to say that the rest shape (shape of an object in a frame where that object is at rest) is objective then that is also true. So is the shape in any other reference frame, you just have to supply the velocity of the frame you're in before shape can be defined.