mezarashi

Google as well as other competitors like Yahoo use bots to scour the web for information. They check forums like this one too. All text information is saved on their servers... yes this is ALOT of information, terabytes and terabytes, and stays there pretty much forever. It gets cached. You can read about new privacy issues that have arisen because of this: http://www.reuters.co.uk/newsArticle.jhtml?type=reutersEdgeNews&storyID=746393

I agree with you herme3. The fact that more companies are implementing these so called "automatic updates" makes me very uncomfortable. If in the wrong hands, this can be used to hi-jack a lot of computers. The update process will be able to get past all your other "protections" of course, because originally this process is authorized. The problem is you don't know what it's auto-updating itself to... somehow I just don't trust them. I can imagine one of those Skynet takeover predicaments sometime in the close future where Microsoft or some entity takes over the world by trying to update malicious code onto all computers running Windows.

I'm not sure if this is what you mean, but I think the topic applies for every leaf-growing tree. Generally for everything classified under the Plant Kingdom, "leaves" are important. That green stuff is the stuff that has chlorophyll, used to synthesize basic carbohydrates (energy) from carbon dioxide and water with the helping hand of sunlight. Which is also why leaves need to be broad and thin, to get maximum surface area exposed to sunlight. During winter, I guess the trees go into a plant version of "hibernation" using stored energy. Another interesting thing to ponder though is why trees branch as they do. From trunk to main branches to smaller branches to twigs to leaves.

It looks like you have great enthusiasm for your college studies. Fortunately enough, most college syllabuses prepare you for everything you need to know. In fact, they will start anew as if you don't know any science or math at all (maybe algebra II and geometry needed). For the first 2 or even 3 years pursuing the degree, you will be studying the fundamentals. For the first year, you will probably spend time reviewing the physics, chemistry, and mathematical knowledge required to understand those fundamentals. Electrical engineering currently usually refers to electrical power engineering. I'm not sure if this is the case at your college, as most schools have a separate program for Electronic engineering. I'll mention here only the introductory material subjects that you will need to know in order to pursue the fundamental topics in electrical engineering later on (control engineering, power distribution systems, digital electronics, analogue electronics, high frequency systems & electromagnetics, wireless communication principles, etc). You will prepare for these topics through your first and second year courses. [Mathematics] First and foremost, you will need mathemetics. I understand that calculus is not a requisite in the A-levels, unless you are doing Further Mathematics, so that will be a good topic to study. Knowing calculus will be of great help, especially when your professor starts on things like vector calculus and differential equations. Strong basics will be of great help. Other topics that might be of use: imaginary numbers, complex numbers, series and sequences, and matrices and vectors [science] Subjects that you will be doing in your first year will probably include Physics, especially in the department of electromagnetics, optical, and quantum physics. These subjects should be familiar to you from your A-levels. College level material will add in more mathematical formality and of course that hated calculus. Nonetheless, being aware of these phenomena will give you a good start. Undoubtedly, you will take a course on Electric Circuits. Find yourself a good book that you can understand well, for this material is fundamental beyond fundamental as an electrical engineer. Ohm's law, Kirchoff's Law, complex resistance (impedence), basic electrical components (inductors, capacitors, resistors), and power will be revisited. New topics will include Nodal and Mesh Analysis, superposition, Thevenin's and Norton's Theorem, phasors and AC analysis, complex power, and if you are unfortunate enough Laplace and Fourier transforms. Lastly, you are bound to take a couple of "irrelevant subjects" (you'll know what they are) just for the heck of a "comprehensive education". Books: [serway, Physics for Scientists and Engineers] http://www.amazon.com/exec/obidos/tg/detail/-/0534408427/qid=1117718072/sr=8-1/ref=sr_8_xs_ap_i1_xgl14/104-9450055-9175101?v=glance&s=books&n=507846 [Alexander, Fundamentals of Electric Circuits] http://www.amazon.com/exec/obidos/ASIN/007249350X/ref=sib_rdr_dp/104-9450055-9175101

I'd just like to add an interesting fact. Water molecules H2O consist of 2 hydrogen and 1 oxygen atoms which form covalent bonds at about 104.5 degrees (making it polar as well, a good solvent). It is because of this unique structure, the hexigonal crystal structure forms upon freezing as this minimizes the energy between the hydrogen bonds. Also, it is because this great property of water, expanding upon freezing that makes life on Earth possible (well not to mention the fact that we are 75% water). The coldest depths of the ocean is at about 4 degrees Celsius as this is the temperature in which water is most dense. When it gets any colder, the water slowly rises where sunlight will heat it up again. If water were to increase in density upon freezing (as with a majority of common substances), we will find that ice will start accumulating on the ocean floor and slowly the world will freeze over.

Hi. First of all, I'd like to note that not all results of a fractional power will be rational, meaning you may not be able to solve it by hand, or it would require you to do infinite repetitions. Nonetheless, given that your answer is rational or you don't mind having an irrational component in your solution, then these are the steps. To do fractional powers manually, it is advisable that you first convert them to proper fractions. For example 3^(0.4) can be expressed as 3^(4/10). This is easy to do if the number is non repeating. If the number does repeat however, a universal method to do this which I learnt is to multiply the decimal by 100, subtract by itself, and divide by 99. Formula being: fraction = (100x - x)/99 = 99x/99 Back to the topic. A fractional exponent is basically a root. For example, square root 4 can be expressed as 4^(1/2), cube root 4 as 4^(1/3) and etc. I guess that was the key piece of information you were looking for. In anycase, all the properties used for exponents can be used on fractional exponents as well, so: 4^1.5 = 4^(3/2) = 4^(1 + 1/2) = (4^1)x(4^1/2) = 4*2 = 8 There will be times where the answer is not rational as I mentioned like: 2^(5/2) = 2^(2 + 1/2) = (2^2)x(2^1/2) = 4*(square root 2) If you have a large number, the general strategy may be to factor that number down into something you know, for example if you have: 64^(1/3) = 8^(2/3) = 2^(3*(2/3)) = 2^2 = 4 You will also find cases where you break down the numbers and you get something ugly which is neither irrational nor integer. In these cases, the only way I know is to resort to iteration, meaning guess the answer, check, adjust, check, adjust until your answer is to an acceptable level of precision. There are formulas that can help you get in the right direction and more quickly, but these are usually used in the realm of programming where computers do the work for us. Conclusion is, fractional exponents are nasty, and even more, fractional exponents of fractional numbers !!!

Ah, I see how I may have caused some ambiguity with my post. I'll edit it for more clarity. I was a bit confused for a moment there, because purportedly I supported the same views. Sorry

I didn't take the IB program back in highschool. I opted for the easier AP (advanced placement) stuff instead, but I know plenty of friends who did including my class valedictorian who currently is a full scholarship holder at Stanford (if that's any persuasion to its 'greatness'). I would recommend it. It is a broader and more comprehensive education program. The International Baccalaureate is relatively intense. It's a complete 2 year high school course that covers all fields. Instead of doing your high school classes, you will instead be doing IB Classes for about all your subjects. You need to fulfill the requirements for SL (standard level) subjects which are in themselves harder than the average high school course syllabus and a couple of HL (higher level) subjects for your specialty. The HL courses are very difficult. More than AP I can testify. They will be demanding on you in workload and demand high proficiency in the final exam. AP teaches college level material true, but IB HL will test your understanding of fundamentals to the limit. (If you want to understand what this means, take for example doing decimal division in your head. We all know how to do division, but it can get hard if the question makes it so.) Nonetheless you will learn a very broad range of topics, and most importantly, it is high school material, meaning that if you are liable to taking any entrance examinations etc, this material will be relevant !! Knowing how to do differential kinematics from AP Physics C will get you no advantage. Knowing the broad fundamentals of everything from Thermodynamics to Introductory Particle Physics that you will learn in the IB program will ! When I was studying for an academic examination, I remember referring often to the IB HL and GCE 'A' Levels syllabus. Secondly, while you will not get college credit (CORRECTION: You can in fact earn college credits as well !) as AP can (given you score a 4 or 5 on your final exam), the IB diploma is more widely accepted and usually given priority in college admission. If you score yourself a couple of 6 and 7's in your HL's, this will definitely be a point to boast. Given you can handle the workload, this is a very advantageous program. If anything else, you should see your college counselor... this is an important time in your life for the to be college student. Make your decisions wisely and I hope I've been of some help

I'll give some input from having been a computer overclocker at one point in time.. basically I can say that the computer hardware you have there is relatively rugged, physically atleast. It can resist considerable shock. In terms of electrostatics, I've actually pulled out the soundcard while having the computer running, causing a corona discharge and yet this only blew the PSU (power supply unit). The motherboard still functioned perfectly fine. But most importantly is that unless your planning to put your computer motherboard bare into a airplane test windtunnel, I doubt there will be sufficient electrostatic build up to really damage the electronics. Most of them are not magnetically sensitive (ROM memory is burnt in, RAM memory is flushed/refreshed constantly) so any buildup wouldn't be significant enough in my opinion. Even tasing a magnet around an off computer (not your harddisk though) will generally cause no harm. Lastly, what I WOULD be worried about in the case that you want to produce serious windflow is the dust build up. They will jam fans, reduce the dissipating capability of heatsinks and in rare cases possibly cause a short circuit on your motherboard, which can be very bad.

The quantum model of the atom is based on electron's probability distribution. Why the distribution curves look like so is a mathematical task that I must admit has eluded me. It is derived by that horrendous looking time and space variant Schroedinger's equation. Back to the topic however, the funny looking shapes you refer to are known as the orbitals in the subshell. For each principle quantum number, there exists a number of available orbitals. The higher the principle quantum number, a higher a Azimuthal quantum number it will have. The Azimuthal quantum number basically describes something like its orbital momentum. In addition, there is a concept known as "spin" or magnetic spin. Hund's Rule and the Pauli Exclusion principle both indicate that for each orbital there can be only 2 electrons in orbit. I found this link that shows the relationship between the probability distribution and the "funny shapes" they take on in 3D space. http://www.chem.ufl.edu/~chm2040/Notes/Chapter_9/quantum.html There is also an interesting phenomenon that happens to these orbitals when atoms get close together, called hybridization. The orbitals will actually combine to form a new hybrid orbital. This is usually studied in chemistry for chemical bonding. Interestingly, if you understand how these wave functions interact, it will better your understanding of the quantum atomic model. The interaction of atoms through quantum orbits is also used in semiconductor and nano-engineering, take for example carbon nano-tubes ^^

I feel sorry for urza, so I thought I'd give it a try since I've not much better to do "and you can't do it using simple triangle rules..." - urza This statement isn't true at all. Most of these geometry problems require a very good understanding of the basic rules. This problem turns out to be quite challenging and I did enjoy thinking about it for a bit. Very tedious nonetheless, and I'll leave the 2 later parts for you to try out. It is solved using "basic" intuitive rules, but if you feel insecure, you can try adding in some of the trigonometry if you are math savvy (e.g. sine rule) and you will see that the two approaches yield the same answer. (a)i. Prove that MD + 4MF = 0 This is the same as saying that MD = 4FM, since FM = -MF. They are collinear so all we have to do is prove that MD is 4 times the length of FM. For the rest of my explanation, I will not take the direction into account. So I'll say things like MD = 4FM as well as MD = 4MF interchangingly. I've drawn a diagram of an arbitrary triangle. Note the congruency indicated by the colors due to the parallel line QE. In addition, you will notice that triangle EFM is congruent with triangle BDM. This can be verified by showing that all the internal angles are the same. <FME = <DMB : this is due to the reflection rule of two intersecting lines <AFQ = <ADB (<MDB) : Congruent triangles (green) because of the parallel line. <AFQ = <EFM : reflection of two intersection lines again <FEM = 180 - <EFM - <FME : Sum of triangle's internal angles always = 180 But we know that <EFM = <MDB and <FME = <DMB, so <FEM = DBM. Having established the congruency of EFM and BDM, we can then move on to to proving that MF (a side of the smaller triangle) is equal to 1/4 MD. To do this we simply need to prove that ANY side of the smaller congruent triangle is 1/4 the length of the corresponding side in the larger congruent triangle. My approach was with using the blue congruent triangles (ABC) and (AQE). Since AE = 1/2 AC, then by congruency, QF = 1/2 BD and most importantly FE = 1/2 DC. But we know that DC = 1/3 BC. We also know that BD = 2/3 BC = 4/6 BC. With that, FE = 1/6 BC. So we see here that the corresponding sides FE = 1/4 BD, as FE = 1/6 BC and BD = 4/6 BC. Since FE = 1/4 BD, then ME = 1/4 MB, and ultimately MF = 1/4 MD. If you don't trust your instincts here brought to you by the congruency rule, then apply the sine rule. This completes the explanation for 4MF = MD. (a)ii. Show that MB + 4ME = 0 This was shown in part (i) " Since FE = 1/4 BD, then ME = 1/4 MB, and ultimately MF = 1/4 MD. " (a)iii. Deduce that 2MA + 3MD=0 or that 2MA = 3DM This requires just a bit of arithmetic, no trig involved (I suck at trig). Equations we have from the diagram. AD = MA + MD AF = 1/2 AD AF + MF + MD = AD; MF + MD = AF = 1/2 AD From part (i), we know now that MD = 4MF (right hand side) 3DM = 3(4MF) = 12MF (left hand side) 2MA = 2(DA - DM) = 2DA - 2DM = 2( 2(DM + MF) ) - 2DM = 4DM + 4MF - 2DM = 2DM + 4MF = 2(4MF) + 4MF = 8MF + 4MF 2MA = 12MF =D And we're done =D The rest of the question relies on the same elementary principles. Keep looking for clues in the congruencies, good luck and have fun

Haha, I'm sorry to really cut your argument short but your very first assumption there is not correct. The fact is: Magnetic fields (part of the electromagnetic field) along with all other fields (i.e. gravitational field) propagate at the speed of light. That means that if the sun were to right now just disappear, the Earth would remain in normal orbit for another 8 minutes (approx. the time it takes for light to reach Earth from the Sun), and so there is no way we can know about it until those 8 minutes come. There are many other real life examples I'm sure the other forumers here can give to help reinforce this fact.

Weight in a more formal definition is the "normal force" acting on an object given that the object is placed on a surface that is perfectly parallel to a plane tangent to the Earth's surface at the point of weighing. This normal force will ensure that the object remains basically unmoved. It will fluctuate depending on the forces on it that are trying to make it move (gravity mainly since we've taken out vibrations and wind). As mentioned, the precision in which you can measure an object's weight will be hindered by the gravitational fluctuations present. If you are going to precision approaching infinity, you will need to take into consideration the presence of the Sun, planets and possibly all stellar objects. Oh, and the bulging of the Earth as well as other assymetries (non-homogenous density) that will cause the gravitational field to vary on the Earth's surface.

Hotmail, the offline non-webbrowser version uses a protocol called HTTPMail. Although it has been around for many many years, Microsoft still considers it a trial product and apparently they have not documented the protocol, so up to date, I only know of Hotmail and MSN that use it (if anybody knows of another mail service that provides this). HTTPMail is built on WebDAV (http://www.webdav.org) as an extension to HTTP/1.1. I doubt that any non-Microsoft product would have extensive support for this protocol as it is still a "beta" product. However, as far as I know, there are still currently a few independent "projects" on HTTPMail, and you can learn more here (they seem to be non-ASP however): http://httpmail.sourceforge.net/

When you get a problem like that where the equation is as you mentioned: x(squared) + y(squared) = 10x-8y+6 The strategy is basically to factor the x and y so that it takes the form of the equation mentioned by yourdadonapogos. Once the equation is of that form, then information like it's center and radius are only but obvious

Saraisme, well first off you have to understand that currently our technology only permits us to generate electricity through some sort of AC generator. Well there are things like solar power and stuff, but those aren't the main power sources we use. AC generator means having some kind of electrical induction going on between a coil(s) and a magnetic field that are moving relative to one another. A DC generator is actually just an AC generator with brushes and slip rings at its electrical connection to reverse polarity every 180 degrees. While the model that you are proposing for generation is not the type commonly used in real industry/commercial applications, it is a much easier model to understand. If you want to charge a battery for example, you will definitely need DC current. Conventional batteries don't work on AC. They have a clear + and - sign on them if you've observed carefully. Creating DC current from an AC generator is not easy with the use of just mechanical parts. The use of the suggested commutator will result in your waveform being a series of bumps. However, since the advent of power electronics and highspeed switches, this task has become trivial. A more simple method would be the aforementioned rectifier circuits. Try the full-wave rectifier with capacitor filter. This configuration works with a direct AC input

Yes ^^ only e is true. Follow the arguments. See if the changing the variables in the equations for a, b, c, and d will make the assumptions in the questions hold true.

I believe that you can do your calculations as if the reactances were resistances, which results in simple voltage division to find the potential difference across each component. When working with the current and voltage values, you should use RMS values to be consistent (they are convenient when you want to do power dissipation calculations) and thus the results of the calculations will also be in RMS. This is the electrical average used in AC circuits, because when you have sinusodial waveforms, the current and voltage changes all the time. Sometimes it will go from left to right, then it will go from right to left and at certain times it will be zero. You will also find that at any instant, the instantaneous voltage across the capacitor or inductor may be 100 to 1000 times higher than the driving source. There are many many other things to note about resonant circuits, but I don't know where to start ~_~

No, the sphere of light does not "instantly concentrate" on your one pixel. It obeys classical power dissipation laws as with any other wave. It is difficult to give light a universal visualization. The way I like to look at it is to imagine it to be "localized possibility". I consider them to be more particle than wave. They are called photons after all. As for their wave and quantum properties, imagine a small probability wave function. Each little wave function is like a photon of light. Because the wave smears out at the edges, it is never certain where exactly this photon of light is. It also interacts with itself (or possibly the electromagnetic field) in such a way that it behaves like a wave around corners, tight spaces, or when changing mediums (refraction, diffraction).

It'd be nice if someone could show us some simple calculations using tensors For some practical application of course

Hmmm, I think the coordinate transformation here is simply the intuitive definition. I'll be drawing something on paint now to illustrate this. Though I hope I'm right. Rotational dynamics has never failed to stump me time and again It would be more easy to comprehend if we refer to the change in vector A now as some velocity vector. The absolute velocity of A with reference to some inertial frame is the velocity of A with respect to the body f in rotation, and the absolute speed of the body at which point A is at (w x r).

I'm really sorry, but that last comment you made about the sink and brain cleaning just reminded me of this >.< The aluminum oxide layer that forms around aluminum parts is pretty much non-reactive. Other than physically etching it, I have no idea though x_x

Yes, we do need rain. For the world's vegetation relies on rain to redistribute water to where it would not usually occur. And of course, all other living things live off of plants. So if all plants die, so would they. If humans can live off a barren wasteland of a planet (I don't think we're ready yet technologically), then living on Mars wouldn't be a big deal would it. In addition, evaporation helps to maintain humidity in the air. Most importantly I think it may have eluded you. Rivers flow because of rain!!! When it rains, the water collects together at a river and eventually drains out to the ocean where it re-evaporates and the cycle begins. Without rain, all the rivers in the world would dry up. And that would be bad

LoL, my bad completely. I would say that you will have a difficult time differentiating acceleration and jerk. The only possible sensible parameter I'd say our human bodies can judge is like Kygron said, force. Since acceleration is proportional to force, any change in the force exerted on you (most probably by your seat) can indication this change in acceleration.

I agree with (e) too Here are some mathematical aids. See for yourself which one is true Using the formal definitions introduced in electric circuits: (a) i = C dV/dt Cdv = int (i dt) CV = Q where C = capacitance, V = voltage across capacitor = Vcapacitor, Q = charge on capacitor. Note that Vcapacitor is always less than Vbattery, and so Q/C is always less than Vbattery. i = V/R (Ohm's Law) i = (Vbattery - Vcapacitor)/R i = (Vbattery - Q/C)/R (b) Voltage drop across resistor: delta V = Vbattery - Vcapacitator delta V = Vbattery - Q/C © i = (Vbattery - Q/C)/R (d) Q = charge on capacitor = charge across capacitor = Qcapacitor (e) delta V = Vbattery - Q/C