mezarashi

This is something you probably have noticed, but if you have an incoming call while your mobile phone is close to some speakers, you will hear some jitter in the speakers, which can get relatively violent depending on their quality and magnetic shielding. You can conclude that the presence of strong electromagnetic fluctuations in the radio frequency range will induce electrical fields in the electrical-magnetic system. The question is (knowing that light and radiowaves are both electromagnetic waves simply of different frequencies), why is it that if you shine visible light, infared, or UV light onto the speakers, you will definitely get no response, even if the light is so intense it will melt the speakers away. This is suggesting that there is some special "cut-off" frequency in which electromagnetic waves no longer can induce currents in the speakers.

A projection is like a shadow. A projection of A onto B would be like the shadow of A on B, which is different from the dot product, because there is no multiplication with the magnitude of the B vector. You are simply outlining how much of the vector A is in the same direction as B. The trickey thing to do is usually finding the angle between A and B. The cross and dot product are mathematical operations that you can perform on vectors. There are many physical interpretations. As such they are useful in helping us find solutions to physical problems which are accurately modeled by them. For example the Lorentz force (a force vector) on a moving charge in a magnetic field is the cross product of the charge's velocity vector and the magnetic field vector.

I've been reading a bit on advances in optical->electrical energy technologies and as Locrian mentioned, it seems promising. If our solar technology can advance to the point where we can get reasonable voltages and current from a roof-top installation, I think that we will start seeing commercial production. They would need to work in non-direct sunlight too I suppose (i.e. cloudy weather) There is one other factor however that continues to stop us from using it as the only source of energy, that is, our inability to save this energy as it is a fact that sunshine doesn't shine all the time. Along with solar technology needs to come battery technology. We could always of course have a complex power grid so that we can relay power from areas in sunlight. Either this or we develop efficient wireless energy transmission systems to beam in solar energy harvested from space. It's exciting to speculate how our harvesting of energy will morph over time. Considering that fossil fuels run out in 50 years, I guess we will start seeing major changes in our lifetimes.

Well true. It's difficult to go into a career hating a subject that is fundamentally important in understanding your trade. However, if you can get over "hating" it and understand it whilst hating it, this is all right. Take for example math. Many probably hated math when you were in elementary. But as adults, all of us (I hope) know how to do simple arithmetic, and we can use it to perform our daily chores to go on with life comfortably. . Something similar happens in the more "difficult" sciences in university. Once you get so used to a certain subject, whether you hate it or not, you are able to do it and use the knowledge to achieve your ultimate goal. I remember myself and my hate for mathematics, Fourier and Laplace transforms particularly. They have become but common terms I take intuitively in my trade of electrical engineering. As a doctor though, I doubt you'll be adding chemical formulas for a living. The chemical and biological processes are just fundamental knowledge you'll need for diagnosis and most importantly the proper prescription of drugs while understanding their effects on the human body. I know many doctors who more or less do this on experience.

I guess your ultimate question here is from the photosynthesis formula: 6H2O + 6CO2 -> C6H12O6+ 6O2 which oxygens on the left end up in the sugar and which end up in the O2 which is released as a gas. This would need understanding of the sub mechanisms for the reaction as this reaction will not occur spontaneously. From what I understand, photosynthesis occurs in two stages, the first which is the light striking the chlorophyl creating energy, a reaction with water (H2O) and will produce oxygen as a by product. The second stage is the Calvin cycle which can occur independently given the energy and H atoms to turn carbon dioxide into sugar. So my conclusion is that, if the radiactive oxygen is in the carbon dioxide, you will produce radiactive sugar, while if the radioactive oxygen is in the water, you will get radioactive oxygen gas but "normal" sugar. I'm sure a chemistry/biology expert who knows more details of the photosynthesis cycle would be glad to verify if this is incorrect

Oh how I wish I had that great book of "They Said That!?" by famous celebrities. My local newspaper hosts a section for recent "updates" if you will, and they are hilarious. Just a few from the internet: "The internet is a great way to get on the net" -Republican presidential candidate Bob Dole "My parents have been there for me, ever since I was about 7." David Beckham "We're going to turn this team around 360 degrees." -Jason Kidd, upon his drafting to the Dallas Mavericks I haven't committed a crime. What I did was fail to comply with the law. - - David Dinkins I do not like broccoli and I haven't liked it since I was a little kid and my mother made me eat it. Now I'm President of the United States and I'm not going to eat any more broccoli. --George Bush "I would not be bothered if we lost every game as long as we won the league." Mark Viduka

Interesting question. Well I guess this reinforces the understanding of how the summation formula for arithmetic series works. Remember the story about how Gauss added 1 to 100 in 5 seconds for his teacher although it was meant as punishment. So we have an arithmetic series: a1 + a2 + a3 + a4 + a5 + ... an Normally, you would add this through the formula Sn = (n/2)(2a+(n-1)d) or more simply Sn = (n/2)(a1 + an) which is basically how Gauss did his consecutive integer calculations. You add the last term to the first and multiply by the total number of terms divide by 2. 1.Given the conditions you stated, the first condition: a1 + a3 + a5 + a7 +... an You can see that the summation here will be Sn1 = ((n+1)/4)(a1 + an) = 320 If this isn't clear. Write it out, perhaps to n=9. Write out all a1 to a9. You won't be adding all numbers this time, but how would you find the sum quickly? You still continue to add (a1 + a9), (a3 + a7) + a5. Note that (a1+a9)=(a3+a7). Instead of (n/2), we have here (n+1)/4. 2. Doing similar analysis for the second condition, we would get Sn2 = ((n-2)/4)(a1 + an) = 224 Just assume that the last term will be part of the series, (i.e. if you have n=9, you will notice that the formula doesn't work, but n=10 will) In anycase, this fraction will cancel out later on. So we have 2 equations Sn1 and Sn2. Let's group (a1 + an) together as a constant and let it be x. We then have two equations to solve with 2 unknowns. Solve for n. When you have n, you can then solve for (a1 + an). Knowing both n and (a1+an), we know that the proper formula for the entire summation is: Sn = (n/2)(a1+an) as I mentioned, so plug in the numbers and you have your answer. [Additional] Further calculations for the values of a1 and d can be made as well, now that you know n. Just expand it accordingly and equivalate. I've put my answer below as a series that complies to all the conditions. Highlight. 192 160 128 96 64 32 0 -32 -64 a1 = 192, d = -32, n = 9, S9 = 576

I'd think google could probably help you here. A search for "helium balloon lift" would probably do the trick ^^. In principle however, the helium balloon works on the principle of bouyancy the same way certain things float in water. You can take the weight of the air displaced by the helium balloon as the bouyant force on it, causing it to "levitate". Of course, since we must take the weight measurements in air itself, you'll probably have to use the density of air vs density of helium instead.

Yes, we say that there is a voltage drop across the resistor. Voltage is also known as electric potential. Yes. Because ideally you don't want to disturb the circuit. You want to be able to probe any two points on your circuit without changing the circuit dynamics. This is only possible if there is a high input resistance.

I remember learning this one in 5th grade. It's "con" + "science" to spell conscience. However, you'r conscience makes you conscious . Makes you wonder if the whole concept of "conscience" is actually true or just a con... okay bad joke. About quantum consciousness, it sounds like a religion meet science type of topic. I've seen multiple books published that would fall under this category. They are not "hoaxes" and at least philosophically and spiritually they give you something to think about.

ln is log base e. It's just used so often in natural sciences that it has it's own symbol. ln y = x, is the same thing as saying: e^x = y

To your first question. No, it's not a loss in velocity. Actually the drift velocity of electrons in a copy wire isn't all that great. Consider how many electrons are in a centimeter of wire. How much fast does charge need to flow to get 1A? The loss is the loss of electrical potential. You can imagine yourself on an elevator coming down from the 80th floor. When the elevator passes by the 10th floor, you are at the same speed, but you have lost gravitational potential. For question 2. The voltmeter (atleast in theory) does not affect the circuit due to its very very high input resistance. Meaning no current will go into the voltmeter. Modern voltmeters will probably use a ADC (analog to digital converter) somewhere to detect the charges on internal capacitors.

I'm sick and tired of reading new news about how some company or committee has come up with a new way to encrypt or protect their content from piracy. The truth is that piracy will continue to happen, because there is nothing fundamental you can do to protect media and software from being pirated. This leads to a couple more issues. 1. Who are you actually hurting? The bootleggers who know how to hack and get around your "protection" technology and are selling your DVDs for a dollar a disk down in China or HongKong. Or is it actually the innocent consumer who is becoming increasingly hassled with all the protection unlocking steps (like for Windows XP even a hardware detection system). This completes another cycle: I get the bootleg because the installation and usage is more convenient than the original. Accept it, the majority of software, media piracy is done by people who know how to get around your technology. Once they create a "ripped" version, this version becomes easily distributable. Often, these rips are not any lower in quality than the original. 2. Is it even possible to protect media? Once the original copy is in my hands, then I can watch it. What I can watch and hear I can record. For games, what I can install I can copy. If you try to disable something, then I disable whatever it is you are using to disable. It's always a losing game for the guy on defense in the world of computer programming. 3. Do you think that those people who watch pirated movies and download pirated software would buy otherwise? If they can't get their hands on Norton, there are also free anti-virus scanners. If they can't get MS word, there's also Open Office. If I watched that pirated movie, it's because I didn't have any more money to spend at the cinemas. I think this third issue has been shown to be imprecise time and again. Just because someone has $30,000's worth of mp3's on his/her computer doesn't mean that the MPAA has lost that much in revenue, nor do I think it would've been a significant amount ( an average college student can barely earn 1/10 of that per annum in free allowance ) So, after saying all that, WHY DO THEY DO IT? Why put millions and billions of dollars into investing something as absurd as this. WHY is it that everytime I play Warcraft 3 using my original copy, I have to put in that annoying CD while my friend who got the bootleg for free doesn't!

Googling "brain-freeze": -7-Eleven registered the term "brainfreeze" in 1994 to communicate the painful joy of drinking a frozen Slurpee beverage. -Brainfreeze is also known as an "Ice Cream Headache", and "Frozen Brain Syndrome" Nice. Thanks guys ^^;

I wanted to ask about the science behind the feeling you get when you drink a cold liquid too quickly. You feel this rush of "coldness" up to your brain, and you simply must stop drinking until the "coldness" goes away. And is there even an official name for this?

Well hmmm, maybe I'm looking for a bit of both ^^. I've seen Lagrange's equations being derived from Hamilton's and D'Alembert's principles. I guess I'm just confused on who got who's idea from who, because they seem all interlated, I'm not sure which is the more fundamental or atleast original of ideas.

1. It is really neat how the Lagrangian Equations of Motion can be derived from Hamilton's Variational Principle, and solve so many classical systems much easier than from a Newtonian approach. Does anybody have any leads to how Lagrange came up with his "Lagrangian", specifically the equation: [math]L = T - U[/math] where T is the kinetic energy and U is the potential The equation seems somewhat arbitrary and made up yet it works! 2. The second question is, who really came up with this formalism? The Lagrange equations are derived from Hamilton's principle (or can they somehow be derived from each other?), but then again, Hamilton's principle has the Lagrangian in it, so... who came first and who really put it together. [math]J = \int_a^b L dt[/math]

Try using netsend. This is a default service on most XP machines which is god darn annoying and will at times frighten your friends who don't know about it because it pops right up. Go to the DOS console by typing... Run -> cmd. Then type in: net send and follow the instructions, generally you can try it on your own computer. If your computer is called: mycomputer, then type: net send mycomputer hello1234 message Here's a site on it. Note that some people have developed GUI interfaces to accomodate this net send feature to make sending and receiver much more user friendly. http://www.lantalk.net/net-send-command/

I think the hint was in the question. Break the motion into the two components, x and y. Remember that only the y component is affected by gravity. The velocity in x will be constant.

Actually I think you're a bit confused on the part that relativity doesn't suggest that time travel is possible. This means going to the future or into the past. The consequence of assuming that the speed of light is constant (a postulate for special relativity) is that time is relative (variant). The way you phrased it: "I fail to see how any one object in space, going at any speeds, could change time itself" is a bit misleading, because there is no universal time with relativity, as with there is no universal frame of reference. The analogy with the grid is usually used to explain general relativity from a qualitative point of view. The time dilation effect (slowing down of time) suggested by special relativity is better shown using the light clock analogy. Special relativity says that two observers moving relative to each other will observe time slowing down in the others' frame of reference.

I'm not sure what you mean by the definition of a derivative. Do you mean find the derivative using first principles? Or else you can use the power rule f(x)=x^a, where a=1/3.

Seems like it. Simple test, plug in 7.999999 or 8.000001, and see what you get It's a bit strange that your teacher would give you this problem before teaching how to differentiate in the first place but oh well... maybe there's another way to do it.

Unfortunately, that cannot be answered with the information given in the original question ;;; which seems to be concocted through a lack of understanding of the ideas he/she wished to ask.

L'Hopital's Rule: Basically you differentiate the numerator term and differentiate the denominator term SEPARATELY. Then do substitution again. If you still get the indeterminate form of 0/0 or infinity/infinity, then you may repeat the rule again.