Sensei

Take sqrt() only from positive numbers.. So use fabs()/abs() or equivalent code, to get rid of negatives... BTW, y[] is defined as float, so complex class sqrt() is never executed, only regular one.

sqrt() the most likely is returning NaN (Not A Number) for negative input. The same happens with regular hand calculators (E letter appears etc.) http://en.wikipedia.org/wiki/NaN It's mentioned on wiki page "There are three kinds of operations that can return NaN: [...] The square root of a negative number."

value ^ power is not working in C/C++ the way as in math. It's bitwise xor operator.. http://en.wikipedia.org/wiki/Operators_in_C_and_C%2B%2B

It's better described on Electronvolt wiki page, than I can express it.. http://en.wikipedia.org/wiki/Electronvolt eV is simply alternative unit of energy, instead of Joules. eV/c^2 is alternative unit of mass.

I found in shop such accumulator http://sklep.avt.pl/akumulator-zelowy-6v-4-5ah.html "pojemność (25 stopni C) 20 godzinna: 4.50Ah 10 godzinna: 4.28Ah 5 godzinna: 3.78Ah 1 godzinna: 2.70Ah" It means that capacitance is variable, and depends on how fast or slow, you will be utilizing accumulator. If you will be utilizing it for 20 hours, you will have capacitance 4.5 Ah = 16200 C. And you need I=0.225 A to last it that long. The faster you will use accumulator, the smaller capacitance. There is also showed dependency of capacitance on temperature in page that I linked. There is also showed spontaneous discharge of accumulator in 3,6,9 months..

It won't affect functionality of your code, but better place prototype declaration outside of main() function (before it). In large sources, prototypes are usually put to separate files with .h extension (h = header). Then you can include them in multiple source files.

calculate_area() doesn't mean cal_area()... They have to have the same name for both. You should have functions declared ABOVE main() where they're used. Otherwise you have to make prototype. Prototype for your function would be line int cal_area(int); Get rid of this "int" where is cursor, it's causing error.

If that's all what you really have, the most visible error is lack of bracket. It should be int area(int m) { And error is obvious - you have uninitialized variable i Shouldn't you multiply by m? int calculate_area(int m) { return(m*m) } Better don't use the same name for variables and function name! Difference between uninitialized variable: int a; and initialized: int a = 0; or int a = b; Initialization means assigning some value to it. Uninitialized variable has random data so using it might result in crashing computer in the worst scenario.

In star core there is happening nuclear fusion, which releases a lot of energy. You should read this article http://en.wikipedia.org/wiki/Nuclear_fusion

Because molar mass was earlier. The idea came from Amedeo Avogadro in 1811 http://en.wikipedia.org/wiki/Avogadro_constant He didn't calculate Avogadro const, he just came up with idea. Scientists calculate it hundred years later. f.e. take 18 grams of water, and 58-59 grams of NaCl, then pass current through it, and you will produce equal volumes of Hydrogen and Chlorine gases on electrodes, and 40 grams of NaOH will remain in container. You don't even have to know what is Avogadro const. Doing electrolysis of various materials/metals, you can calculate how much they lost mass. Electrodes are dissolved in solution, joining with oxygen and/or hydrogen. Containers which are normally collecting hydrogen and oxygen, will have different volumes of gases than in pure water electrolysis. f.e. electrolysis of iron made electrodes gives no oxygen on positive electrode - it's entirely immediately reacting with iron, and it's visible as yellow-reddish water. Electrode is literally disappearing. 1 hour is enough for me to wipe it out entirely. And water is totally opaque, yellow-reddish not transparent. But hydrogen container will be filled by gas - so we can easily calculate how much of oxygen reacted with iron. After heating such solution, to get rid of entire water, there will remain small amount of powder. We can also calculate its mass.

In high energy physics electronvolt unit is simply useful and convenient. You can say "electron at 2 GeV" and everybody know what you had in mind. You can see example of usefulness of using eV in this thread http://www.scienceforums.net/topic/81036-nuclear-properties-of-isotopes/?p=788316 I don't want to hijack your threads. We would have to speak off-forum.

I showed you how to calculate stuff in #3 post..

P = I * U so in units you have 1 W = 1 A * 1 V but 1 W is also 1 J /s so E/t = I*U 1 J/1s = 1 A * 1 V which gives E = I*t*U 1 J = 1 A * 1s * 1 V (I am using 300 kWh of energy per month - so it's like saying: 300,000 W*h = 300,000 J/s*3600s = 1,080,000,000 J of energy) 1A*1s = 1 C so again 1 J = 1 C * 1 V But 1 C is quantized by e const 1.6*10^-19 C 1 C / e = 6.25*10^18 charges/electrons. so 1.6*10^-19 J = e*U (U=1 V) And this is exactly 1 eV energy unit http://en.wikipedia.org/wiki/Electronvolt If you divide 1.6*10^-19 J by Planck const 6.626*10^-34 J*s you will get the same if you would divide 1 eV / 4.135667*10^-15 eV*s 1.6*10^-19 J / 6.626*10^-34 J*s = 2.414*10^14 Hz 1 eV / 4.135667*10^-15 eV*s = 2.414*10^14 Hz You can go back and forth from eV to J, dividing by Planck const in one unit, and then multiplying by Planck const in other unit. f.e. Electron has 510999 eV energy 510999 / 4.135667*10^-15 = 1.23559*10^20 Hz (it's Compton frequency BTW) 1.23559*10^20 * 6.62607*10^-34 = 8.187*10^-14 J m=E/c^2 so m=8.187*10^-14 J / 299792458^2 = 9.11*10^-31 kg and we have our mass of particle in kilograms matching data..

Psi http://en.wikipedia.org/wiki/Psi_%28letter%29

Kramer, build particle detector, Cloud Chamber, like I said many times in this and other forums. It costs $20-$30 and $3-$4 for single run. Then you will be able to see electrons, positrons, muons, pions, kaons. Effects of ionization from x-rays, gamma photons, etc. etc. Use magnet/electromagnet below cloud chamber and it will reveal charge of particles. http://www.youtube.com/watch?v=Efgy1bV2aQo http://www.youtube.com/watch?v=ZLiXgdymIYE

Pion 0 has mass ~135 MeV/c^2 In kg it would be ~2.40676*10^-28 kg ~264.2 higher mass than electron.

50 bar is almost ~50x higher pressure than standard pressure. It's almost pressure that's at 500 meters below sea level in ocean. Water mass is 500 tons per m^2 at such depth. Submarines from II war were squashed at 200... 280 meters, at twice smaller pressure. Pressure will be so high, it's unlikely to push piston back... Normal piston is releasing gas that's inside when it'll go to upper est position, then piston goes in reverse direction, because it's empty and nothing disallows going it in that direction. Then there is injected fuel, and spark is igniting it, and changes liquid to gas which pushes piston again. And everything is repeated. You want to push piston back, without releasing CO2 from inside.

If you want practical chemistry, I would rather recommend electrolysis of water.. http://en.wikipedia.org/wiki/Electrolysis_of_water You can produce hydrogen, oxygen from pure water. If it's solution of salt, you will be also able to produce chlorine and sodium hydroxide. From previously made hydrogen and chlorine you will be able to make chloride acid. After using metal electrodes you will be able to produce various oxides, and hydroxides. This way you can calculate masses of the most common chemical elements. Also ideal gas law is essential thing to learn. http://en.wikipedia.org/wiki/Ideal_gas_law

Under standard pressure, normal temperature (on Earth), CO2 dry ice goes directly from solid state to gaseous state skipping liquid state (sublimation) at -79 C. So "bottom liquid CO2" from image C, the most probably, will never happen. Yes, piston would be pushed up. But why on image D it would go back to "squeezed"?

1 A*h = 1 A * 3600 s but 1 A*s = 1 C so 4.5 Ah = 4.5*3600 = 16200 C One electron has charge 1.6*10^-19 C so 16200 C / 1.6*10^-19 = 1.0125*10^23 electrons If you will have resistance 1.33 ohm, I=U/R=6/1.33 = 4.5 A, your battery will be working just 1 hour with such current. With I=0.6 A (3.75*10^18 electrons per second) it'll be working ~7h 30m. I don't know where you get 20h.. You would need 0.225 A to work it 20h.

People start programming hardware from buying equipment for programming EPROM etc. PIC chipsets. On eBey etc. there is plenty of these, and in your local electronic shops probably should have them too: http://www.ebay.com/itm/New-Genius-G540-USB-Universal-Bios-GAL-Programmer-EPROM-FLASH-51-AVR-PIC-MCU-SPI-/310852711326 You program it in its machine code. Java, Obj-C is way too high level.. Read about PIC microcontrollers http://en.wikipedia.org/wiki/PIC_microcontroller After programming chip is doing what you told it to do, like you would build whole chipset using logic gates etc. elements. This item has nicer photons. http://www.ebay.com/itm/High-speed-true-USB-Universal-Programmer-TL866CS-Full-Pack-include-7PCS-adapters-/320967107440?pt=LH_DefaultDomain_0&hash=item4abb216370

Do not you have electroscope? It's just $18.. Building your own, nothing (everybody has in home some aluminum foil and glass) Machine for making high voltage costs $160 (Wimshurst machine) Device for visualizing magnetic field lines? Costs $37 You can build your own using magnet and a lot of needles. How about laser and some diffraction material, polarization filters, lenses etc. ? Your school really doesn't have the simplest cheapest the basics equipment? You can order them through Internet from EU/USA/China..

Pion 0 can also decay to gamma photon, electron and positron. It's just one of decay modes. They can be really complicated.. Gamma photon will collide sooner or later with charged particle and accelerate it, and new photon will be produced with lower energy. See Compton scattering. http://en.wikipedia.org/wiki/Compton_scattering Without collision of photon with something, you have no idea that photon was there..

You put a lot of effort in preparing your pdf, and we appreciate it. But you should first learn how to calculate stuff. Otherwise it's quite wasted time (but wasted on good thing). I am pretty confident that you "forgot" to subtract energy of electrons from isotope energy-mass that you took from database of isotopes prior doing calcs. You should calculate just energy of nucleus, not nucleus with electron cloud! Let's analyze case that's showed in this page: http://en.wikipedia.org/wiki/Positron_emission 11C → 11B + e+ + νe + 0.96 MeV Carbon-11 has mass 11.01143361 u, you have to multiply it by 931.49406121 MeV and receive energy 10257.0850131232 single electron has energy 0.510999 MeV Carbon has them 6 so E-Eelectron*6 is 10254.0190191232 MeV (approximate energy of nucleus of C-11) Boron-11 has mass 11.00930544 u *931.49406121 = 10255.1026354069 MeV -0.510999*5=10252.5476404069 MeV Subtract them: 10254.0190191232-10252.5476404069 = 1.4713787162 MeV energy of positron that is escaping nucleus is 0.510999 MeV so subtract it as well will give 0.9603797162 MeV And it's decay energy of this isotope. Pretty matching value from article. It's combined kinetic energy of positron plus energy of neutrino. If I wouldn't subtract electrons cloud energy, I would receive 1.47 MeV instead of 0.96 MeV, giving impression that nucleus is absorbing electron from cloud.. No. In beta decay+, which is positron emission, nucleus is not absorbing shell electron, nor it doesn't need to import electron from environment.. Rather reverse. Carbon-11 is emitting positron, and losing electron from electron cloud because it's no longer hold by electrostatic forces (less protons in nucleus after decay). Positron will eventually annihilate with some electron ionizing other element and producing gamma photons which can ionize even more elements.

One way is to fill hermetic container (with little hole on bottom) with distill water. You have to place it in even bigger container with water, so water will be able to escape on demand. Then put pipes from setup where is done f.e. electrolysis of water, where is produced oxygen (on + electrode) to container. Oxygen will push away water from container. After that container is closed from bottom and contain very pure oxygen. We now need to get ride of oxygen by burning some metal inside of this container. Pass large current through wires, and metal will react with oxygen and burn. Oxide of metal will remain inside hermetic container as dust.