Sensei

How come? It's just a piece of plastic.. Reasonable price IMHO is $10-$20. Personally I would prefer to have 32 columns (not special separated region for Lanthanides & Actinides). How are you going to keep gases? They will be empty boxes? It would be fun to have them hermetic and filled with gas, and just connectors exposed, which after providing high voltage would start emitting light like discharge tubes.. Although it's way way beyond simple plastic container.

If you're the real scientist, in quantum physics, then you should no problem to answering following questions: - what is energy (in eV) emitted by Hydrogen atom going from n=5 to n=2 excitation state? - what is decay energy of Thorium-134, and what is decay mode? Either in eV and Joules. If you would provide velocity of products of decay, you would be master. Please attach calculations for verification for either question. Are you ready for it? Then answer above questions..

Why are you saying so.. ? @ 1 minute 27 second is hit. @ 1 minute 26 second we see object flying (just one frame). As we can see CCTV camera is taking 1 FPS (frame per second). This can be estimated using car a few seconds prior crash. American Airlines Flight 77 https://en.wikipedia.org/wiki/American_Airlines_Flight_77 was Boeing 757-223 as we can read on airplane website, max cruise speed is 850 km/h = 236.11(1) m/s. Although they probably flied much faster (because of rapid drop of altitude prior hit). Distance = Velocity * Time, Time is quantized in steps 1s, 2s, 3s, etc. That's why object is visible just on one frame. If it would be moving slower, or CCTV would be working at higher FPS, it would be recorded on more frames.. The real investigator could even estimate speed of object prior crash from such video (for higher precision there would be needed experiment in real world near Pentagon, to calculate distance visible from camera (it's fisheye-like, so it's distorting view)).

PIC means Peripheral Interface Controller. You need to have electronics that you will control by your chip. The simplest case is control of LED. Say you write code for turning on light for 1 second, then turning off light for 1 second. Upload it to chip. And LED is starting blinking. If you want 2 seconds delay, you need to upload new program. Or complicate electronics and code to allow user to change it by potentiometer. Chip won't do anything by itself. It needs stuff it will control.

Total energy of n photons is given by: [math]\sum\limits_{i=0}^{i<n}{h*f_i}[/math] or [math]\sum\limits_{i=0}^{i<n}{\frac{h*c}{\lambda_i}}[/math] For visible light: [math]400nm\leqslant\lambda_i\leqslant700nm[/math] For pretty monochromatic laser light with peak at 532nm (+-10nm): [math]522nm\leqslant\lambda_i\leqslant542nm[/math] Do you see light.. ?

Control electronics that you will make. f.e. electric engines, LEDs, light. Robotics. Automation. I don't think so. As long as you won't be blowing them everyday. $50-$100 for a start. Take a look also on Arduino: https://www.youtube.com/watch?v=b2CFKx5ASUY search for Arduino @ YouTube videos.

PICs are programmed in their own assembler. f.e. http://www.microchip.com/forums/m466343.aspx It's nothing familiar to anything that you know like high level language such as Pascal.

How come? f.e. acetic acid is organic polar solvent. And with many metal it's going into reaction creating acetate. f.e. Zn + 2CH3COOH-> Zn(CH3COO)2 + H2 (This one is pretty slow at room temperature) If there is reaction between metal and solvent, then how come it's not "damage of metal".. ? If you will polish, or do whatever you can to remove oxide layer, and leave on air again, it will go into reaction with oxygen sooner or later.. Quote from https://en.wikipedia.org/wiki/Neodymium "Metallic neodymium has a bright, silvery metallic luster, but as one of the more reactive lanthanide rare-earth metals, it quickly oxidizes in ordinary air. The oxide layer that forms then peels off, and this exposes the metal to further oxidation. Thus, a centimeter-sized sample of neodymium completely oxidizes within a year" ps. There is needed such solvent which reacts with oxide layer, but not with metal alone.

Level nine.. ? Are you scientologist.. ? It's very easy: build one.. If you manage to do so..

You mean that near death experience, inspired you to become scientist.. ? IMHO you should start from buying equipment. For a start: - distillation set up: Graham condenser, Dephlegmator, retort stand, few clamps different models (to hold condensers, to hold round-bottom flask etc.), hot plate for heating. - round-bottom flasks 50mL-1000 mL, - beakers 50mL-1000 mL, - graduated cylinders 50mL-250 mL. - pocket jeweler's weigh 100 g/+-0.01g, 500 g/+-0.1 g For learning about quantum physics, spectral lines of different gases etc., behaving of light, optics: - discharge tubes: Hydrogen, Helium, Neon, Krypton, Xenon, Nitrogen, Oxygen, CO2 f.e. - source of high voltage, such as Van der Graaf generator, Cockcroft-Walton generator. - prism - Young's slits different models, diffraction grating, polarization filters (at least two) - at least three laser models, red, green and blue. You will also have to have source of CO2, dry ice to buy, or from reaction. The easiest one is to take baking soda NaHCO3 and acetic acid CH3COOH. Reaction NaHCO3 + CH3COOH -> CH3COONa + CO2 + H2O will release plentiful of carbon dioxide for further experiments. Additionally you will have sodium acetate for other experiments. It's interesting substance. Search YouTube for details. Fill flask by water, connect to distillation set up, turn on hot plate, and distill water. You will have any amount of it. And gain experience in distillation process. Once you will have 1L distilled water in beaker, place in it electrodes and turn on electricity. And you will see nothing happens. No bubbles of gas from either electrode. Ampere meter will show 0 A current. Add salt, and bubbles will start appearing and ampere meter will show I>0 A. Clean up beaker, and use tap water - there will be gas. Not as much as with salted water. But enough to be visible. Clean up beaker, and use again distilled water. Turn on electricity. And put in it piece of dry ice.. and tell us, or better record on video all experiments, what happens.

In what way is electro-magnetic energy (i.e. light) "associated" with a water molecule? And why would you expect this to increase? Feel free to use mathematics in your reply. Well, temperature of water is increased while passing current through it. So it'll emit photons in not visible range (as always with such low temperatures as 273.15 K-373.15 K) microwave/infrared region of spectrum. Use any IR thermometer on boiling water to read temperature change from distance (because of emitted radiation from body).

This must be real world experiment or can be 3D simulation.. ? Plenty of 3D applications have physical particle simulation systems built-in. https://www.youtube.com/watch?v=0Drl97e8vP4 There are calculated collisions between particles each other, and normal objects in scene (collision objects, the one that you specify).

I really would like to see it. It might be accurate enough to measure +-0.050 J difference in burned sample, but it won't measure it by itself. If you turn off stirring, temperature won't be uniform anymore, but create gradient of temperatures inside of device (sealed) water container. Making reading from thermometer not reliable anymore. Even my hot tea that I just measured has 57.5 C at near top, and 54.5 C at bottom (measured by my -50 C...+300 C thermometer with +-0.1 C precision) If stirrer engine has U=5 V voltage drop, and I=0.05 A, that's 0.05 J/s introduced to system at worst maximum. (just example data for electric engine, the real data would be from 2nd voltage/ampere meters) If volume of water in container is 100 mL = 100 g, 4.1855 J/K*g * 100g * 1 K = 418.55 / 0.05 = 8371 seconds wait for increase water temperature by 1 C.. But is it really needed? You should rethink it. If stirrer is turned on, its engine has pretty much the same power when stirred liquid has temperature T0=25 C, and when T1=50 C, or when it has T2=75 C. It's always introducing exactly (+-) the same amount of energy to the system, regardless of whether heating element is turned on or off. You also misunderstood what I am interested in measuring in my device, in the first place. 4.1855 J/K*g is just example. I don't need to measure it. But can measure it, if I would like to. But what I wanted to really measure is amount of energy radiated away by body. Again. We have T0=25 C initial temperature. Device is turned on. Voltage on voltage regulator slider is adjusted to reach temperature f.e. 50 C. If T is still going up, voltage is adjusted to smaller value, if T is going down too fast, it's adjusted in reverse direction. Adjust until we reach equilibrium T=const, and some U on regulator. When it does not change, we can read amount of energy radiated away by body. Repeat the same with T=70 C, or any other. The key is to have constant temperature reading. Amount of energy introduced to system by stirrer is constant, regardless at which T we're currently working. Amount of energy from environment introduced to system is constant, regardless at which T we're currently working.

Sort of. Quote from link: "It does not account for the heat loss through the container or the heat capacity of the thermometer and container itself." Mine device is designed for measuring this loss. Typical calorimeter is insulated as best as they can make it. To keep heat inside of device. While in mine device insulation is not needed. Because what is radiated away is what we want to measure. Not energy released during burning f.e. (as there is nothing to burn in it).

Thermodynamics. How to precisely measure energy released by body. Imagine circuit with resistor (heating element) with resistance R. There is also voltmeter, ampere meter connected. [math]I=\frac{U}{R}[/math] and [math]P=I*U=I^2*R=\frac{U^2}{R}[/math] R is constant (would be hard to control it in heating element underwater). Heating element is placed in water (or other liquid) with mass m, and volume V. There is also precise thermometer, stopwatch, and magnetic stirrer to uniformly spread energy in liquid. At the beginning temperature is ambient T0 read from thermometer. Say 25 C. (ambient temperature is when amount of energy released by body equals amount of energy received from environment) We're heating it to T1, say 50 C. Electrons are accelerated, and while passing through heating element, they lose their kinetic energy and resistor is heated, and indirectly liquid. We read voltage drop on heating element from voltmeter. And read quantity of electrons Q/e, Q=I*t from ampere meter and stopwatch. Einput=I*t*U We can make equation to calculate energy needed to heat liquid with mass m, from T0 to T1: [math]4.1855 \frac{J}{K*g} * m * (T_1-T_0) = ~ I*t*U[/math] Calculate amount of energy needed to heat 1 gram of substance by 1 K: [math]4.1855 \frac{J}{K*g} = ~ \frac{I*t*U}{m * (T_1-T_0)}[/math] Calculate amount of energy needed to heat 1 mole of substance (water in example): [math]4.1855 \frac{J}{K*g}*18.015\frac{g}{mol} = ~ \frac{I*t*U}{\frac{m}{18.015\frac{g}{mol}} * (T_1-T_0)}[/math] [math]75.4 \frac{J}{K*mol} = ~ \frac{I*t*U}{n * (T_1-T_0)}[/math] (etc. etc.) It's idealized equation: when no energy is lost by emitting photons (in this case in microwave and infrared region of spectrum), nor by air gas particles taking part of energy. I was thinking how to precisely measure amount of energy released by body to environment and influencing our idealized equation. And came up with idea to use voltage/current regulator. If we supply more energy to our system, than is released, temperature will increase. Tcurrent must by higher than ambient temperature T0. Einput>Eradiated, Tnew>Tcurrent, Tcurrent>T0 If we supply less energy to our system, temperature will decrease (only to T0). Einput<Eradiated, Tnew<Tcurrent, Tnew>=T0 But if we use voltage/current regulator to precisely set value, with precision counted in mV, or uV, or better, temperature will be constant. Neither grow, nor fail. Einput=Eradiated, Tnew=Tcurrent=const And we will know exactly how much is radiated as photons or heat. This experiment could be performed at couple different temperatures, increased by f.e. 10 C. And receive set of data f.e.: Einput(T0+10) Einput(T0+20) Einput(T0+30) .... Einput(T0+70) Which will be placed on graph with temperature in x axis, and Energy/Current/Voltage/Power in y axis. Second experiment: change area which has contact with air, while mass and volume remain the same. To compare how results changes with area exhibited to air. Third experiment: compare plain liquid graph, with graphs obtained with liquids which have metal radiator different shape and size. Thus we will be able to measure efficiency of different radiator models. Fourth experiment: repeat everything, with different liquid than water. And compare data with water on graph. Best Regards!

Not sure whether Windows drivers allow multiple applications to record what's being said to microphone. You would have to check OS API specification. But as usual in Windows there is probably couple different API (old legacy, less old, modern). OTOH, if they have admin access they could (theoretically) replace drivers or part of OS to modified version, which would include such functionality. Or modify any file on disk such as Skype.exe or its libraries.

But there are such processes: redshift or blueshift. If somebody travels at speed close to speed of light, he/she just needs to analyze microwave background radiation. If it's relativistic Doppler shifted, such person should be able to detect it.

You must start from disallowing this machine connection to Internet. Because they're probably still monitoring what you're doing (observing screen-shots taken every second or few seconds, recording keys that you press, listening microphone if you have it, observing room by net camera etc. if you have them built-in, like in top end modern models, you can't even disconnect them).

One note. It's very simplified version. It's true for low voltages. But when voltage is measured in kilo volts or mega volts, exceeding voltage breakdown, even good insulator will start conducting electricity.

[math]E=m_0*c^2*\gamma[/math] m0 is rest mass. [math]m=m_0*\gamma[/math] m is relativistic mass. So when somebody writes: [math]E=m*c^2[/math] It can also mean the same as the first equation, depending on interpretation of m. where gamma is: [math]\gamma=\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}[/math] When v=0, [math]\gamma=1[/math] (the same frame of reference as we are) The first equation simplifies to: [math]E=m_0*c^2[/math]

Maybe I should explain a bit to UK/USA members what is happening (although I doubt you can help without even being able to replicate issue): If we have textView control, with inputType=numeric or inputType=numericDecimal etc, there is showed special smaller keyboard with just digits, and dot keys, so we can enter f.e. 3.141265, but half of world is using comma as floating point separator, not dot. When using Double Double.parseDouble( String string ); dot is treated as separator (even if locale is using comma). (there is exception while trying to parse string with comma, or it's cut in half) When using String.format( "%f", double ); where is used locale, to print double. So, if we print double to text, with comma as separator, and then try to parse it, with dot as separator, it will obviously screw up float/double. If we try to use String string = textView.getText().toString(); try { NumberFormat format = NumberFormat.getInstance(); Number number = format.parse( string ); value = number.doubleValue(); } catch ( NumberFormatException e ) { } catch ( ParseException e ) { }parsing and printing will be fine, using comma (locale),but (virtual) keyboard on smartphone (at least mine) is showing digits keys and dot key, and not comma key, so not being able to enter correct separator character, and not being able to enter fraction..

Hehe. It won't stop you from eating radioactive Potassium-40. Or from radioactive Carbon-14..

Lol. If you would win lottery (if you would play at all), you could buy whole island and build new CERN on it.. I would certainly do it this way.. Island on which you need to have PhD/significant scientific achievements to even land on it.. Yes, he is very similar with Planck. (and me, if I cut beard) Hehe. I meant, our own forum scientists conference..

Oh, really? How come so.. ? If we measure now object with mass m to be accelerated by force F to speed v, will it change? Only Jesus, would say so.. Are you Jesus? Anyway, it was very nice. We should love even people with who we disagree.

Hello! Android and entering floating point numbers. With inputType=numericDecimal Has anybody found any sensible solution to this madness.. ? Google has not done anything since at least 2009 AFAIK. This for instance http://stackoverflow.com/questions/12780125/soft-keyboard-with-comma-instead-of-dot-for-decimal-separator-in-android doesn't work on my Samsung Galaxy S5.. Tried input.setKeyListener(DigitsKeyListener.getInstance("0123456789.,")); and the only thing it did, changed "." (dot) key on keyboard, to ".-,+", but it's just visual (!), not having any impact on entered characters in text view field.. (still not able to enter comma character on the phone). The only thing that really works is ignoring "numericDecimal" and going with "text" inputType... Best Regards! ps. If you're UK/USA citizen you will probably have no idea about what I am talking about... But as you can see on the map (middle of page) https://en.wikipedia.org/wiki/Decimal_mark https://en.wikipedia.org/wiki/File:DecimalSeparator.svg It's affecting at least half of human kind..