What is Maxwells Constant and how to use it?
370472
Check this site: http://dgleahy.com/dgl/p35.html\n\nMaxwells constant is actually the velocity of the electromagnetic wave,including light at 3 times 10 by the power of 8 metre per second. Use it with Einstein's famous equation e=mc square, m as mass of photon or particle and c the velocity.
807253
nu = c / lamda\n\nsolve for lamda - you want wavelength.\nc is the speed of light = 3 x 10^8 m/sec\nnu is your frequency = 3.2 e-2\n\nplug into the calculator: 9.375 e9 meters
972525
Calculate it yourself, using the equation: c = f x w\nc = velocity of light in vacuum = 3 x 10^8 m/s\nf = frequency of the light\nw = wavelength (convert 4000 A to meters; 1 A = 10^-10 m)
748559
The dielectric constant of a material can be found by a variety of static electrical measurements. The complex permittivity is evaluated over a wide range of frequencies by using different variants of dielectric spectroscopy, covering nearly 21 orders of magnitude from 10−6 to 1015 Hz. Also, by using cryostats and ovens, the dielectric properties of a medium can be characterized over an array of temperatures. In order to study systems for such diverse exciting fields, a number of measurement setups are used, each adequate for a special frequency range.\n\n    * Low-frequency time domain measurements (10−6−103 Hz)\n    * Low-frequency frequency domain measurements (10−5−106 Hz)\n    * Reflective coaxial methods (106−1010 Hz)\n    * Transmission coaxial method (108−1011 Hz)\n    * Quasi-optical methods (109−1010 Hz)\n    * Fourier-transform methods (1011−1015 Hz)
737785
I'll let you do the calculation, but the equation you need is:\n\nE = hc/lambda \n\nwhere h is Planck's constant, c is the speed of light in m/s and lambda is the wavelength of the light (in meters). You can find the rest of the numbers in your text book, I'm sure...
692212
f = v / lambda -> lambda = v / f\n\nv = velocity\nf = frequency\nlambda = wavelenght\nv in vacuum = c = 3 x 10^8 m/s\n\nlambda = 3 x 10^8 / 6 x 10^14 = 0.5 x 10^-6 m= 500 x 10^-9 m = 500 nm
851776
E=mc^2\n\nwhere \n E is in joules or kg·m²/s²,\nm is in kilograms, and \nc equals speed of light, in meters/sec (approx 3*10^8)
819160
em energy radiated is proportional to T^4, so it would be : ((273.15+9627)/ (273.15+5227))^4 =10.5
937514
wavelength = speed of light / frequency\nfrequency = speed of light / wavelength\nf = (3*10^8 m/s) / (615*10^-9 m)\nf = 4.878*10^14 Hz\n\nwavelength = planck's constant * speed of light / energy\nenergy = planck's constant * speed of light / wavelength\nE = (6.626*10^-34 J*s) * (3*10^8 m/s) / (615*10^-9 m)\nE = 3.232*10^-19 J
944667
Electromagnetic waves travel at the speed of 300,000 km/s.  Just divide your distance by that speed.
872993
E=m c^2\nEnergy=Mass &#92;times (speed of light)^2
922532
E = 423.6 kJ / mol ; the amount of energy needed, but this is per mole, you want it per atom, so you use Avagadro's number:\nE = 423,600 J/ mol / (6.02E23 atoms/mol) = 7.036 E-19 J/atom\n\nh is planck's constant; 6.626068 × 10-34 J * s\nc is the velocity of light in a vacuum 3.00 E8 m / s\n\nThe E__ is a shorthand way of writing * 10^__, so E8 means * 100,000,000 and E-34 means * 0.0000 0000 0000 0000 0000 0000 0000 0000 1 (which makes it a very small number)\n\nYou can rearrange the equation like this:\nE = hc/λ\nλ * E = h * c\nλ = h*c / E\n\nλ = 6.626068 E-34 J * s * 3.00 E8 m/s / (7.036 E-19 J/atom * 1 atom)\n\nλ = 2.825 E-7 m = 28.25 E-6 m = 28.25 μm
222936
Consider a conductor of uniform area of crossection A, length L. Let resistivity of conductor be S and resistance of material to current be R.\n\n  Then,  R = SL/A\n\n     Multiply and divide by mass of conductor say m and L\n\n          R = SL*L*m/L*A*m\n\n But L * A  = Volume = V\nAnd m / V   = Density = d\n\n  Hence,  R =  S*L*L*d/m\n          S =  (m*R/L*L)/d\n\nThus, Resistvity (S) is inversely proportional to density(d) when other quantities are kept constant.
232906
The energy is a single photon of a wave is given as,\nE = h * c / lambda\nwhere E is the energy, h is plank's constant, c is the speed of light in c a vacuum and lamba is the wavelength.\n\nSince h and c are constants, If you are trying to find the wavelength, you can calculate the energy using the following equation, then plug it back into the first and solve for lambda,\nchange in E = b[1/n_i ^2 - 1/n_f ^2]\nwhere change in E is the change in the energy level of the electron (how much energy it gives off) when the electron moves from its initial energy level (n_i) to its final energy level (n_f). b is a series of constants with an over all value of 2.18 E-18 J.\n\nI couldnt really type the formulas very well due to the limitations I have here, but there are nice versions show on this web site along with an explanation,\nhttp://www.amazingrust.com/Experiments/background_knowledge/Spectroscopy.html
566230
Set up the equation\n\nP = dRT\n\nAssuming P to be constant,\n\ndT = d'T'\n\nor d' = 0.76*273/298 = 0.696 g/l.
833112
First, I'm assuming that your question is "what do i do to make sure the length and mass per unit length of the wire is constant AS tension changes?" (note change of 'and' to 'as') and the rest of the material is just for explanation.\nThe answer is, you can't keep M/L constant. If your setup actually uses two fixed supports, L will be constant, but tensioning the wire stretches it, removing some M from between the supports. You can measure the amount of stretch by making two benchmarks (maybe small pieces of masking tape) in the unstretched wire and measuring their distance as you change the tension. (Doing the appropriate scaling between the baseline distance and the length of the stretched wire.)\nFor the theory on wire tension and resulting dimension changes, search Poisson's ratio. Basically stretching a length L of wire by epsilon*L (epsilon is a small number) would result in a thinning of radius R of 0.5*epsilon*R if the material were incompressible (thus inexpandable). However all solids including steel expand under tension to some extent, so the Poisson ratio of 0.5 above is more typically 0.25 to 0.35 for metals. See the ref.
1076485
i dont know much about physics but i did a search and this is what i got hope it has what you're looking for:\n\nhttp://science.electronicsinfoline.com/Physics/Electricity_and_Magnetism/