propagation constant, gamma. Is the numerator amplitude and the denominator distance??
100480
That would be my best guess too.
51150
UGH!!!  Thx for the effort (I was the one to post that question, as a part of a solution to another problem I found here on Yahoo! Answers.)  Incidentally, I made a mistake when simplifying the other problem and it should be xln(2)-2ln(x)-ln(86400) = 0\n\nI don't know any way to get rid of the ln(x) unless maybe it's using calculus techniques?  I'm just glad this problem is making other ppl trouble about it as much as I did :-).
866698
Check out this website. They discuss both direct and reverb loudness. It seems the steady-state reverb level depends on surface area and absorption coefficient while direct is 1/R^2 dependent. Unfortunately the symbols in their equations are not always defined right below the equation but they will be somewhere on the page. If you see a symbol that looks like 2 vertical sides and the top of a square, it's pi. Also they're not too careful about parentheses. One eq. you'll notice says\nWreverb = Wsource * (1 - a / Sa) but I'm betting it should be (1 - a) / Sa. (Otherwise they're using the fraction a/Sa which reduces to 1/S with 'a' irrelevant). Good luck. Maybe you'll find a more readable site.\nP.S. Of course one major effect you'll notice if your program material extends into the deep bass is a falloff of low-frequency response below about 22 Hz. From the ref., Fcutoff = c / 2 * longest dimension, should be c / (2 * longest dimension).
695323
Hi. From the web, they use the constant 'k' :    http://answers.yahoo.com/question/;_ylt=Aku8q5AETPeTKbquDgodzrYezKIX?link=answer&.crumb=xa6shFEtoH4&qid=20061103074030AA2hrrt&.done=http%253A%252F%252Fanswers.yahoo.com%252Fquestion%252Findex%253Fqid%253D20061103074030AA2hrrt\nThermal conductivity = heat flow rate × distance / (area × temperature difference).
363774
Radio propagation\nRadio propagation is a term used to explain how radio waves behave when they are transmitted, or are propagated from one point on the Earth to another.\n\nIn free space, all electromagnetic waves (radio, X-rays, visual, etc) obey the inverse-square law which states that an electromagnetic wave's strength is proportional to 1/(x2), where x is the distance from the source. Doubling the distance from a transmitter means the strength is reduced to a quarter, and so on.\n\nRadio propagation on Earth is not only affected by the inverse-square model, but by a number of other factors determined by its path from point to point. This path can be a direct line of sight path or an over-the-horizon (see also radio horizon) path aided by reflection from the ionosphere. A variety of phenomena make radio propagation more complex.\n\nSince radio propagation is somewhat unpredictable, such services as emergency locator transmitters, in-flight communication with ocean-crossing aircraft, and some television broadcasting have been moved to satellite transmitters. A satellite link, though expensive, can offer highly predictable and stable coverage of a given area.\n\nA sudden ionospheric disturbance is often the result of large solar flares directed at earth. These solar flares can disrupt HF radio propagation.\n\nPropagation modes\nRadio waves at different frequencies propagate in different ways.\n\n\nThe Antenna\nThe beginning and end of a communication circuit is the antenna. The antenna will provide gain and directivity on both transmit and receive. The take off angle of the antenna is based on the type of antenna, the height of the antenna above ground, and the type of ground below and in front of the antenna. The take off angle will determine the angle of attack on the ionosphere, which will effect if and where the signal will be reflected by the ionosphere.\n\nGround wave Propagation\nGround waves are radio waves that follow the curvature of the earth. Ground waves progress along the surface of the earth and must be vertically polarized to prevent short circuiting the electric field through the conductivity of the ground. Since the ground is not a perfect electrical conductor, ground waves are attenuated as they follow the earth’s surface. At low frequencies, ground losses are low and become lower at lower frequencies. The VLF and LF frequencies are mostly used for military communications, especially with ships and submarines.\n\nEarly commercial and professional radio services relied exclusively on long wave, low frequencies and ground-wave propagation. To prevent interference with these services, amateur and experimental transmitters were restricted to the higher (HF) frequencies, felt to be useless since their ground-wave range was limited. Upon discovery of the other propagation modes possible at medium wave and short wave frequencies, the advantages of HF for commercial and military purposes became apparent. Amateur experimentation was then confined only to authorized frequencies in the range.\n\nLine-of-Sight Propagation\nGround plane reflection effects are an important factor in VHF line of sight propagation. The interference between the direct beam line-of-sight and the ground reflected beam often leads to an effective inverse-fourth-power law for ground-plane limited radiation.\n\nSkywave Propagation\nSkywave propagation is any of the modes that rely on reflection of radio waves from the ionosphere, which is made up of one or more ionized layers in the upper atmosphere. These layers are directly affected by the sun, and its varying activity (sunspot cycle) determines the utility of these modes. Forecasting of skywave modes is of considerable interest to amateur radio operators and commercial marine and aircraft communications, and also to shortwave broadcasters.\n\nTropospheric Scattering\nAt VHF and higher frequencies, the atmosphere at a height of around 6 miles (10 kilometres) can scatter some of the norm
307713
11.6317283966\n\nThere is the notion of extending the Factorial (n!) function into the Reals, the Gamma function.\n\nThis website will compute gamma for you:\nhttp://mathworld.wolfram.com/GammaFunction.html\n\nThis one explains more details about it:\nhttp://mathworld.wolfram.com/GammaFunction.html
100023
The  Faraday constant divided by the Electron charge  =\n\n Avogadro's number
971127
The sound intensity due to a point source radiating equally in all directions (spherically symmetric case) is proportional to 1/r^2, where r is the distance from the source of the sound:\n\nI = k/r^2, where k is a constant of proportionality.\n\nIf we know the intensity (I1) at one distance (r1), then the intensity (I2) at a second distance (r2) is given by:\n\nI2 = I1*(r1/r2)^2\n\nIn Q1, we are told that the sound intensity at 1 m is 10^-5 W/m^2, and we want to know the intensity at 5 m.  Plugging these values into the above equation, we have:\n\nI2 = 10^-5 W/m^2 * (1m/5m)^2 = (10^-5 W/m^2)/25 = 4*10^-7 W/m^2\n\nThe intensity of a sound wave is proportional to the square of the amplitude of the displacement of a particle in the medium through which the sound is propagating:\n\nI = m*x^2,\n\nwhere m is a constant of proportionality, and x is the amplitude of the displacement.  Plugging this into the equation above that relates intensity and distance, we have that:\n\n(x2)^2 = (x1)^2 * (r1/r2)^2\nx2 = x1*r1/r2\n\nThe amplitude of the displacement is inversely proportional to 1/r.\n\nFor Q1 we are told that x1  = 70 microns at 1 meter, and are asked what x2 is at 5 meters.  Plugging these values into the above equation, we have:\n\nx2 = 70 microns * 1/5 = 14 microns\n\nFor Q2, we are told that x1 = 10^-7 m at r1 = 3 m, and are asked what the amplitude is at 5 m.  Again plugging these into the above equation, we have:\n\nX2 = 10^-7 m * (3/5) = 6*10^-8 m
241348
try:\n\n  \nFaraday constant  over Electron charge = Avogadro's number\n\n96,487 / 1.6022*10-19 = 6.022*10^23
370398
O.K.\n\nUse the substitution:\n\ny = uv , x = u/v\n==>\n(1+u^2)(u/v)(udv + vdu) + (1-u^2)uv(du/v - udv/v^2) = 0\n==>\n(1+u^2)(udv + vdu) + (1-u^2)(vdu-udv) = 0\n==>\n2vdu + u^2(2udv) = 0\n==>\ndu/u^3 + dv/v = 0\n\nIntegrating gives:\n\nv = (constant) e^(1/(2u^2))\n==>\nv^2 = A e^(1/u^2)\n\nReverse substituting x,y gives the answer as an implicit function:\n\ny/x = A exp(1/xy)
697180
Which previous question are you talking about???\n\n3e8 is the speed of light. (3 x 10^8 meters per second)\nThe symbol for the speed of light is usually  "c".\n\nPerhaps they meant Velocity of Propagation??
419817
deviation= (refractive index -1)*angle of prism\n\nif prism angle is small...\nand\ndependence of rf with wavelength is given by cauchy's formula\n\nrf= A/wavelength....  A is constant\n\nso dependence (deviation) proportional to [(1/wavelength) -1)]
226422
This one is trickier - ln-2/x wasn't allowed on the source. I tried the following:\n\nln[e]-2/x ==> 1-2/x ==> 2/x^2\nln[2/x] ==> -1/x\n\nI hope this helps.
796420
length = 3.5m (appr)\n\nn=6\n\n(hint: find GCF of frequencies)
729175
A wave with y-values going from 1 to -1. An x-intercept exists at pi/2
834583
You can use the Rydberg formula, which relates the reciprocal of the wavelength of the photon absorbed or emitted to the squares of the reciprocals of the principal quantum numbers.\n\n1/wavelength = R(1/n_l^2 - 1/n_u^2)\n\nSo, for the listed values of n, \n\nA.  1/w = R(1/ 2^2 - 1/4^2) = 0.1875R\nB.  1/w = -0.75R\nC.  1/w = 0.21R\nD.  1/w = -0.04861R\n\nSince energy and frequency are proportional to 1/wavelength,\nlooks like the order is D, A, C, B.
104650
Let f1 = 35.0 kHz\nvbat = 12 m/s\nvsound = 330 m/s\n\na) f2  = vsound*f1/(vsound-vbat) = 330*35/(330-12) = 36.32kHz\n\nb) For bat f = (vsound+vbat)*f2/vsound = 342*36.32/330 = 37.64kHz\n\nc) It will obviously affect the answers