Extinction Coefficient?
929883
You are correct on both counts!
144611
You might find this paper useful\nhttp://arxiv.org/abs/math-ph/0106027
243416
loss= coupler loss + fiber loss + detector loss
921732
If you look at a picture (aka a graph) of the function, it will all become clear...\n\nhttp://upload.wikimedia.org/wikipedia/en/3/3e/1-over-x.png
620893
The best way to work this out would be to plot a speed time graph for the information.\n\nOr you could express it as a simultaneous equation.
250483
Try using a non dimensional correlation for the Bio number or possible Prandtl number. I can't remember off the top of my head, but one of them should relate your Mach number to pressure or viscosity allowing you to have only one unknown. Another idea would be to model the situation with a known boundary layer pattern and find where the boundary layers meet. Check the solution and see if it is reasonable.
608010
epsilon=0.6\nsigma=5.6704*10^-8 [J/(s*m^2*K^4)]\nP=Q/A=120 [W/m^2]\n\nWe know:\n\nP=epsilon*sigma*T^4\n\n120=0.6*5.6704*10^-8*T^4\n\nSolving for T we get:\n\nT=(120/(0.6*5.6704*10^-8))^(0.25) = 243.7 [K]\n\nThen, using Newtons cooling law:\n\nT(t)=T(environment)+(T(initial)-T(environment))*e^(-rt)\nT(t)=293+(243.7-293)*e^(-r*t)\nT(t)=293+(-49.3)*e^(-r*t)\n\nNow, I'm stuck, because I don't know the rate of change of temperature (r).  Maybe you try a different formula!?
432008
t=rho/(sqr((1-rho^2)/(n-2)))\nwhere:\n\nrho: correlation coefficient\nn: number of observations
912617
P=P(0)*exp(-1.1e-4*H)\n\nP(0) is the pressure at sea level. Note it's temperature dependent and it assumes a stable Boltzmann type distribution.
1014916
i second the a answer.  the field diameter decreases.  even if you solve this problem by process of elimination b c and d are all ridiculous answers if you give it any thought.  e not correct either, because it would occupy a larger portion since you are "zoomed in" and focusing on a smaller section of the slide or whatever so it would occupy a larger portion of the field.
933542
No.  But it's wavelength decreases.\n\n\nDoug
702843
Fringe width β= λD/d\nIf the wavelength is decreased,  the fringe width also decreases.\nSo, if you consider any order fringe, say 5th order, it would form at a lesser distance from the centre as compared to the previous case(i.e.when red light was used)
664722
http://www.madsci.org/posts/archives/mar97/853956505.Bt.r.html\nhttp://www.chariho.k12.ri.us/faculty/dmmor/scifair02/sf023.html
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
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).
1125729
The index-of-reaction equation (i.r.e.): dF/dtZ = kZ(l - βF)m is compared with a generalized Prout-Tompkins equation (g.P.T.e.): dF/dt = kn(l - βF)y With the boundary condition F = 0, t = 0 the equations are equivalent for z = 2 or m = 0 or β ≤ 1 and the i.r.e, is closely approximated by the g.P.T.e. for other values of the parameters by use of the following approximation ∫(1- F)-m dF ≈ AF(1-BF)5 The parameter s is a function of m. Both A and B are close to unity for 1/3 < m < 3. The i.r.e. is recommended as being more useful than the g.P.T.e. because the i.r.e. has a better theoretical basis and because it is more readily integrated and fitted to experimental data.
875835
OD=E*d*C\n\nWhere OD=optical density\nE the molar extinction coefficient\nd the length of the path of the light through the solution (width of the cuvette)\nC the concentration in mole/L of the solute.\n\nYou need d and C as well to get E.\nThis is a fair approximation especially for student problems.\n\nIn reality if you wanted to find E, you would do a standard curve by measuring the OD of samples of known concentration.\nThen you would use linear regression to find the equation of the curve and from its slope you would calculate E (slope=E*d)\n\nA good standard curve and thus accurate calculation requires care to ensure the accuracy of the standards, several points (ideally 9 at least) and being within the linear are of the method.\nThe region of highest accuracy is 0.2-0.8. You can go even below or above this range (though not above 1.0) but this is the range where the photometric error is minimal.