It's due tomorrow, so (obviously) I'm getting desperate. The question I'm stuck on is about the superposition of two plane waves... I've found the probability density of the superposition, but the next part of the question is asking for a calculation of the time-average of the probability density as a function of position. I have NO IDEA what this means. Does anyone know what a "time-average of the probability density" is?
210989
time average means sum of hte probability density over a certain time interval divided by the time interval. and it will be a function of position since the wave function is a function of position AND time.\nso u have W(x,t) where W is the wave function and x is position t is time. you integrate it with respect to time from a starting time to an ending time (not that hard, its just integral of e^-iwt, i think) and then divide by the time interval. and ya. i dunno.\n\nhope that helps. let meknow if u need more help. later
112827
One cool way that I have seen it derived is based on Jayne's Formalism.\n\nAssume we sample process, and measure the mean and standard devation of the samples. There are many possible distributions that will produce those exact mean and standard deviation. \n\nJayne's Formalism chooses the distribution which maximizes the entropy of the distribuiton. This is viewed mathematically as the distribution that makes the minimum number of assumptions regarding the distribution.\n\nI always thought that was kinda cool.
497266
a.\nThere are few observation in X(t)\n-constant w set\n-cos and sin are periodic functions, this means cos(t) and cos(t+T) would behave identical over the time period T, 2T,3T... [consider sameargument for sin too]\n\nFor your refference let us consider the indices as n=1,2,..\n\nSo, for each n, X(t) will have a joint distribution with independent unit normal random variables Y1,Y2. And, this would be nothing but linear combination of two unit notmal distirbutions.\n\nSince, the behaviour if sine and cosine remains same over time intervals (nT), this joint distribution will remain same for all n=1,2,...\n\nThis indicated that the process {X(t)} is stationary.\n\nSince, the mean & variance of the each distribution occuring at eact time interval may not remain constant or same, This is the reason why {X(t)} is a wekly stationary process.\n\nI hope now you understand.\n\nAll the best.
304528
Maybe this explanation can help? http://en.wikipedia.org/wiki/Spectral_density
470009
Use the certical component to find the time in the air and since the horizontal velocity does not change; Dx = vx * time\n\nDx = v^2 / g\n\nheight = v^2 / (2g)\n\nHope this helps without actually solving it for you.
344610
Write down all relevant variables, and units:\nV = volume in mm^3\nt = time in min\nx = edge length in mm\n\nWhat down what you are told:\ndx/dt = 0.49\nV = x^3\n\nWrite down what you want:\ndV/dt\n\nTo get dV/dt, we need dV/dx and dx/dt. The second one is 0.49, and the first one can be gotten by differentiating the volume equation: dV/dx = 3x^2. Substitute in x = 6.7, and the final answer is \n3*6.7^2 * 0.49 = 65.9883 mm^3 / min (decreasing)
425520
I don't quite understand your notation, but here is how I would do this:\nThe general equation is y(x,t) = A sin 2 pi (t/T - x/L), where\nA = amplitude = 0.12 m\nL = wavelength = 0.3 m\nT = period\nThe latter is not given, but you can get it from the wavespeed v, via the formula v=L/T. Hence T = L/V = 0.3/6.4 = 0.0469 s.\nPlug A, L and T in the general equation and compare the result to the expression given in the problem.\nLooking at your expression, I would guess \na = A = 0.12 m \nb = 1/T = 21.3 s^-1\nc = 1/L = 3.33  m^-1
394234
T = 2pisqrt(l/g)\nl = g(T/(2pi))^2 = 9.80(1.0/(2pi))^2 = 0.25 m\n\n(The object's weight is irrelevant for small oscillations.)
172938
Tthe moving object (mass) travels a distance of one meter in one second and that is an average velocity;that means the speed was not constant as it was moving. So dividing the distance travelled by the period of time traveled would give you only the average velocity.\nThe proof of this was given by a frenchman by the name Paris.\n\n However the final velocity would be twice the average velocity which would be two meters per seconds.dont feel bad I learned that in the 8th grade.
397848
I can tell that nobody wants to just do this problem.  There's nothing interesting about it.  It's a typical problem that a stat student does to show understanding.  Get out your textbook and study up how to solve it.
260803
I think what you're looking for is the "equation of time", which is the difference between local mean time and local apparent time.  The term "difference" is used because "variance" has a very specific mathematical meaning.   Local apparent time is determined by the position of the sun in the sky.  local mean time is based on Greenwich Mean Time.  The difference occurs because GMT is based upon a fixed rotational rate, and local apparent noon is determined is by the maximum altitude of the sun in the sky.\n\nThe two time scales differ by as much as 15 minutes.  The web page below has more details.\n\nThere are formulas for this, but they are complex calculations.  The equation of time changes constantly, although the change is very slow, so a graph of the difference in 1980 is still mostly correct for 2006.  You'll be accurate to within a few seconds.
380048
Yes...\n\nThe refraction indecces\n\nn =c/v, but c = wave lenght * frequency
412470
Suppose X is a continuous random variable whose density function f is defined by f(x) = x/2 whenever 0<x<2 and f(x) = 0 otherwise. Then,\nE[X] = integral (x * f(x), x=0..2 ) dx\n     = integral (x * x/2, x=0..2 ) dx\n     = (1/2) integral (x^2, x=0..2 ) dx\n     = [(1/6)*x^3, x=0..2] \n     = (1/6)*2^3 - (1/6)*0^3 \n     = (8/6)\n     = (4/3).\n\nNext, \nE[ | X - E[x] | ] = E[ | X - 4/3 | ] \n     = integral ( | x - 4/3 | * f(x), x=0..4/3 ) dx + integral ( | x - 4/3 | * f(x), x=4/3..2 ) dx\n     = integral ( - (x - 4/3) * f(x), x=0..4/3 ) dx + integral ( (x - 4/3) * f(x), x=4/3..2 ) dx\n     = integral ( (4/3 - x) * x/2, x=0..4/3 ) dx + integral ( (x - 4/3) * x/2, x=4/3..2 ) dx\n     = integral ( (2/3)x - (1/2)x^2), x=0..4/3 ) dx + integral ( (1/2)x^2 - (2/3)x, x=4/3..2 ) dx\n     = [(1/3)x^2 - (1/6)x^3, x=0..4/3 ] dx + [(1/6)x^3 - (1/3)x^2, x=4/3..2]\n     = ( (1/3)(16/9) - (1/6)(64/27) - 0 + 0) + ( 8/6 - 4/3 - (1/6)(64/27) + (1/3)(16/9))\n     = ( 16/27 - 32/81 ) + ( 16/27 - 32/81 )\n     = 2(16/27 – 32/81)\n     = 32/81.\n\n--- \n\nSome notation used:\nintegral (f(x), x = a..b) dx means we evaluate the definite integral of f(x) bounded by x = b and x = a.\n\n[ F(x), x =  a.. b ] means we evaluate the difference, F(b) - F(a).\n.
550649
The answer has to be a speed!  It involves seeing that dy/dt = 1/2 x^(-1/2) dx/dt in this case.  Also, the distance from the origin is given by s^2 = x^2 + y^2; to get ds/dt (the required result), differentiate that wrt "t".  Put this all together yourself; it's the only way to learn rather than simply parrot.\n\nWithout a cast-iron guarantee, I get the value 27/(4 x sqrt 5) \ncm/s.\n\nLive long and prosper.
459283
P=W/t\nP=Fdcosθ/t         (as we know W=Fdcosθ})\nP=Fvcosθ           (as we know v(avg.)=d/t)\ntherefore\n25=F3cos50\nF=12.96 N   (ans)
428609
57) P1V1/T1 = P2V2/T2\n(temp in kelvins)\n140 x 8.00 / 293 = 1 x V2 / 273\nV2= 1043.55 L but sig figs is 3 so 1040 L\n\n93) uhh I'm just learning this myself.. very confused.. something about london dispersion, hydrogen bonding, dipole-dipole forces? perhaps someone else can take this part... and give them the 10 points I don't really care
531144
a.dy/dt=(3/2)(0.2)t^1/2.\nwhen t=1\n=(3/2)(0.2)(1)^2/\n=0.3 cm/sec\nwhen t=2\n=(3/2)(0.2)(2)^1/2\n=0.4242 csps\n\nb.at t=3 it will be the thickest