Is there a formula to convert Hertz frquency of sound to Nanometers of light?  If so, How many nanometers is 528 Hertz?
210718
All light has a velocity of "c."  c is ~ 3*10^8m/s.  With this knowledge you can say that frequency is the inverse of time/distance.  Thus, you can use the formula C(meters/sec) / Freq.(cycles/sec) = Wavelength(meter/cycle) to solve the question.\n\nDo your own work from there.
346119
use E=(hc)/lambda\nwhere h is plank's constant 6.63e-34\nc is the speed of light 2.99e17 nm\nE is energy\n\nlambda is the resulting wavelength.
188316
the wavelength is half...\n\nvavelength and frequency are inversely proportional
399587
the wavelength will be speed/frequency=300000/100100000\n=0.002997002997002997002997002997003 kilometers \nin aproximation\n=2.997002997002997002997002997003 in meters\nor 3meters
251025
hi dude!\ni'll give u only the unworked answers\nwavelength of light i the cornea is 193/1.376 nm\nand the frequency would be 3*10^8*1.376/(193*10^-9) hz\ni hope i'm not too late!!
442169
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.
370137
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...
377359
c) the frequency of incident radiation\nThis is according to the Einsteins photoelectric equation\nKmax = hf - phi\nwhere phi is the work function, h is the planck's constant and f is the frequency
66957
This is ultrasonic, not electromagnetic, so C is irrelevant.  The equation is:\nF=V/L where V=sound velocity in water (~1475 m/s), F=frequency in Hz, L=wavelength (0.0003 m)\nResult is you need 4.9 MHz or higher.
380048
Yes...\n\nThe refraction indecces\n\nn =c/v, but c = wave lenght * frequency
50724
E=hc/lamda  (h: Planck constant, h=6.626*10^-34 J.s^-1\n             c: speed of light,  c=3.00*10^8 m/s\n             lamda: wavelength, lamda=478.0nm=0.478*10^-6 m\n\nE= (6.626*10^-34*3.00*10^8)/(0.478*10^-6)\n = 4.157*10^-19 J
179143
Wave length and frequency are inversions.\n\nThe average speed of sound is 343 meters per second.\n\nA 1 meter wavelength then would be a frequency of 1/343 hertz.\n\nA tube will produce most easily a wavelength equal to it's length.  It can also produce a wave length of twice it's length.\n\nAll other tones will be some division of these lengths.\n\nA 12 meter tube would produce wave lengths of 24m (twice the tube) 12m (equal), 6m (half), 4m (a third), 3m (a quarter), 2m (a sixth) etc.  They will all will be completely divisible into the tube.\n\nThat makes frequencies of 12/343 Hz, etc.
341956
ok, you have the frequency and what you want is hte wavelength. You need this equation. The wavelength multipled by the frequency gives you the speed of light. (2.9979x10^8) \n\nThis will give you the answer and if you want to convert from m to nm, you have to know that the relationship is 10^9. There are 10^9 nm in one meter.
442011
I'm stepping waaaayy out on a limb because I deal with radio frequencies but, \n\nif it takes .5 ns to process from memory, that leaves 1.5 ns for pulse to travel there and back.  \n.75 ns to travel to.\n.75 ns to travel from.\n\nAssuming velocity times are in free space (vacuum), which your statement doesn't say, then:\n\nd=c*t\n\n(c) 299,792,458 km/s  x( t) .75  = \n\nFree Space = .2248443435 m/ns (.225 m/ns)(22.5cm/ns)\n\nIf you take into account the velocity factor of silicon \n(said to reduce the velocity of light by a factor of 200 in first of article source, and 300 in second article of source)\n\nI think it's next to impossible to come up with a figure other than in a perfect medium, until scientists decide a consensus of opinion on the velicity factor of silicon.\n\nClose?\nI tried.
413370
1. b\nBecause \n'1/s' is inversly related to 's'\n\n2. c\nBecause\n1/s AKA Hertz\n\n3. d\nBecause\nN.P,As,Bi are all in the same chemical group\n\n4. b\nBecause\n(3*10^8)m/s /  (2.50 * 10^13)Hz = 1.20 x 10^-5 m\n\n5. b\n\n(I am 100% sure for every question accept the last one. I am only 50% sure on 5)
344207
Well this is just like sound measurement. First of all you can calculate the amount of energy released from the eq. That would be equal to 2000000*(pi*48000^2)J/s = 1.45 x 10^16 J/s\ntherefore at a distance of 1km, it would be 1.45 x 10^16 J / (pi*1000^2)sm^2 = 4.61 x 10^9 J/sm^2. \n\nRate of energy passing through 5km^2 would be (4.61 x 10^9)x(5x1000)^2 = 1.152 x 10^17 Watts = 115200 Tera Watts
509739
c = lambda x nu\n\nc is the speed of light 3.0 x 10^8 m/s\n\nlambda is the wavelength\n\nnu is the frequency