i have two physics questions relating to intensity of wave?
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
131583
i would think that the radiation from the freqs listed above should all cause excite the water molecules in the same way but to differing extents. that's a very good question; i think i'll research it.
339135
first one is quite correct but frequency is in per second\n\n47 waves/min = (47/60)Hz   (1 Hz= 1wave/sec)
772799
Ok i'll help with your homework. Even though you may want to try a little harder.\n\nIt's not that difficult, you just have to remember that the oscillations ideally are represented by a sine or cosine function.\n\nlet's say    x = f (t)\n\nx is position and t is time. so we know f involves a 'sin' or 'cos'.\n\nThen you have to increase the amplitude of the wave to reach the numbers 55 and -55 (multiplying the sin or cos by some factor).\n\nYou can also notice that the sin or cosine wave doesn't start at time t=0. Therefore you may have to use t+something instead of just the variable "t".\n\nThe answer that involves the sine function is:\n\nx = 55 sin ( (π/30)*(t+10) )\n\nFigure out the calculations for yourself. You might even try finding the OTHER answer which involves the cosine.
778569
This is a classic problem in 1-D wave propagation.\nWhen the medium is limp (i.e. you can ignore its stiffness) then the speed of propagation of a transverse wave is sqrt(T/mu) where mu is the mass per unit length and T is the tension.\n\nT is the same for all segments. So in the first section the wave propagates at sqrt(T/mu), in the second segment at 1/sqrt(2) as fast, in the third segment at 1/2 as fast.\nThe time to traverse each segment is the length of the segment divided by the speed, and all segments are L/3 long.\nTherefore the time to reach the wall is approximately\n1.47 L * sqrt(mu/T)\nif I have done the algebra correctly.
646637
This question can be answered quite easily, if you can consider this idea: Suppose you have some amount of water which is contained in a tub. You bury your hand inside the water and make some forward movement, obviously, the waves of the water flow directly in one direction. Whereas, if you oppose the motion of these moving waves, by burying another hand of your's and pushing some water against an already moving set of waves of water, the first set of waves defintely will decrease their speed, as being opposed by an other ENERGY. \n               So, considering this example, sound waves' problem can be solved: If you yell against the wind, your sound waves are being OPPOSED, hence their speed slows down, and they tend fade off travelling LESS distance. If the sound is going in the direction of the wind, it will definitely tend to move FASTER(when compared to it, moving against the wind)as it is NOT BEING OPPOSED by any other form of energy. \nThat's it mostly,\ncheers,\nDa Vinci.
750132
The major differences are two concepts: 1 Wave mechanics.\n2. Probability, that is , uncertainty.
910771
"Both light and sound are waves." "and is there a physical analogous comparison of light waves to sound waves?"\n\nYes they are both waves. However, light is known as an electromagnetic wave which is a propagating disturbance of both the electric and magnetic field in space. Sound however is a pressure wave transmitted through a medium. Light waves are transverse meaning the direction of travel is perpendicular to the direction of oscillation. While sound waves are longitudinal meaning the direction of travel is parallel to the direction of oscillation. Another difference is that light waves DO NOT need a medium of travel (they can go through a vacuum) however sound waves exist only in a medium.\n\n"Sound is measured in Hz. how are light waves measured?"\nYou are correct, one measurement of sound is in Hz. However, waves have 4 unique properties. Velocity the rate that it propagates which is measured in m/s. Frequency, how many times the wave repeats itself in a period of time measured in Hz. Wavelength, the distance between complete cycles measured in meters. And finally, Amplitude or Intensity which is measured in decibels or W/m^2 which measures how high the crests become and how low the troughs become. The larger the amplitude the higher the peaks and lower the valleys. This is also measured in meters. Higher intensitys is proportional to the amount of energy the wave has. \n\nAs far as measuring a light wave, we usually measure the intensity and either the wavelength or frequency of the light wave.  The wavelength of light and sound is inversely proportional to the frequency because the velocity is fixed to the speed of sound 343 m/s or the speed of light 3.00 x 10^8 m/s.  \nThe higher the frequency of light, the more the light moves to the purple end of the spectrum (think about the rainbow). The human eye typically picks up light between 390 (purple) nm - 970 (red) nm\n\nvelocity = wavelength * frequency.
835410
You need to get your units in a consistent form.  I suggest changing the velocity of the electron to 2.5E6 m/s so when you divide it into c-----------WAIT A MINUTE!-----Yor formula is NOT right, because you're dividing a speed by a speed, which will give a dimensionless answer.  Wavelength is a LENGTH, so the answer must come out that way.  I forget the formula you're refering to, but what you've typed isn't it................
923764
a) F = ma\n    F = 0.0412 kg * 9.8m/s^2\n    F = 0.404 N\n\nb) F = 0.0370 kg * 9.8 m/s^2\n    F = 0.363 N\n\nc) The difference in "weight" is equivalent the weight of the displaced volume of water...  So...\n   0.404N - 0.363N = V * 1000 kg/m^3 * 9.8 m/s^2\n   0.041N = V * 9800N/m^3\n   V = 4.18 * 10^-6 m^3   or  4.18 cm^3\n\nd) Oh Juliet you are my density.... (bad joke!)\n     Density = m/V\n     D = 0.0412 kg / 4.18  * 10^-6 m^3\n     D = 9848 kg/m^3
1009361
y dont u do ur homework urself .ok heres abrief intro.\n\n1.see the level of frequency changes when beam of ultra violet light rays enter to one medium from another medium(to glass from air).\n\n2.when light undergoes refraction it frequency changes its speed acc. to the medium such as(glass,air and water)\n\n3.yes there will be,as light enters to one medium from another medium.
437661
This requires the use of doppler shift formula for a moving source and stationary observer:\n\nf = f0 ( v / ( v + u ) )\n\nf0 = actual frequency = 5\nv = speed of wave = 20\nu = relative velocity of source away from observer = -10\n(since travelling toward)\n\nf = observed frequency = 10 Hz\n\nNow velocity = v = fl = frequency * wavelength\n\nso wavlength l = v / f = 20 / 10 = 2m
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Intensity.\n\nThats the word that describes "how much" light there is, or as you put it, how many light lines there are at a point at the same time.\n\nThe analogous property for materials is density, as you have described for water. true you cannot put 2 tons of water in the same space occupied by 1 ton of water, however if you cool water to 4 degrees celcius or increase pressure, you can hold slighty more than 1 ton in the same space as you are increasing the *density* of water\n\nYou can increase the intensity of light by throwing more light in the same area, either by increasing the number of sources or by increasing the power of the one source, (increase current in that light bulb or throw in more firewood for eg)\n\nIf you thing of light as photons, then intensity is defined as the number of photons striking a surface per unit area per unit time. you can also think of Intensity as brightness, they are directly related.\n\nas you go away from a source the photons spread out and there are fewer photons per unit area: reduced intesity/brightness, infact Intensity falls off as the square of the distance from the source.\n\nIntensity is not limited like density is, as in: in the same amount of space, you can have any intensity you like, there is no limit, just keep adding light sources. however if you try it practically, you will hit a limit, when you are not able to fit anymore light sources into the space, or when you are unable to supply any more energy.\n\nThe standard unit for measuring intensity is candela(cd), which is defines as lumens/steridian, which can roughly be translated as energy per unit volume in a particular direction.
179131
Try this:\nhttp://kr.cs.ait.ac.th/~radok/physics/f3.htm
809913
1. Momentum=Massxvelocity=3600\n2.3600=.2KgxVelocity=18000\nThats all i know
27171
Its actually derived from Maxwell's Wave Equations, see source. Its too complex and requires mathematic symbols, so I can't type it here.
827543
The equations involved are:\na) freq*wavelength = velocity of wave\nb) energy of radiation E = h*freq, where h is Plank's constant.\nYou can work out from these that the choice should be a.