whats the amplitude of resultant wave>?
447635
the maximum amplitude is 10 times square root of 2. if the first wave is a sine wave, then any wave shifted by 90 degrees is the same as a cosine wave. so you have to find the maximum value of sin x + cos x. to find the maximum value of a function, set its derivative equal to zero, so you get cos x - sin x = 0. this is true when x = 45 degrees and sin x = cos x = 1 / square root of 2. when you add the two, you get square root of 2, which you multiply by the initial amplitude.
788532
Is it x*(e-2) or x^(e-2)?\n\nI don't think it matters anyway, because the bottom fluctuates between 3 and 5 while the top increases indefinitely, i.e. approaches "infinity" if you like.  So there's no limit.
913325
If we assume SHM, then\n\n(a) for a given amplitude A, the max velocity and max acceleration is greater for higher frequency sounds, because SHM is given by the following equations:\n\ny(t) = Asin(2πft + γ) where γ is a phase factor\nv(t) = dy/dt = 2πfAcos(2πft + γ)\na(t) = dv/dt = -(2πf)²Asin(2πft + γ)\n\n(b) For A = 1.0 x 10^-8 m and f = 20.0 Hz\n\nv(max) = 2πfA = 1.26 x 10^-6 m/s\na(max) = -(2πf)²A = - 1.58 x 10^-4 m/s^2
577560
http://en.wikipedia.org/wiki/Sound_waves
928305
The volume V=abc\nSurfce area S=2(ab+bc+ca)\nTherefore,2/S(1/a+1/b+1/c)\n=2/2(ab+bc+ca )](ab+bc+ca)/abc\n=1/abc=1/V proved
290580
It is the same as the graph of -cos(2 theta).\n\nIt starts at (0, -1) , has an amplitude of 1, and has a period of pi radians which means it passes through (pi/2, 1)\n\nOr in degrees, it has a period of 180 and passes through (90, 1)
880047
I can find this on a TI-84+, is that how you want to get it or do you need an algebraic solution?
334268
331 = 20.0(lambda) ----- (the "upside down y")\n331/20 = lambda\n16.55 m = wavelength\n\nMuch of understanding science is in understanding the concepts - sometimes, you have to solve problems (using math skills) to see numerical proof of the concepts.\n\nIn this particular problem, you have a velocity measured in meters per second (m/s), and a frequency measured in hertz (Hz), which is also cycles ("times") per second, sometimes noted as sec^(-1). Thus, you know what you need to find is the distance/length in meters (m) as stated in the problem. This is what's known as "dimensional analysis" - you're looking at the dimensions you know in order to calculate the one you don't know as a check to make sure you're doing the problem correctly.
1078871
you have not given the distance on screen D\nso i took D=5cm\n \nso displacement y=m*wavelenght*D/d\n                             =0.4152 cm\n\nand angle q=4.7469 degree\n\nok if you don understand then go on this link and put the values
279026
The energy transported by a wave is directly proportional to the square of the amplitude of the wave.\nSo if a wave has twice the amplitude of another wave, it will have 2x2= 4 times more energy.
970598
Try this website for information on transverse and longitudinal waves:\n\nhttp://www.kettering.edu/~drussell/Demos/waves/wavemotion.html
362195
Wavelength= 1/wavenumber\n\nWavelength= 1/1700cm-1\nWavelength= 5.882 x 10^-4 cm\nWavelength= 5882nm
741319
The phase relation is constantly changing at 10 Hz, or 3600 degrees/second or 20*pi radians/sec. If both sinewaves are of equal amplitude A and you sum them (mix them together) you will have an amplitude modulated wave that varies in amplitude between 0 (the value when they are out of phase) and 2*A (the value when they are in phase) at 10 Hz.
939597
measure the distance between the beginning of the positive transition to the end of the negative transition.\nThis will be the wave length.
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
711054
It's two.  Energy goes as the square of amplitude.
482358
I admire mathematician's concise answers, but in this case some additional explanation may be due:\n\nWhen a wave is partially transmitted and partially reflected the two outgoing waves are shifted in phase from the incident wave. If energy or probability is to be conserved, the difference between the two phase shifts is pi/2 (90°). In this diagram each of the two possible outgoing  beams consist of components from two paths (R=Reflection,T=Transmission):\n\nOutgoing Horizontal: TRR + RRT\nOutgoing Vertical: TRT + RRR\n\nIn the first case the relative phase shift between the two components is 0 and in the second case it's pi (180°) ==> the two components are negatives of each other and interfere destructively.