Please tell me everything you know about sine waves; from how they're graphed to whatever. Please.
126556
go to this site:\nhttp://en.wikipedia.org/wiki/Sine_wave
73250
Assuming you mean over R, solve the equation s*t + r*sin(t) = 0 for r and s for all t that is not the trivial solution (r,s)=(0,0). \nIn other words, if f(t)=sin(t) and g(t)=t were linearly dependent, then there is some non-zero slope s, s*g(t) = s*t and some non-zero amplitude r, r*f(t) = r*sin(t) such that the graphs of these functions completely coincide for all t. Since f is non-linear and g is, this is nonsense and the functions are linearly independent.
467770
You will find all details with calculation and experiment on:\nhttp://www.jgp.org/cgi/reprint/1/4/387.pdf
519216
divide the voltage by R\nI=60sin(120(pi)t)\n\nThe amplitude is 60 amps\n\nThe period is 1/120 seconds\nj
400135
The voltage in alternating current oscillates up and down above and below zero, ideally with an average voltage of zero. Now using sense you could describe the voltage as zero, but this would not describe the signal properly; you need to know where you are on the wave (picture a sin wave) to know what the voltage is at a particular instant. A system called phase is used for this, describing how far you are into the wave pattern by degrees. The distance (in time) from one wave crest to the next is 360 degrees for example. An alternating system is ony fully described by magnitude AND phase. A three phase system actually has three waves equally spaced (120 degrees apart).
17468
The last answer is correct.  The equation:\n\nA*sin(bx+c)\n\nCan be graphed.  A is the amplitude of the wave (how high along the y axis).  b is the frequency (how fast do you need to complete one wavelength).  c is a wavelength shift, how far away from x=0 do you have to start the wavelength.
94378
You may never have to know the sine of an angle in real life, but the point of your math classes is to teach you how to think logically. What the different subjects show you is their applications in the real world. If you become an English teacher, you won't need to know the area under a curve.\n\nBut the thing with math is that the numbers provide a numerical visual for the concepts. When you get the right answer, you can see how the concept works via that example (sometimes called a "problem" or a "question"). And that's it. It's somewhat akin to writing a paper - you make a thesis statement, and then follow by offering proof/support of your statement. You're showing why your statement works by using words. Math often uses numbers for the same purpose.
134098
Such a pulse produces a standing wave. In such standing wave energy is not transferred from one place to another.\n\nAt each instant we see a wave, but the waves do not move.\n\nThere are certain points which never move, they are called nodes.\n\nBut between nodes the points are displaced.\n\nAt one instant, all points are motion less.\n\nAfter this instant, points that previously went up will now come down and vice versa.\n\nExcept the nodal points, all points are oscillating between potential and kinetic energy. \n\nThe maximum displacement corresponds to potential energy and the mean position corresponds to kinetic energy.\n\nAt the instant when all points are motion less, ALL POINTS EXCEPT THE NODAL POINTS have kinetic energy, by which they continue their oscillation.
442017
The maximum value of sin(x) is 1.\nV = 120(1) = 120.\n120 = I*20\nI=6 at Max\n\nThe sin function has a period of 360 deg.\n2pi = 306 deg.\n120(pi)t = 60 cycles of the function.\n\nIs any of this right?
486851
Use a cyclical function, like a sine function, but add a third dimension, z, and have the value of z oscillate with the same wave length at which y oscillates.  Like y=sin x, and z=cos x.  This will produce a spiral around the x axis.
527662
first divide [0,2pi] into 6 intervals [0,2pi/6],[2pi/6,4pi/6],[4pi/6,6pi/6] amd so on\n\nNow let's look to [0,2pi/6]. Since cos is periodic the graph on other intervals will be the same as on [0,2pi/6]\nEven on [0,2pi/6] we have to restrict more and exclude the case when cos(6 theta) is negative since R is always non-negative.\n Now if we take R=2cos(theta) you will obtain a circle, since it's R=2cos6(theta) this circle would be compressed, you obtain a "petal"\nNow since this repeats 6 times you get 6 "petals"( but on the graph you see only one)
652265
http://science.howstuffworks.com/electricity.htm
371797
By definition, something that has a defined frequency has a period. For the purposes of electricity, alternating current occurs as a sine wave pattern.\n\nThis means that the value of a 60 Hz signal added to a 100 Hz signal gives a waveform that is the compound of the two. Basically, it would appear to be a 160 Hz signal superimposed on a 40 Hz carrier.\n\nThe correct way to determine a value for the signal at any given point in time is to add the values of the two curves based on their frequency ... X sin (pi/3) + Y sin (pi/1.8), where X and Y are the amplitudes of the original two waveforms (X fo 60 Hz, Y for 100 Hz).
202627
this might prove to be useful:\nhttp://islab.oregonstate.edu/koc/ece575/02Project/Mor/
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.
221888
If one wave equals one cycle, then the answer is:\n\nA. 6 Hz
522809
This seems to be equivalent to saying what is the maximum frequency of a sine wave with a maximum amplitude of 1.1e-6 metres and  maximum acceleration of about 9.8m/s^2.\n\nLets have a think , the area under 1/2 a cycle of sin is a/2*period/2 and period is 1/f so max v = a/2/2/f  - ha!  similarly max_v/2/2/f= max acceleration = a/2/2/2/2/f/f , that's better.  a/16/f^2 > g  gives max acceleration > g  so \nf^2 > g/16/a  or  f > about 750 Hz.