Doesnt inertial acceleration of a wave  change from maximum, to minimum repeadetly?
821401
yes, it's a sine wave, and velocity is integral of that, ie a cosine\n(this is considering only fist order effects)
619882
frequency doesn't change, wavelength increases, velocity increases
792607
0 km/hr^2 .... because your speed doesn't change with time due to constant as you mentioned
1185891
Yeah it provides\n\nAnd since I1* W1 = I2* W2\n\nIntroducing load will increase the Moment of inertia and hence the angular speed will decrease to keep product constant
3944
The Trolley + Child being one single system, the speed of the Centre of Mass would not vary. It would remain the same velocity V.
943735
a) T = dL / dt = 16\nL = 16 t \n\nI don't think you can do any better than that since they don't ask for the angular momentum at a particular moment in time, and it certainly is not constant.\n\nb) L = Iw\nw = L / I = 16 t / I = 16(33 * 10^-3) / (1.2 * 10^-3) = 440 rad/s
738834
d is the most correct answer, yes.  \n\nHowever keep in mind that despite g being the "gravitational constant", acceleration caused by gravity DOES change depending on the the distance from the earth.  When we're talking about the difference of a table, there is practically no difference, however keep in mind that it does change.
1121006
Gravity waves have a frequency.  That's a good point about destructive interference.  I hadn't thought about that.  However, from what I understand, there should be gravity waves generated by two co-orbiting large bodies.  If you think about it, their gravity would increase and decrease if that was the case, but always remain far above zero.  So....if you could cancel it with a gravity wave generated by another source that was identical and just out-of-phase, you should still have a baseline gravity that was nonzero.  \n       In other words, no negative gravity (that I know of) so when it cancels, it's not zero.
372422
If each bug moves towards its target at constant rate r inches per second, then I think "infinitesimally speaking" because each target is moving at right angles to the direction of its seeker that the distance between seeker and target will lessen at the same constant rate of r inches per second.\n\nThus it will take 10 inches / (r inches per second) = 10/r seconds until all four meet in the center, during which time each will have traveled 10/r * r = 10 inches.
45821
are you sure that you want acceleration instead of velocity??\n\nwell, the acceleration is going to be 0.  since the speed is going to be the same the whole time, there won't be any acceleration.
998322
Lets make an enormous assumption to fill in the gaps.\nYou have a thing on a string of length L and spin it round at a certain rate of rotation. Keeping the rate of roataion what would the tangential velocity of the thing be if the string was shortened to L/2?\nIn this case the tangential velocity (= the radius x the angular velocity) will also halve.
396138
a=F/m, lim x->inf x/x = 1. The object will accelerate.
751077
Yes...\n\nThe refraction indecces\n\nn =c/v, but c = wave lenght * frequency
863361
x = -11 + 12t + 3t^2 + 4t^3 - 4t^4\n\nTaking the 1st derivative gives the velocity.\nv = dx/dt = 12 + 6t + 12t^2 - 16t^3\n\nTaking the 2nd derivative gives the acceleration.\na = dv/dt = 6 + 24t - 48t^2\n\nTo find when the acceleration is maximum, we have to take the derivative of the acceleration equation and set it equal to zero.\n\nSo, da/dt = 24 - 96t = 0.\nThus, t = 1/4.\n\nThe location of the particle at this time is then found by substituting t = 1/4 into the location equation.\n\nx = -11 + 12(1/4) + 3(1/4)^2 + 4(1/4)^3 - 4(1/4)^4\n= -11 + 3 + 3/16 + 1/16 - 1/64\n= -497 / 64\n\nThe velocity at this time is found by substituting t = 1/4 into the velocity equation.\n\nv = 12 + 6(1/4) + 12(1/4)^2 - 16(1/4)^3\n= 12 + 3/2 + 3/4 - 1/4\n= 14
679906
or is it maximun voltage?\n\nlike terminal velocity is maximum speed...
1082252
It does not differ!\n\nAccording to Cavalieri's principle, if the peak of that pyramid moves on a surface parallel to its base, the volume will not change at all.
805281
OK, first we need a formula for the motion, which, if it's simple harmonic motion is\nf(x) = Asin(kx)\nwhere A is 1.8*10^-4\nand k = 2000*2*pi\nSo f(x) = 1.8*10^-4*sin(4000*pi*x)\n\nTo compute the maximum force, we need the accelleration f''(x)\nf'(x) = 1.8*10^-4*(cos(4000*pi*x))*4000*pi, or \n1.8*4*pi*10^-1*cos(4000*pi*x)\n\nf''(x) = 1.8*4*pi/10*(-sin(4000*pi*x))*4000*pi\nor\n1.8*16*pi^2*100*(-sin(4000*pi*x))\nthe MAXIMUM acceleration is then 1.8*16*pi^2*100.\nThis is because you can find a value for x where -sin(4000*pi*x) is 1.\n\nUsing F=ma, the maximum force is 50 *1.8*16*pi^2*100 KG m/s^2, or approximately 1.42*10^6 KG m/s^2