Simple Harmonic Motion?
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
85418
dx/dt+ax=b :P\n\n\nmathematica had some online find-the-derivitive program, last I remember
362696
The only link you need:\n\nhttp://hyperphysics.phy-astr.gsu.edu/hbase/hframe.html\n\nHint:  look under "Mechanics"
565790
g=m*a\nGravitational Force=Mass * Acceleration
256104
Read Newton's 3rd law of motion.\nAlso, use F=ma to figure out your acceleration.
804155
a) T = 0.5 sec\n\nb) freq = 1/T = 2 Hz\n\nc) w = 2pi*freq = 4 pi\n\nd) w^2 = k/m\nk = m*w^2 = 0.5*16 pi^2 = 8*pi^2\n\ne)  V_max = A*2*pi*f = 0.35*2*pi*2 = 1.4 pi m/sec\n\nf) A_max = A*(2*pi*f)^2 = 0.35*(2*pi*2)^2 = 5.6*pi^2 m/sec^2
756774
P=P (momentum before = momentum after)\nmv=mv\n(55kg)(4.6m/s) = (230kg)v\nv= ((55kg)(4.6m/s))230kg\nv=1.1m/s
415527
First of all, the sum is incompleted.  Is it for maximum height, or for range?  Secondly, let us assume you are asking for maximum height.  Then differentiating the given equation of motion, d = -5(t+1)^2 + 258 with respect to 't', we get dt/dt = -10(t+1).  At maximum height, the velocity is zero. \nTherefore, -10(t+1)=0, or t = -1.  This is absurd time can not become negative once the motion is initiated.  Check up there is something wrong in the problem.
569320
v=vo + at\nv=0 + (9.8)(8)\nv=78.4\n\ny=1/2(vo + v)t\ny=1/2(78.4)(8)\ny=1/2(627.2)\ny=313.6m\n\nB
75525
a) 72.072 cm/sec^2\n\nb) 82.63 cm/sec^2
865054
Let u=ln t, e^u=t, du=dt/t, dt=t du\nThen:\n∫ln² t dt = ∫t u² du = ∫u² e^u du\nIntegrating by parts:\nu² e^u - ∫2u e^u du\nu² e^u - 2u e^u + ∫2e^u du\nu² e^u - 2u e^u + 2e^u + C\nt ln² t - 2t ln t + 2t +C\nAnd we are done.
808209
Using one of the equations of motion\n\n(v*v) - (u*u) = 2*a*s\n\nwhere \nv is final velocity\nu is initial velocity\na is accleration\ns is displacement\n\n\nOR \nuse \n\nv * (dv / ds) = a
855447
if it is equations of motion then \nv=u+ft\ns=ut+1/2ft^2\nv^2=u^2+2fs\nhope this is what you wanted
666681
1/2 * mass * velocity^2.
307841
One-dimensional motion is motion along a straight line, and cen be represented by four variables, velocity (v), time (t), displacement (x), and acceleration (a).  There are five equations that can solve for any unknown variable, assuming constant acceleration:\ndelta x=v(delta t)\ndelta v=a(delta t)\ndelta x=v initial*delta t + 0.5a(delta t)^2\ndelta x=v(delta t) - .5a(delta t)^2\nv^2=v inital^2 + 2a(delta x)\n\nAlso, keep in mind the derivative of displacement is velocity and the derivative of velocity is acceleration.  Likewise, the integral of acceleration is velocity and the integral of velocity is displacement.
74552
these are data from end fixing of the structure you should introduce to solve the system of equations, otherwise you would get infinite solutions or (rigide body motion)-Good Luck
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