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Velocity extreme has the smaller cams than just velocity and i know the string is 57 3/4 on the extreme. I think 59 1/8 on reg. but not sure. No clue on cables mine are broke and i bought it that way.

Q: What is the string and cable lengths for an Xi Velocity?

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parabolic graphs where f(t)=xi + vt + 1/2at^2 f(t) = distance travelled, xi = starting distance, v = starting velocity, t = time elapsed, and a = acceleration.

Never.Average velocity is total displacement (final position minus initial position) divided by the total time: vave = (xf-xi)/tAcceleration is the rate at which your velocity is changing or change in velocity over time: a= (vf-vi)/tThese two quantities may have the same numerical value but will never have the same units.Average velocity for a trip can equal instantaneous velocity at a certain point during the trip, however, at any time during a trip in which the velocity is constant or at half way through the total time of a trip where the acceleration is constant.

v2 = v02 + 2a(delta x)v = sqrt(v02 + 2a(delta x))This is based on the assumption that there is constant acceleration.Another way to find velocity is using a little Calculus, this method is better since it does not assume constant acceleration, therefore this would work even if there is a change in acceleration.Since v = dx/dt, you can differentiate position with respects to time to find instantaneous velocity, given that you do know the time.And dv/dt = a, therefore dv = a x dt, integrate both sides you get velocity. This approach again requires the knowledge of time.

This site might help clear things up: http://id.mind.net/~zona/mstm/physics/mechanics/kinematics/ EquationsForAcceleratedMotion/Origins/Displacement/Origin.htm(Note: The site is all one line)

The speed and direction of a wave

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Summation of Xi Xi Xi Xi Xi Xi Xi Xi Xi Xi Xi Xi Xi Xi Xi Xi Xi Xi Xi Xi Xi Xi Xi Xi Xi Xi Xi Xi Xi Xi Xi Xi Xi Xi Xi Xi Xi Xi Xi Xi Xi Xi Xi Xi Xi Xi Xi XiXi Xi Xi Xi Xi Xi Xi Xi Xi Xi Xi Xi Xi Xi Xi Xi Xi Xi Xi Xi Xi Xi Xi XiXi Xi Xi Xi Xi Xi Xi Xi Xi Xi Xi Xi Xi Xi Xi Xi Xi Xi Xi Xi Xi Xi Xi XiXi Xi Xi Xi Xi Xi Xi Xi Xi Xi Xi Xi Xi Xi Xi Xi Xi Xi Xi Xi Xi Xi Xi XiXi Xi Xi Xi Xi Xi Xi Xi Xi Xi Xi Xi Xi Xi Xi Xi Xi Xi Xi Xi Xi Xi Xi XiXi Xi Xi Xi Xi Xi Xi Xi Xi Xi Xi Xi Xi Xi Xi Xi Xi Xi Xi Xi Xi Xi Xi XiXi Xi Xi Xi Xi Xi Xi Xi Xi Xi Xi Xi Xi Xi Xi Xi Xi Xi Xi Xi Xi Xi Xi XiXi Xi Xi Xi Xi Xi Xi Xi Xi Xi Xi Xi Xi Xi Xi Xi Xi Xi Xi Xi Xi Xi Xi XiXi Xi Xi Xi Xi Xi Xi Xi Xi Xi Xi Xi Xi Xi Xi Xi Xi Xi Xi Xi Xi Xi Xi XiXi Xi Xi Xi Xi Xi Xi Xi Xi Xi Xi Xi Xi Xi Xi Xi Xi Xi Xi Xi Xi Xi Xi XiXi Xi Xi Xi Xi Xi Xi Xi Xi Xi Xi Xi Xi Xi Xi Xi Xi Xi Xi Xi Xi Xi Xi XiXi Xi Xi Xi Xi Xi Xi Xi Xi Xi Xi Xi Xi Xi Xi Xi Xi Xi Xi Xi Xi Xi Xi XiXi Xi Xi Xi Xi Xi Xi Xi Xi Xi Xi Xi Xi Xi Xi Xi Xi Xi Xi Xi Xi Xi Xi XiXi Xi Xi Xi Xi Xi Xi Xi Xi Xi Xi Xi Xi Xi Xi Xi Xi Xi Xi Xi Xi Xi Xi XiXi Xi Xi Xi Xi Xi Xi Xi Xi Xi Xi Xi Xi Xi Xi Xi Xi Xi Xi Xi Xi Xi Xi XiXi Xi Xi Xi Xi Xi Xi Xi Xi Xi Xi Xi Xi Xi Xi Xi Xi Xi Xi Xi Xi Xi Xi XiXi Xi Xi Xi Xi Xi Xi Xi Xi Xi Xi Xi Xi Xi Xi Xi Xi Xi Xi Xi Xi Xi Xi XiXi Xi Xi Xi Xi Xi Xi Xi Xi Xi Xi Xi Xi Xi Xi Xi Xi Xi Xi Xi Xi Xi Xi XiXi Xi Xi Xi Xi Xi Xi Xi Xi Xi Xi Xi Xi Xi Xi Xi Xi Xi Xi Xi Xi Xi Xi XiXi Xi Xi Xi Xi Xi Xi Xi Xi Xi Xi Xi Xi Xi Xi Xi Xi Xi Xi Xi Xi Xi Xi XiXi Xi Xi Xi Xi Xi Xi Xi Xi Xi Xi Xi Xi Xi Xi Xi Xi Xi Xi Xi Xi Xi Xi Xi

Impulse XI 53 7/16 on a 28-30 draw length 56 3/8 on a 30-32 draw length

Divide the change in position, (total distance covered) by the time it took. Xf = xi +at a = xf-xi / t That is the definition of velocity, not acceleration. Acceleration is rate of change of velocity. (vfinal - vinitial)/t for constant acceleration so vf equals vi + at. Or a equals dV/dt otherwise.

Xf= Xi + Vxi(t) + (.5)(Ax)(t)2 where: Xf is the final position Xi is initial position Vxi is initial velocity Ax is average acceleration t is time Xi and Vxi will both be zero since the initial velocity and position are both zero. Xf = (0) + (0) + (.5)(9.8m/s2)(20)2 Xf = 1960 meters

Xf= Xi + Vxi(t) + (.5)(Ax)(t)2 where: Xf is the final position Xi is initial position Vxi is initial velocity Ax is average acceleration t is time Xi and Vxi will both be zero since the initial velocity and position are both zero. Xf = (0) + (0) + (.5)(9.8m/s2)(20)2 Xf = 1960 meters

You can answer this question using the following kinematics equation:v2 = vi2 +2a(x-xi)where v is the final velocity, vi is the initial velocity, a is the acceleration, x is the final position and xi is the initial position.The displacement (x-xi) is given as 0.5m, and since the person is jumping upwards, the acceleration (a) is the acceleration due to gravity, or 9.8 m/s2. The acceleration is negative because it is pointing towards the earth, opposite the direction of her jump. The final velocity (v) is 0, since it is being considered at the peak of the jump. Using this information, solve for vi.v2 = vi2 +2a(x-xi)0 = vi2 + 2(-9.8)(0.5)0 = vi2 + (-9.8)9.8 = vi2sqrt(9.8) = sqrt(vi2)3 m/s = vi

parabolic graphs where f(t)=xi + vt + 1/2at^2 f(t) = distance travelled, xi = starting distance, v = starting velocity, t = time elapsed, and a = acceleration.

body is projected with a velocity 3o m/s at an angle 30 degree with vertical find maximum height time of flight and range

XI-XI-LXXXVIII 11-11-1988 = XI-XI-MCMLXXXVIII

the piecewise linear chaotic map is defined as follows: xi+1=Fpi(xi)= xi/pi if 0<=xi<pi (xi-pi)/(0.5-pi) if pi<=xi<0.5 Fp(1-xi) if xi>=0.5 where 0<=xi<1 and the control parameter 0<pi<0.5

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