Two particals located at (8i+7j+2k) m and (11i+j+2k) m respectivly, start moving with constant velocities of (
829059
I hope I did this correctly.  Would you let me know?\n\nI set up the equation of the relative distance the particles will have from eachother at time t:\n\ndistance equals location of particle one minus location of particle two:\n\nx=\n(4t-3)i+\n(6-3t)j+\n(5t)k\n\nthen I distributed the i, j and k to get\n4it-3j+\n6j-3tj+\n5tk\n\nThen I took the derivative dx/dt treating i, j, and k as constants to get\n4i-3j+5k\nthis is the minimum separation\n4i-3j+5k\n\nThis I then used to set up the following three simultaneous equations to find the time that occurs:\n\n4t-3=4\nt=1\n\n6-3t=-3\nt=1\n\n5t=5\nt=1\n\nj
700987
The energy taken to slow the object from v1 to v2 is\n\nE=FS=(1/2)m(sqr(v2)-sqr(v1))\n\nand given v2=(1/2)(v1)\n\nWe get  v1=sqrt((8/3)(FS/m)) = 12m/s\n\nAlso from v2=v1+at     and          F=ma\n\nt=v1/(2a)=mv1/(2F) = 1 second
413991
They are moving towards each other at an increasing rate.
925426
one half and seven eightths
493729
almost right ...\n\nc1* v1 = c2 * v2\n18M (V1) = 6M*0.25L\n\nBut that makes V1 = 0.083L
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
874949
c. Velocity will have size and direction.
494819
MORE OR LESS 135 KM/H.
1012426
Using 9.8 m/s^2 for gravity, the answer is 3,234 J.
740205
They both travel at the exact same speed.  This was recently verified here : http://www.newscientist.com/article.ns?id=dn3232
834761
Lets call the distance of the first part S1 and the total distance S2, and the corresponding times T1 and T2.  We know:\nS1 = 40\nS2 = 100\nT2 = 9.85\n\nIf V is the maximum velocity we know S1 = 0.5VT1 (const acc.)\nFor the second part we know S2-S1 = V(T2-T1) = VT2 - VT1\n\nVT1 = 2S1 so S2-S1 = VT2 - 2S1 or V = (S2+S1)/T2 = 140/9.85\n\nSo a) - V = 14.21 m/s\n\nWe know V^2 = 2AS1 where A is acceleration so A = V^2./2S1\nor A = 14.21^2 /(2x40)\n\nSo b) A = 2.525 m/s^2\n\nWe know V = AT1 so T1 = V/A = 14.21/2.525 = 5.62 s\n\nSo c) the two parts are 5.62s and 4.23s
916585
Conservation of momentum so mv is constant.\n\nTotal final mass = 5333+14.6 = 5347.6g\n\nTotal final momentum = 5347 x 1.7 gm/s = 9090.92 gm/s\n\nBullet's initial momentum = 9090 gm/s\n\nMomentum/mass = velocity\n\n9090.92/14.6 = 622.666m/s
790665
starting from the very beginning\nv ≡ ds/dt\na ≡ d²s/dt² = dv/dt\ndv = adt\nif a is constant,\n∫dv = a∫dt\nv = at + C\ncall C v0, or initial velocity:\nv = v0 + at\nv = 0 +(9.80662 m/sec^2)(2 sec)\nv = 19.61324 m/s
912927
1D\n2D (acceleration can be negative in which case you mihgt call it deceleration)\n3B
611971
Mass is a constant; weight varies by gravitational pull.
864791
A.) 3.02891 seconds\nB.) 290.1 m/s\n\nTrust me... =)
819996
T1=m1*a, no matter are there m2 & m3 or not.\nT2=(m1+m2)*a, no matter whether m1 & m2 are connected with a string or are in one lump, the presence of m3 being unimportant too.