Velocity of Billards Balls?
413326
Depends how hard you hit it and at what angle, quality of baize etc.
901904
I think here is a mistake:\n\n(1500 * 36) = 1000v\nv = 55.8 m/s\n\nyou have to put 500 instead 1500, because this is what we have after 1000 left us!\n\n(500*36)=1000v\nv=15 m/s
843122
good start.\n\ntotal inital momentum = 480-397.5\n\nthis is an inelasitc collision, and aferwards you have \nboth masses travelling at velocity v:\n\n(75+120)v = 480-397.5\n\nsolve for v.  This is the velocity at which they both slide.\nThe inital energy now is 1/2(75+120)v^2\n\nThis energy is converted to frictional energy:\n\ninital kinetic energy = total work of friction = Fr(distance)\n\nFr = uN = u(120+75)g = .3(195)(9.8)\n\nso we have\n\n.3(195)(9.8)(distance slided) = 1/2(195)v^2 \n\nYou know v form above.  So find distance slided.
752814
is the scoring player 'dragging' the 73 newton defender? then the defender is a frictional load NOT a horizontal load\nALSO remeber what is dynamic and what is static?\nWhich is acceleration and which is velocity\nThat always give me the hardest part of these questions
195717
dba_gregg has missed the point that both are same weight\n\nThe moment of Inertia of a hollow sphere  is 1.6 times greater than the solid sphere and so it rolls down slowly.\n\nThe speed of solid sphere will be 1.09 times the speed of hollow sphere.
816469
well F=mv/t given: m=78kg, v=24m/s, t=60s\n\nF=78(24)/60=31.2N
821062
I think you are supposed to assume the minimum velocity possible, which means tossed at a 45 degree angle. Otherwise he could throw it farther by using the same velocity at 45 degrees.\n\nvelocity = v, y velocity at time t = v/sqrt(2) - 10t so y velocity is 0 at \n\nt = v/(10sqrt(2))\nIn time 2t the distance travelled is 96.90, so\n\n96.90 = 2tv/sqrt(2)\n\nThese are 2 equations with 2 unknowns, solve for t and v (there may be a careless mistake in the details since I did this fast, but the technique will work)
341043
Previous answerer had it correct.  Falling-ball viscometers are based on Stokes' Law.  See sources below for a complete description.
251303
The car has almost 700 times the momentum of the ball.\n\nJust to be standard, let's convert everything to kilograms, meters and seconds.\n\nStandard baseball equals 5 ounces or 0.14 kg\n\n150 (km / hr) = 41.7 m / s\n\n41.7 x 0.14 = 5.8 kg·m/s for the baseball\n\n--------------------------------\n\nPT Cruiser is 1446 kg\n\n10 (km / hr) = 2.78 m / s\n\n2.78 x 1446 = 4,020 kg·m/s for the PT cruiser
803921
Use conservation of energy.\n\na.  1/2mA*vA^2 - mA*g*h = 0 \n\nwhere h =  L - Lcos60   assuming L is the length of whatever is suspending the two balls.\n\nb. Use:  1/2mA*vAi^2  + 1/2mA*vAf^2 = 1/2mB*vBi^2  + 1/2mB*vBf^2 \n\nc.  1/2mA*vAf^2 - mA*g*ha2 = 0 \n     1/2mB*vBf^2 -  mB*g*hb2 = 0 \n\nsolve for ha2  and hb2
1171482
tehre's a formula for it...\nit depends how much you run druing the game
127834
My husband was playing racket ball with a guy, who slammed the ball into the front wall only to have it come back and hit himself in the balls.  Have you ever saw the velocity on a racket ball?
578599
Don't you just hate it when your walls aren't ball-secure?\n\n40mm eh -\nso that's 12000/40 = 300 balls width\nand 25000/40 = 625 balls length\n300*625 = 187500 balls would form a layer 1 ball thick\n\nand 14000000/187500 = 74.6667 layers of balls\n\nSo they'd stretch up to 75*40mm = 30m high in places. BUT that's not allowing for the fact that they're balls and would hence tessellate with one another, so it wouldn't probably be quite as much as that.\n\nEDIT: ag_iitkgp, you've got a decimal place wrong. It's 75 not 7.5 layers.
904552
Both balls would hit the rocks at the same speed.\nBall #1 would leave your hand travelling straight up at speed 'u'. It travels up, stops, and starts plummeting back to earth. When it reaches the height at which it left your hand, it is travelling at speed 'u' straight down. The speed that it hits the rocks at is u+(time x 10m/s/s).\n\nBall #2 would leave your hand travelling straight down at speed 'u'. The speed that it hits the rocks at is u+(time x 10m/s/s). It is the same speed.
1069779
Draw a free-body diagram of the ball and examine it's accelerations - if you assume wind and air resistance are negligible and that the ball is accelerated straight up, then you're left with its initial acceleration (68.57 m/s^2) and gravity (9.8 m/s^2 if I remember correctly) \n\nThe ball will decelerate by 9.8 m/s every second, you can divide to find the amount of time the ball takes to reach its apogee (point of zero acceleration), then integrate the acceleration curve to find the velocity curve, which can be integrated again to find the distance traveled.  I'm not sure it can be done without the calculus.
666681
1/2 * mass * velocity^2.
711024
Look, task 1,3,4 are very alike. We shall use the same formulas there:\n1. let us find energy of the body at A, at B, and at C\nat A:   mV^2/2 + mgH = m(V^2/2 + gH)\nat B H=0, so the energy constitutes the kinetic energy mV(b)^2/2\nat C: mV(c)^2/2 +mgH(c) = m(V(c)^2/2 + gH(c))\nthe energy does not change (not dissipates), so energy in A equals the energy in C:\n\nV^2/2 + gH = V(c)^2/2 + gH(C), so \nV(c) = sq.root {V(a)^2 + 2g(H(a) - H(c))} = \n = sq.root{64 - 2x2x9.81} = 4.976 m/s\n2. The force of gravity for the mass 100 kg will be \nF = mg = 100kg x 9.81 m/s^2 = 981N\nAs the mass is moved along steep surface, some part of the gravity force will hinder the movement. We can fin the component of the gravity force along the surface using the triangular 20m and 5 m proportion, (because the triangles are similar), so\nFg(along) = 981N x 5/20 = 981N/4 = 245,25N\nWe have to apply the force 300N to move the body along the surface, hence we overcome the gravity component 245.25, and the rest is for friction.  \nSo the Ffriction = F - Fg(along) = 300N - 245.25N = 54.75N\n\n3. The formulars are as in problem 1. \nat the start   Es = Ekinetic + mgH\nat the highest (final) point of the bounced ball\nEf = Ekineticf + mgHf,\nHere we have to make some remarks (very important)\nIf the question would be "How high the ball can bounce",\nthen we could assume that the highest point would be when the ball bounces vertically up, and at its highest point it has zero velocity, no vertical one, no horizontal one. Than we can makethe energies equal (assuming no dissipation) and find Hf:\nmV^2/2 + mgH = mgHf\nso,  Hf = H + V^2/2g = 4(m) + 36(m^2/s^2)/2x9.81(m/s^2) = \n             = 4(m) + 1.835(m) = 5.835 (m)\nIn case the ball bounces not vertically, but along the steep line, at his highest point it will have zero velocity too, so the answer will apply to the case. \n\n4. The velocity in B will be found using the formula for energy at starting piont and at B:\nmV^2/2 + mgH = mV(b)^2/2,\n\nV(b) = sq.root{V^2 + 2gH} = sq.root{16 + 12x9.81} = 11.56 (m/s)