A 0.70 kg rubber ball has a speed of 2.0 m/s at point A and kinetic energy of 6.5 J at point B.?
962433
The  kinetic energy of the ball is increased from 1.4 J to 6.5J by the work done on it.\n\nThe gain in kinetic energy is 6.5 - 1.4 = 5.1 J\n\nWork done on the ball = gain in kinetic energy = 5.1 J.
1010100
ok, so if we're assuming that it was a head on elastic collision, and all the energy was transferred then there are two possible situations: if the 25lb ball was sliding then the total energy transferred will be equal to the amount it had originally (since we assumed elastic collision) that is equal to 1/2mv^2...where mass would usually be measured using the metric system. \nif the ball was rolling then it has not only translational energy (1/2mv^2) but also rotational energy which is equal to 1/2Iw^2 where I = inertia of a ball...if it is a solid sphere the moment of inertia is equal to 2/5mr^2 (m = mass, r = radius) and w (omega) = rotational speed in radians per second...which is equal to velocity/radius. so the amount of energy transferred would be equal to 1/2Iw^2 + 1/2mv^2.
840035
Ok, we're going to assume the 4.6268 cm is in addition to the 58.8cm, so the ball doesn't stop until it has fallen 63.4268 cm.  Therefore it has lost .0833*.634268*9.8 = .51778 Nm of energy.\n\nSince all this energy goes into compressing the spring, and its energy is .5kx^2, k = 2*.51778/.046268^2 = 483.74 N/m
816469
well F=mv/t given: m=78kg, v=24m/s, t=60s\n\nF=78(24)/60=31.2N
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
149886
Hope this helps,  first remember that energy must be conserved so E initial will equal E final so you use the equations for potential and kinetic energy.  mgh for potential and 1/2mv^2 for kinetic.  You know the initial velocity, gravity and the height of the ramp so you simply break it into components and get the answer of 3.15 m/s.  HOpe this helps
864791
A.) 3.02891 seconds\nB.) 290.1 m/s\n\nTrust me... =)
1149825
I'll give you a head start, and we'll see if you can figure it out from there:\n\nFirst, the givens:\n\nd = height from the ground to the point at which the two balls are released = 122.5 meters\n\nv1 = initial velocity of the first ball = 0\nv2 = initial velocity of the second ball\nt1 = time ball 1 is dropping (seconds)\nt2 = time ball 2 is dropping (seconds)\nt2 = (t1)-2 seconds\n\na = gravitational acceleration = 9.8 m/sec^2\n\nFor both balls:\n\nd = vt + at^2\n\nFor the first ball, v1 = 0, so d = at1^2\n\nFor the second ball, d = v2t2 + at2^2\n\nThe distance in both case is the same, so:\n\nat1^2 = d = v2t2 + at2^2\n\nOkay, so here you go:  You have the basic equations,so solving is trivial arithmetic:\n\nd = 122.5 meters = (9.8 m/sec^2) (t1)^2\n\nand solve for t1.  Once you know t1, then you know that t2 is two seconds less, so with that value you can solve:\n\nd = 122.5 meters = v2t2 + (9.8 m/sec^2)(t2)^2\n\nI leave the rest to you!  (And I have calculated the answer if you want to check it)
83305
I suppose the problem was already due so I'll finish it for you now.\n\nF=m*a\na=F/m= 2*10^6N/1.2*10^6kg = 1.67m/s^2 * g/9.81m/s^2 = 0.17g
931347
x =70m, t = 7s, Vf = 15m/s\na.)\nx = 1/2(Vi + Vf)t\nVi = initial velocity\nVf = final velocity\n\n70m = 1/2(Vi + 15m/s)(7s)\n10m/s = 1/2Vi + 7.5m/s\n2.5m/s = 1/2Vi\nVi = 5m/s\nb.)\nVf = Vo + at\n15m/s = 5m/s + 7s(a)\n10m/s = 7s(a)\na = 1.43m/s/s\nbasic kinematics\nyou can't use distance = rate x time because there is acceleration and the speed is not constant.
771612
hello\n\na)take the momentum of the sports car and divide by the total mass of  both cars. 25×920÷3220=7.14m/s\n\nb)using the same method as a), we can find the answer. but use a little logic here, if the total mass doubles then the final speed must be halved for conservation of momentum to be followed. v=12.5m/s\n\nc)KE=0.5mv². speed is halved. v² means kinetic energy should be quartered (0.5²). but mass was doubled due to there being 2 cars now. so overall kinetic energy has only been halved (0.5²x2). the other half is lost.
829880
the equation is:\nF=G*m1*m2/(r*r);\nso, F=6.67*10E-11*0.14*0.17/(0.65*0.65);\n\nthis is the force that either will exert on the other
495603
Assuming that the energy of the ball remains constant (i.e. no air resistance), the speed will be the same. The VELOCITY, however, will be -35 m/s.\nSpeed and velocity are NOT the same!
888217
v=w*r, so 24pi/s*6cm=452.4cm/s
816137
a) v^2-u^2=2as\n    v^2-52.4^2=2*(-9.8)*28.5\nv^2= 2745.8-558.6\n     =2187.2\nv=46.76m/s\n\nKinetic Energy=1/2mv^2\n                     =1/2*46.9*46.76^2\n                      =51,290 Joules\n\nb)height=28.5-5.06=23.44m\n\nv^2-u^2=2as\n\nv^2-52.4^2=2*(-9.8)*23.44\nv^2=2745.8-459.42\n     =2286.4\nv=47.8m/s
487670
None. It takes no power to maintain that speed (on level ground, no friction). The object has kinetic energy (0.5 * m * v^2). Power was required to put it in motion and power can be recovered by decelerating it. If you do the math, the energy is 3551.86 ft-lb, equivalent to 1 horsepower for 6.458 seconds. Answer 1 neglected the fact that weight must be converted to mass by dividing by g, and energy is proportional to v^2.
805324
Part a)\n\nVf^2 = Vi^2 + 2 * a * d\na = (Vf^2 - Vi^2) / (2 * d) = -4,489/162 = -26.7 m/s^2\nm = 220*10^3 / 9.81 = 22,426 kg\nF = m*a = -598,777 N\n\nW = F * d = 50,297,245 J\n\nPart b)\n\nF (from above) = 598,777 N\n\nI hope this helps!