What is the kinetic energy of an object that has a mass of 30 kilograms and moves with a velocity of 20 m/s?
486085
KE=1/2mv^2\n=1/2(30)(20)^2\n=1/2(30)(400)\n=1/2(12000)\n=6000J\n\nC
276917
KE = (1/2)m*v^2 where m is the mass of the object in kilograms and v is the velocity of the object in meters/second. Plug and chug, answer will be in Netwon-meters or Joules.
866141
You have to calculate p=massXspeed before the "collision".  The quantity remains constant, so if the mass increases, the speed must decreases by the same ratio for the p to stay the same.  (First, you have to convert the 8N weight to a mass using the 9,8N/kg gravity constant)
716765
The potential energy at the top is mgR = 2 x10 x R = 20 R.\n\nKinetic energy at the bottom of the curve = 0.5 m U ^2 = 0.5 x 2 x U^2. = U^2.\n\nAs there is no friction  K.E = P.E\n\nU^2 = 20 R.\n\nOn the horizontal path, \n\nInitial speed is U.\n\nFinal speed is V = 0.\n\nDistance traveled = 8 m.\n\nAcceleration = Fricttional Force / mass = -8/ 2 = - 4 m/s^2. ( Negative sign since the force acts opposite to the motion.\n\nUsing V^2 - U^2 = 2as, and also noting that V=0,\n\nU^2 = 2 x 4 x 8 = 64.\n\nU = 8m/s.\n\nThe time of travel is found using V= U+ at.\n\nt = -8/ (-4) = 2 second.\n\nUsing U^2 = 20 R.\n\nR = 64/ 20 = 3.2 m.
528388
You can go about this problem a few ways. A force exherted on an object will give it a certain velocity. If F=m*a, and you have the force along with the mass, you can determine the acceleration. The acceleration, of course, is the change in velocity/change in time. You'll get a velocity change per second. If you know the ball is in contact for 0.03 seconds, you can determine the final velocity, v. \n\nKinetic energy is just 1/2 * m * v^2 so that will be a simple calculation once you figure out v.
906577
Hi!\n(A) \nBy definition di=dm * X^2 , where di is the amount added to I when the mass dm is added to the system. I~Inertia . In this case for the particle X=d.\n\nEach mass m=0.85 kg add to the system (delta I)1 = m * d^2.\n\nEach rod has an Inertia of (delta I)2 = (1/3) M * d^2 . This is because for a rod you have to calculate:\n             S { dm * x^2 = (M/d)*dx * x^2 } ==> (M/d) * (1/3)(X^3) with X=d. => (1/3)M * d^2  \n\nso the total inertia of the system is I= 2*[ m*d^2+ (1/3) M*d^2] because we have 2 rods and tho particles.:\n    I= 7.84 * 10^(-3) kg / m^2.    Where ^ is used for power!\n\n(B) kinetic energy= (1/2) Inertia  * (angularspeed)^2 =3.528 /10^4\n\nIf you want to know more try the link below!
860183
This is a momentum question.\n\nMomentum = mass x speed\n\nSo, the original momentum is contained solely withing the moving engine and equals 4 units of mass times 5 km/h.  This equals a 'momentum' of 20.\n\nThe final momentum equals the combined mass of the engine and freight car, which is 4 units plus 1 unit for a total of 5 units of mass.\n\nSo, if Momentum = mass x speed, then since our final momentum is 20 (the same as the original momentum) and our final mass is 5, then 20 = 5 x Speed\n\nDividing both sides by 5 yields a final speed of 4 km/h.\n\nBest wishes and good luck.
686704
The increase of the gravitational energy (m1+m2)gh is the loss of kinetic enery m(v^2)/2.\nThe bullet gets the same speed as the bob.\nUse conservation of momentum.
860846
Force = 60 N.\nMass = 3 kg\nTime = 10 s\n\nAn object's change in momentum, Δp equals the Impulse, I on the object. I = ∫F dt = Ft. \n\nΔp = I = Ft = (60 N) * 10 s = 600 Ns\n\nAssuming the object started at rest,\np (initial) = mv = (3 kg) (0 m/s) = 0 Ns\nΔp = 600 Ns\np (final) = 600 Ns\n\np (final) = mv = (3 kg) * v(final)\nv(final) = (600 Ns) / (3 kg) = 200 m/s\nΔv = v(final) - v (initial) = 200 m/s - 0 m/s = 200 m/s\n\nThe object's change in momentum was 600 Ns (kg * m/s).\nThe object's change in velocity, assuming it started from rest, was 200 m/s.
837367
Kinetic Energy\nKinetic energy is the energy of motion. An object which has motion - whether it be vertical or horizontal motion - has kinetic energy. There are many forms of kinetic energy - vibrational (the energy due to vibrational motion), rotational (the energy due to rotational motion), and translational (the energy due to motion from one location to another). To keep matters simple, we will focus upon translational kinetic energy. The amount of translational kinetic energy (from here on, the phrase kinetic energy will refer to translational kinetic energy) which an object has depends upon two variables: the mass (m) of the object and the speed (v) of the object. The following equation is used to represent the kinetic energy (KE) of an object.\n                     \nke=1/2*m*v^2\n\nwhere m = mass of object\n\nv = speed of object
596819
Kinetic Energy = mass x velocity^2 divided by 2\n\nKE = 12 x 12^2 / 2
825498
Kinetic Energy = 1/2mv^2; Potential Energy = mgh\n\n1/2mv^2 = mgh or 1/2v^2 = gh\n\nv^2 = 2gh = 2 * 9.81m/s^2 * 0.49 m\n\nv = 3.1 m/s
751049
First find Vo:  Vo^2 = Rg/sin(28)  Where R = range & 8 = theta\n\nUsing your numbers, Vo = 41.523 m/s\n\nImpulse = change in momentum ----> Ft = mVo---> F = mVo/t\n\nSo, F = (.0431kg*41.523m/s)/.00438sec = 408.59 N\n\nVoila!
1054528
The AVERAGE kinetic energy of the atoms/molecules an object is composed of.
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
354450
WORK IS EQUAL TO ZERO (READ ALL FOR EXPLAINATION)\nmass of body=30kg\nits weight=30kg X 9.8m/s^2\nweight= 294N\n\nBUT HERE WE DON'T HAVE TO DEAL WITH ITS WEIGHT.\n\nAs the body is moving at constant speed, it means that it's not accelerating and thus no force is being applied to it.\n\nOr in other way\n\nthe frictional force which is 20N is canceling the force with which the boy is moving the body. Thus the boy is also exerting a 20N force (20N - x = 0, of course x=20N)\n\nTHE WORK IS ZERO BECAUSE THE FORCE IS ZERO TOO.\n\nsince the frictional force is cancelling the force of the boy.\n\n20N(frictional force)-20N(force applied)=0\nResultant force = 0\n\nFIND WORK NOW\n\nwork =  net force X distance\nwork =  0  X  5m\nwork =  0 joule\nNo work is done\n\nLISTEN\nThe person below me by the name of sblogs is incorrect as I think. Let me prove this by giving some EXAMPLES\nPUSH THE WALL BUT THE WALL WILL NOT MOVE FROM ITS PLACE THOUGH YOU ARE WASTING ENERGY BUT THE WORK DONE IS ZERO.\nBECAUSE THE WALL DOESN'T MOVE THROUGH SOME DISTANCE.\nWORK= FORCE X DISTANCE\nWORK= (FORCE APPLIED ON THE WALL) X (ZERO METERS)\nWORK= F X 0\nWORK= 0 JOULE
905951
First of all I assume the height of \nbasket above ground is 2+3.05=5.05\nLet v be the required velocity.\nvertical component=vsin40\n=v*.766\nfrom conservation of energy\nheight gained\n=[v*.766]^2/2g=3.05\nv^2=3.05*2*9.81/.766^2\n=101.986\nv=10.09m/sec