how can you calculate kinetic energy?
608223
For moving objects, calculate kinetic energy using the formula:\nKE = (mass x velocity2)/2 or 1/2 mv2\n\nhttp://www.gmhsscience.com/problems/kineticpotential.htm
955797
Use the Arrhenius equation to predict the rate:\n\nhttp://en.wikipedia.org/wiki/Arrhenius_equation\n\nK = Ae -E/RT\n\nIt all boils down to your reactants, the activation of energy, and the product that is being synthesized.
298488
Energy Unit = Joule = 1 kg x m^2/s^2 = 1 N x m\n\nHow can work transfer into keinetic energy?\nThink in a pendulum in which these two are constantly tranfered into one another.\n\nWork = Force x Distance \n\nwork = 10N x 2m = 20 N x m\n\nKinetic energy = Mass x Velocity^2 / 2\n\nkinetic energy = 2kg x (4m/s)^2 / 2 = 16 N x m
967786
I'm not going to tell you but look use Saha's equation to find out the energy in ev of each level. Then you can work out the energy value of the jump. It's just a simple bit of maths. But you are wrong so have a think
962133
E potential = mgh\n= 4 kg ( 9.81 m/s^2) ( 2.5)\n=98.1\n\nPotential energy was not lost....it transformed into Kinetic Energy\n\nKinetic + Potential = Total Energy\n\nLaw of Conservation of Energy
1131325
Maybe this can help...all you have to do is a search in Yahoo or Google...\n\nhttp://www.think-energy.com/ThinkEnergy/11-14/activities/EnergyTrans.aspx\n\nTry searching for "energy" or "kinetic energy"
167143
Start with formula for total energy (the derivation of this is pretty awful and requires lots of diagrams to explain it)\n\nSo total energy is g*m*c^2 [g is gamma factor 1/sqrt(1 - V^2/c^2)]\n\nThe kinetic energy is total energy minus internal energy of particle = g*m*c^2 - m*c^2 (1)\n\nNow expand the gamma factor as a binomial series:\n\ng = 1/sqrt(1 - V^2/c^2) = 1 + (v^2/2*c^2) + ...\n\nsubstitute this into the above equation (1) and you get\n\nkinetic energy = m*c^2(1 + v^2/2*c^2) - m*c^2\n\n= 1/2*m*v^2\n\n(This is an approximation - so this only applies for low velocities. For relativistic velocities use formula (1) or expand the binomial expression to higher degrees of 'v')
1133901
Use the equation U = 0.5*k*x^2\n\nU = potential energy (joules)\nx = compression distance (meters)\nk = force constant (N/m)\n\nThis gives us the potential energy of the toy.\n\nU = 0.5 * 296 * 0.0122^2 = 0.0220 J\n\nThen calculate the Kinetic Energy:\n\nKE = 0.5 * m * V^2\n\nKE = Kinetic Energy (joules)\nm = Mass (kg)\nV = Velocity (m/s)\n\nKE = 0.5 * 0.153 * 0.299 = 0.0229 J\n\nTotal Energy = 0.0229 + 0.0220 = 0.0449 J\n\nFor part 2, assume total energy has been converted to potential energy and solve for compression distance:\nx = sqrt (2 * U / k) = sqrt (2 * 0.0449 / 296) = 0.0174 m\n\nAmplitude is twice the compression distance (for positive and negative)\nAmplitude = 2 * 0.0174 = 0.0348 meters\n\nFor part 3, assume total energy has been converted to Kinetic energy and solve for speed:\nV = sqrt (2 * KE / m) = sqrt (2 * 0.0449 / 0.153) = 0.766 m/s
606680
If by force you are refering to the kinetic energy of a moving object, then the equation is as follows: kinetic energy = one half momentum squared/mass.  Momentum of course can be calculated by mass times velocity.
530124
Since kinetic energy is proportional to the square of linear momentum, a 300% increase in kinetic energy(which meanse that kinetic enrgy becomes 4 times its initial value),results in doubling of momentum.This meanse that momentum increases by 100%.
934211
Ok, so you have two simple equations to keep in mind:\n\n1)  KE = (1/2)*m*v^2\n(kinetic energy equals one half mass times velocity squared)\n\n2) Paverage = E/t\n(average power equals energy divided by time)\n\nFirst you find the kinetic energy (convert that km/hr speed into m/s).  Now you have the kinetic energy for the car after the acceleration period (in Joules) which can then be divided by the time to get the average power (in J/s or Watts) required to get that car up to the given speed.\n\nThen we sometimes like to convert mechanical power from Watts into horsepower.  The conversion happens to be 745.7 W/hp.\n\nI will leave the calculations up to you, but those are the equations/concepts you need to know and apply.  Also, take note that this is an idealized solution that does not take into account wind resistance, rolling friction and various other factors that would require more power output from the engine of the vehicle.
895448
Kinetic energy can be expressed as\n\nKE = 0.5*m*v^2\n\nDoubling the kinetic energy requires increasing the speed by the square root of 2. Where doubling the speed increases the energy by a factor of 4.
897524
Kinetic energy is .5 * mass * velocity ^ 2\nSo the change in kinetic energy, or delta K.E, would be:\n\n.5 * mass * initial-velocity ^ 2 - .5 * mass * final-velocity ^ 2\n\nSince Kinetic Energy is lost, the initial velocity goes first.\n\nGood luck on your test!
1001657
V= V(i) + AT Initial velocity plus acceleratin x Time\n\n\n   = V(i) + (F/m)T     Acceleration = Force/ Mass\n\nNote  use same system of units
1108031
Kinetic energy (SI unit: the joule) is energy that a body possesses as a result of its motion. It is formally defined as the work needed to accelerate a body from rest to its current velocity. Having gained this energy during its acceleration, the body maintains this kinetic energy unless its speed changes. Negative work of the same magnitude would be required to return the body to a state of rest from that velocity.
802951
Kinetic energy is energy due to motion and potential energy is energy due to position in a force field.  The formula for translational kinetic energy is KE = 0.5*m*v^2, where m is mass and v is velocity.  The most common type of potential energy is gravitational potential energy, and near the surface of the Earth (or some other planetary body) the formula is PE = mgh, where m is mass, g is acceleration due to gravity, and h is the height above the surface.  Another common one is elastic potential energy in a compressed spring.  In that case the formula is PE = 0.5*k*x^2, where k is the spring constant and x is the compression.  An object can also have electrical potential energy, if it has an electrical charge and is positioned some distance from another charge.
984947
W=F*L – work done by F=145N on distance L=7.2m,\nU=k*m*g*cos(a)*L, work wasted on friction, where k*m*g*cos(a) is force of friction, while\nk=0.31, a=11.7deg, g=9.8m/s^2, m=10.3kg.\nY=m*g*L*sin(a) work done to hoist m to height of L*sin(a) meters.\nE0=m*v0^2/2 initial kinetic energy of m, v0=1.3m/s \nNow energy balance: E0+W=U+Y+E1, E1 is final kinetic energy of m E1=m*V^2/2, V is final speed of m. Thus kinetic energy change dE=E1-E0=W-U-Y.\nAs E1=E0+W-U-Y=m*V^2/2, then V=sqrt[2*(E0+W-U-Y)/m], \nI hope you cope with numeric calculations, all are SI units