a sailboat of mass "m" is moving with a momentum p. what represents its kenetic energy?
445873
K.E is given by m*(v^2)/2\nMomentum p = mv\nSquare both sides you get p^2 = m^2*v^2\nNow divide both sides by 2m and you get p^2/2m = m*(v^2)/2\nClearly the right hand side of the equation represents K.E\nso the answer is K.E = p^2/2m
240568
Conservation of momentum
1045430
Momentum - P = mv (P=momentum, m=mass, v = velocity)\nThree Laws of Newton - F = ma (F=Force; a=acceleration)\nEquations of motion - 1) v = u + at\n                                    2) S = (1/2)ut + at^2\n                                    3) v^2 = u^2 + 2aS\n(v=final velocity; u = initial velocity; S=distance)\n\nIf this can help you please mail me back !\nBEST OF LUCK !!!!
257654
Say the anchor weighs 15 pounds in air and displaces 2 pounds of water.\nBefore the drop, the boat displaces its weight plus the 15 lb weight in air of the anchor. This is the maximum displacement situation, so the water is at maximum height.\nAfter the drop, if the anchor hits bottom, the boat displaces its weight only and the anchor displaces 2 lb of water, the minimum displacement situation, so the water is at minimum height.\nAfter the drop, if the anchor doesn't reach the bottom, the boat displaces its weight plus the 13 lb weight in water of the anchor. The anchor displaces 2 lb water. The total displacement is the same as the maximum displacement situation, so the water is at maximum height.
704333
let us say that she is walking in the x direction (positive) \n\nThis is a question of momentum... \np = mv   \n(momentum equals mass times velocity)\n\nand we have the conservation of momentum \np_1  +p_2 = p_(total)\nThe sum of the momentums is equal to the total momentum of the system\n\nSo in the begining she is not moving... so the total momentum of the system is zero... it will always be zero.  \n\nShe walks... her momentum is p_1 = 45*1.5 = 67.5 kg*m/s\n\nand we know that p_1 + p_2 = 0\n\nso \np_1 = -p_2 \nor \n67.5 = - m v = - 150 *v\nso the velocity of the plank is -0.45 m/s.  \n\nNow we just sum the velocities to get the total velocity and we see that she is traveling at 1.05 m/s relitive to the ice. \n\nGood question ... i have not thought about a problem like this in a long time
36280
Assuming it can't be II and III, I would say it's all of the above. The potential energy is converted into kinetic - the velocity increases and the height decreases. The KE + PE remains constant, because there is no work being done because there is no force here (besides gravity). However, the first depends on how you define mechanical energy - if it's kinetic energy, it's true. If not, I really don't know what it is.
383729
It has a dynamo inside it, when you wind it up, that is the driving force for the energy.
229381
either or--- Its the motion in the ocean.. Not the size of the Boat.... REMEMBER THAT!
873605
If you're sailing and the wind stops, you coast.  Then you're moving faster than the wind.\n\nIf there's no wind and you're using the boat as a surfboard.\n\nIf you're reaching with a light planing hull, your hull speed can exceed the wind speed.  The wind speed relative to the boat is a combination of the wind speed and the boat speed. Conditions would need to be nearly perfect.
666315
Kinetic energy Ek\n\nlet's consider that an object with a mass m moves on a distance d = x1 - x2, with an acceleration a and its velocity changes from v1 to v1, under the action of a force F\n\naccording to Galileo's law:\nv2^2 - v1^2 = 2.a.d\n\nbut according to Newton's second law:\nF = m.a\n\n-> a = F/m \n-> v2^2 - v1^2 = 2.F.d/m\n-> m.v2^2 - m.v1^2 = 2.F.d\n-> m.v2^2/2 - m.v1^2/2 = F.d\n\nbut W = F.d = the work done while the force F moves the object on the distance d, it caracterizes a process\n\nBut energy is the capacity of an object of doing work, it caracterizes a state\n\nthe kinetic energy is due to the velocity of the object:\n\ndeltaEk = E2 - E1 = W = F.d \n-> E2 = m.v2^2/2\n-> E1 = m.v1^2/2\n\nPotential energy Ep\n\nLet's consider an object with a mass m is falling from a height h1 to h2. \n\nThe work done by gravity is:\n\nW = G.(h1-h2) = m.g.(h1-h2) = m.g.h1 - m.g.h2\n\nThe potential energy is depending on the position of the the object as compared to the Earth and its equal with the work done by gravity taken with opposite sign.\n\ndelta Ep = Ep2 - Ep1 = - W\n\n-> Ep2 = m.g.h2\n-> Ep1 = m.g.h1
762043
Use the conservation of momentum:\n\nM * (change in velocity) + (change in mass) * V = 0\n\nM * dv + dm * V = 0\n\nWhere M is the mass of the child and boat system, dv is the boat's change in velocity, dm is the change in mass of the initial system (the package) and V is the speed at which the package is ejected.\n\nSubstituting the values:\n\n(26 kg + 45 kg) * V = - (6.4 kg) * (10 m/s)\n\n(71 kg) * V = - (64 kg-m/s)\n\nSolving for V:\n\nV = - (64 kg-m/s) / (71 kg)\n\nV = -.9014 m/s\n\nThus, the boat travels backwards at a speed of .9014 m/s. That's it! :)
863271
mass doesn't matter here, just the acceleration\nv=u+at\na=(v-u)/t\na=(9-0)/0.15=60 m/s^2
253742
Please take the time to\nvote for the best answer\n\nThank You!!\n\nRecall that an object of a certain mass moving with particular speed will have an associated kinetic energy  mass x speed2. An object spinning about an axis will also have associated with it a kinetic energy, composed of the kinetic energies of each individual part of the object. These individual contributions may be summed up to give an expression for the total kinetic energy of the spinning object: \n\nLook this up for more
216275
I believe momentum equals Velocity times Mass.  So momentum is displayed in kgm/s
985678
Incoming momentum = p1 = mv1\nOutgoing momentum = p2  = mv2\nChange in momentum = p2 - p1 = m(v2-v1)\nThis change occurs in time, t.\nRate of change of momentum = m(v2-v1)/t = impressed force, f.\nm=0.149 kg, v1 = 43.4 m/s, v2 = -48.7 m/s, t = 1.1x10^-3 sec\n\nf = m(v2-v1)/t\nYou do the arithmetic.
596819
Kinetic Energy = mass x velocity^2 divided by 2\n\nKE = 12 x 12^2 / 2
244521
there is an engine i saw that has no moving parts. just magnets that attract water in one end and out the other, but it only works in salt water.