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The equation is for kinetic energy, K, is:K = 1/2mv2Plug in the numbers:K = 0.5 x 9000 kg x 9 m2 s-2So that energy is 40 500 J, or 40.5 kJ.
Let's see. Mass constant. 625000 Joules = 1/2(500 kg)V2 1250000 Joules = 500V2 2500 = V2 50 meters/second = velocity ==================================== The velocity doubles if KE doubles.
Before the driver hit the brakes, the car's kinetic energy was (1/2 m V2) = (500) (20)2 = 200,000 joules.After it stops at the stop sign, its kinetic energy and potential energy are both zero. The brakesturned that whole 200,000 joules into heat, and the wind blew it away.Unless . . .Unless the car is an electric or a hybrid. What they do is to take that kinetic energy and puta lot of it back into the battery, so you can use it again, and you don't completely lose it all.That's what's the big deal about electric and hybrid cars.
The answer to both of your questions lies in the different nature of both quantities, momentum and kinetic energy. Momentum is a vector, kinetic energy is a scalar. This means that momentum has a magnitude and a direction, while kinetic energy just has a magnitude. Consider the following system: 2 balls with equal mass are rolling with the same speed to each other. Magnitude of their velocities is the same, but the directions of their velocities are opposed. What can we say about the total momentum of this system of two balls? The total momentum is the sum of the momentum of each ball. Since masses are equal, magnitudes of velocities are equal, but direction of motion is opposed, the total momentum of the system of two balls equals zero. Conclusion: the system has zero momentum. What can we say about the total kinetic energy of this system? Since the kinetic energy does not take into account the direction of the motion, and since both balls are moving, the kinetic energy of the system will be different from zero and equals to the scalar sum of the kinetic energies of both balls. Conclusion: we have a system with zero momentum, but non-zero kinetic energy. Assume now that we lower the magnitude of the velocity of one of the balls, but keep the direction of motion. The result is that we lower the total kinetic energy of the system, since one of the balls has less kinetic energy than before. When we look to the total momentum of the new system, we observe that the system has gained netto momentum. The momentum of the first ball does not longer neutralize the momentum of the second ball, since the magnitudes of both velocities are not longer equal. Conclusion: the second system has less kinetic energy than the first, but has more momentum. If we go back from system 2 to system 1 we have an example of having more kinetic energy, but less momentum. I hope this answers your question Kjell
We'll assume that the 5,000 joules represents all Kinetic Energy of the car's motion,and there's no potential energy, energy stored in the battery, or gas-in-the -tankenergy included in that number.Kinetic Energy = (1/2) (m) (V2)V2 = 2 KE / mV = sqrt( 2 KE / m ) = sqrt( 10,000 / 400 ) = sqrt(25) = 5 meters per second
500 joules is equal to 368.78 ft-lbf. For example, an object has 500 joules of kinetic energy, when its mass is 10 kg (~22 lbs) and it is traveling at 10 m/s (36 km/h or ~38.2 ft/s). Second example: The muzzle energy of a traveling 9mm bullet is around 500 joules. Third example: An object with mass of 5kg (11 lbs) and which is raised at 10 metres (32.8 ft) has around 500 joules of potential energy. So, 500 joules is quite much. Getting hit with an object which has 500 joules of kinetic energy can be lethal.
1,100 j 110 j
1,100 j 110 j
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Dear Wiki Questioner, To calculate the kinetic energy of an object, we use the following equation: KE=(1/2) m v^2 Where KE is the object's kinetic energy in Joules m is the object's mass in kilograms and v is the object's velocity in meters per second So for your question, we first convert the mass of the bullet into kilograms so we can use it in our equation: 25g (1 kg/ 1000 g) = .025 kg The mass of the bullet is .025 kilograms! Now we plug the numbers into the equation and solve: KE = (1/2) .025 kg (500 m/s)^2 = 3125 kg m^2/s^2 = 3125 J So the kinetic energy of your bullet is 3125 Joules
The equation is for kinetic energy, K, is:K = 1/2mv2Plug in the numbers:K = 0.5 x 9000 kg x 9 m2 s-2So that energy is 40 500 J, or 40.5 kJ.
Let's see. Mass constant. 625000 Joules = 1/2(500 kg)V2 1250000 Joules = 500V2 2500 = V2 50 meters/second = velocity ==================================== The velocity doubles if KE doubles.
62,500j
No, not the way you're trying to describe it. But the kinetic energy of a moving object can appear to have different values, depending on how the observer is moving when he measures it. Example: I'm sitting comfortably, reading a book. I doze off, and the book settles down in my lap. What's the kinetic energy of the book ? To the person sitting next to me, the kinetic energy of the book is zero, because it's not moving. But that person and I are sitting in adjacent seats on a transatlantic flight to Europe, cruising along at about 500 mph when I doze off. The scientist down there in his boat, watching us fly over, would not agree that my book has no kinetic energy.
Kinetic energy is 1/2 mv2 = 500/2 x 32 = 2,250 J (joules). Potential energy is mgh = 500 x 9.81 x 30 = 147,150 J. Total energy at the top of the hill = 2,250 + 147,150 = 149,400 J or 149.4 kJ.
Initial kinetic energy = 1/2 M Vi2 Final kinetic energy = 1/2 M Vf2 Difference = 1/2 M (Vf2 - Vi2) = 2,500(302 - 202) = 2,500(900 - 400) = 2,500 x 500 = 1,250,000 joules
If the question is analyzed by a non-relastivistic approach of the kinetic energy due to the relatively low velocity of the objects, the equation is: K = 0.5mv² where: K: kinetic energy (Joules) m: mass of the object (kg) v: velocity of the object (m/s) The kinetic energy of object 1 at 1000 kg and 1 m/s is: K1 = 0.5(1000kg)(1m/s)² = 500 kg-m/2 = 500 J The kinetic energy of object 2 at 70 kg and 8 m/s is: K2 = 0.5(70kg)(8m/s)² = 2240 J If higher energy is preferable, than select object 2. If lower energy is preferable, than select object1.