how does the kinetic energy of a 750kg car compare with the kinetic energy of a 1500kg  car travelling at thes
370716
well you didnt finish the question, but if u were gonna say the same speed\n\nit would be 50%
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!
454337
750g = 0.75 kg\nDelta T = 100-18 = 82°C (or K)\n\nEnergy required = 0.75 x 82 x 4200 = 258300 Joules\n\n258.3 kJ\n(also 0.07175 kWhr, or 244.8 BTU)
852090
Energy can go from place A to place B (think about sound and light)\nEnergy can go from object A to object B (think about a collision)\n\nTh
346376
Inertia.\n\nA body at rest tends to stay at rest.  It needs energy to get it moving.  Once moving in a vacuum, it will continue on that course at that speed until something acts on it to change it's speed and direction.  This might be a planet or star's gravity, or a rocket,  or some such.\n\nAccelerating hard to a given speed takes more energy than accelerating slow.  It would take more energy to get a rocket going 1000 mph in 30 seconds than it would to get it going that fast in 30 hours.  The same principle governs your car, and is the reason that you'll get horrible gas mileage if you take off like a jackrabbit to get to 60 mph, vs. accelerating slowly and gently.
854225
The extent of damage sustained is directly proportional to the amount of energy consumed. If in case a) the stationary car is allowed to move after the collision, then case a) is where the original car sustains the LEAST damage. \n\nDue to conservation of momentum, in case c) the velocity of the two vehicles will be zero after the collision. So vehicle 1 will have lost all of its energy and the energy will be converted into deformation (damage).\n\nIn case b) the velocity of the car will also be zero after the collision, so all of its enery will be converted into damage also.\n\nSo, in case a) the car sustains the least damage, in cases b) and c) the car sustains the same amount of damage.\n\nI am only looking at the damage to ONE vehicle in each case, not the damage to all vehicles.
1179191
Work done by the force 2.34 N is, ‘force x distance through which the force is moved’.\n\nTherefore, it is = 2.34 N x 1.41m = 3.2994 joule.\n\nWork done by the frictional force 0.55 N is, ‘force x distance through which the force is moved’.\n\nTherefore, it is = 0.55 N x 1.41m = 0.7755 joule.\n\nThe last question is vague. \n\nThe total work is the work done by the force 2.34 N and it is equal to 3.2994 joule. \n\nOut of this total work a part (3.2994 -0.7755 = 2.5239) joule is used to accelerate the book and 0.7755 joule is dissipated as heat energy.
109202
Fnet= MA so you acceleration is 36km/h, so convert that to m/s, which is 10 m/s. To get acceleration, we know that the car started out at 18km/s, which is 5m/s. So the acceleartion = Change of velocity/Change of time. So that comes out 1 m/s/s. So there is 1000 Newtons on the car. The other guy's answer is wrong. I'm an engineering major. The standard Units in science is MKS, meters kilograms and seconds
839257
Potential Energy:  A car is at the top of a steep hill, in neutral, wheels straight,  with the parking brake on.  Release the brake.  The car will roll downhill, converting potential energy into kinetic energy.\n\nKinetic Energy:  The car continues rolling onto a flat stretch of road, which happens to be covered with grapes.  As it rolls over the grapes the car squishes them and slows down.  It's kinetic energy is being transformed into grape juice and heat.
913200
Both their accelerations will be the same and equal to the acceleration due to gravity, g=9.81 m/s/s...This is coz the only force acting on them is the force due to gravity
939941
Yes you're on the right track, you know the momentum of the car along with the energy, both are conserved.  \n\nForce=Mass*Acceleration=Coefficient of Friction*Mass*Gravity\n\nYou can then solve for Acceleration=Coefficient of Friction*Gravity\nVelocity=Acceleration*t\nThen its simply a matter of filling t into this equation:\nDistance=1/2*Acceleration*t^2+Velocity*t
1054432
I don't remember very well I but I red somwhere about a F1 car beating a F16 fighter on a 2km distance.\n\nLet's take an example: \nRoughly a 747 weight about 400tons and has 4 engines of 25tones of thrust. This would lead to a maximum acceleration of 0.25Gs at take off. I'm sure a fast car can do better than that
613443
Niuchemist is mostly right and deserves the vote, but there is one correction I would make-heavier molecules don't necessarily have less "kinetic energy", they move more slowly-it is the speed of movement that is the critical thing, not the kinetic energy, which is a product of speed *and* mass.
123977
Are you asking for the time it takes the car to accelerate 13.89 m/s....well then the answer is in the question.\n\nPerhaps you are asking for the distance  traveled while accelerating?\nchange in distance = 1/2at^2 + v_i*t\nwhere a is the constant acceleration, t is the time, and v_i is the initial velocity.\nthe car's acceleration = change in velocity / change in time
573522
kinetic energy is 1/2 mv² so it depends on mass and velocity
778096
The height of the ramp give potential enery in the following format\n\nE = mgh      where E is potential energy, m is the mass of the car in Kgs, g is the gravitational acceleration of the planet you're on (in meters per second per second (9.8 on earth)) and h is the height of the ramp.\n\nThe angle or slope of the ramp will not affect the potential energy, but will have an effect on the rate of change of potential energy.
981312
310 times faster\n\nif A is going 10mph, and B is going 5mph, the 10/5=2 or twice as fast. so 9.3X10^10 / 3.0X10^8 = 310 times faster.\n\nbut nothing can't go faster than 3X10^8m/s!!!! The normal speed of light means it's going though a vacum (nothing), so if it's going though a gas, it will slow down!! Not speed up! Since the energy will hit atoms and slow down. \n\nIf an object is going 3E8 (short way of 3X10^8m/s) releative to the ground and an object passes it at 3E8, then the object passing the first objects speed was 3E8 releative to the ground... so both were going the same speed, even if some was twice as fast as it!