If a stone is thrown up in the air is its motional energy loss proportional to itself?
915353
To answer your question simply and directly; yes. An object's motional energy loss is proportional to itself; it's a result of the forces of gravity acting directly on objects based on their respective masses and distances.\n\nAs the stone goes up in the air, it loses kinetic energy due to the pull of gravity -- it decelerates until it stops moving upward, then begins moving downward. \n\nThe second you let go of the rock, its acceleration is constant; its kinetic energy decreases as it slows down and then increases again as it falls to earth.\n\nThe kinetic energy of the rock turns into potential energy as it gets higher; kinetic energy being equal to one-half the rock's mass times its velocity squared, and potential energy being the rock's mass times the rock's height above ground times the constant acceleration due to gravity.\n\nIn a vacuum, all the rock's kinetic and potential energy would add up to the same value at any point in its rise and fall; if you threw it hard enough to give it, say, a hundred kilojoules of kinetic energy, when it reached the top of its trajectory and began to fall down, it would have zero kinetic energy but a hundred kilojoules of potential energy. On earth, it wouldn't work like that; some small amount of energy would be lost due to wind resistance, etc.
420366
Yes, that is Newton's first law of motion.  Newton's discovery may seem like a no-brainer to you or I, but at that time many people believed that an object fell to the ground because it was attracted to the earth.  His research opened up many new theories and formulas in the field of physics like work, energy, power, momemtum, and acceleration.\n\n#2 - "The acceleration of an object as produced by a net force is directly proportional to the magnitude of the net force, in the same direction as the net force, and inversely proportional to the mass of the object".  Basically force is equal to mass times acceleration.  F=ma\n\n#3 - "For every action, there is an equal and opposite reaction."  This means when you are leaning against a wall, you're applying 60kg of force against the wall and the wall is applying 60kg of force against you.  Action = Reaction.  When this equation doesn't balance out, something will usually break.
190711
Well my friend, two theories come into play here.\n1) Things in motion tend to stay in motion.\n2) Energy cannot be created or destroyed\nTherefore, one or both of the objects (depending of the structure and integrity of the objects) will suffer severe or absolute damage as the transfer of kenetic energy to solid matter occurs. (Depending on the speed and velocity)Whether you think so or not, one obect will move or one object will stop.
778473
Since the stone did not gain any potential energy, the energy is stored in the slingshot bands\n=F*d\n=30*.2\n=6\n\nAssume all of the energy gets transferred to the stone's kinetic energy in 20 cm\n\n1/2 *m*v^2\nset equal to potential and solve for v\n\n=15.5 m/s\n\nj
914874
yes if it is spinning on its own axis as it moves.  This is called Spin Angular Momentum
329300
It doesn't. When work is done a small amount of the system's energy is converted to entropy (heat) and is lost to the system (usually through friction).
791190
Only if E=MC^2\n\n... it follows that if there is a change in the rest-energy of a body, there must be a corresponding change in its rest-mass. For example, if a body is heated, and thereby absorbs a small amount of energy ΔE (as measured in the frame of reference in which the body is at rest), its rest-mass will increase by a very small amount equal to ΔE/c2. This increase is tiny because of the high numerical value of the speed of light. Indeed, for mid-sized objects, such an increase in mass would be too small to measure with even the most accurate balance. For example, if a 1 kg block of gold is heated so that its temperature increases by 10 °C, then, ceteris paribus, its mass would increase by as much as 1.4 × 10-14 kg; a cube of gold of this additional tiny mass would have sides smaller than one one-thousandths of a millimeter. Similarly, if a body emits an amount of energy ΔE, say in the form of light or heat, its rest-mass will decrease by a tiny amount ΔE/c2. In both cases, the important and novel claim made by SR is that the inertial mass of a body can change depending on whether it absorbs or emits energy. This is true, of course, not just for isolated mid-sized objects but for any physical system.
889920
Yes, if it's converted to energy.  E=mc^2
313762
youre holding a stone on a top of a hill (potential energy here) ---> you drop the stone off the hill (here kinetic energy) ---> the stone is on the ground (here again potential energy)
891480
Calculate the k value for the spring.  The stone exerts a force of 8*9.8 = 78.4 N and a compression of 9 cm is\n78.4/.09 = 871 N/m. (Keep it in the MKS system ☺)\n\nWhen the spring is pushed down another 30 cm its total deflection is 39 cm and it's total potential energy is \nkx²/2 = 871(.39)²/2 = 132.5 J\n\nThe stone acquires 132.5 J of gravitational potential energy when released.\n\nSince gravitational potential is mgh, then\n8*9.8*h = 132.5 or h = 1.69meters\n\n\nDoug
818010
Angular momentum is conserved,\n\nAngular momentum is (inertia * angular velocity)\n\nInertia for a disk = .5*mass*radius*radius = 0.5 * 5.6 * .2 * .2\n\nThe initial angular momentum is then: 0.5 * 5.6 *.2 * .2 * 1.8 (note that is in rev/s)\n\nThe inertia for the clay is: 0.5 * 3.1 * .04 * .04\n\nThe final angular momentum of disk and clay is then: 0.5 * ((3.1 * .04 * .04) + (5.6 * .2 * .2)) * f   where f is the answer you are looking for.\n\nInitial angular momentum = final angular momentum:\n0.5 * 5.6 *.2 * .2 * 1.8 = 0.5 * ((3.1 * .04 * .04) + (5.6 * .2 * .2)) * f or\n\nf = (5.6 *.2 * .2 * 1.8)/((3.1 * .04 * .04) + (5.6 * .2 * .2)) in rev/sec
595852
It would end up as heat. At least that is the usual place you say all the energy went when you can't otherwise account for it. Like where does the energy go when you drop a bean bag on the ground? Yell out loud? Turn on a light? slow your fall with a parachute? And on and on and on.
542734
Meaning moving continually with no additional force needed to overcome frictional or other energy losses.\n\nNo.  It violates the laws of thermodynamics for some process to occur without an energy loss.  If you don't add energy to a system, eventually it will run out of energy.
791539
Wow. Your classmate just asked this question. But here's your answer anyways...\n\nThere are a few things to keep in mind here: all of the energy in the initial system is rotational, and the initial energy is equal to the energy lost to friction. Also note that the energy is lost to friction at a constant rate.\n\n(a) The initial energy of the system is:\n\nE = (1/2) * I * w²\n\nWhere I is the moment of inertia and w is the angular velocity. The power transferred from the system due to friction is equal to the time derivative of the energy loss:\n\nP = dE/dt = I * w * a\n\nwhere a is the angular decceleration.\n\nSince the power is transferred at a constant rate and removes all of the energy from the system, we can take this as:\n\nP = -E/t (negative sign denotes power loss)\n\nCombining the power equations yields:\n\nI * w * a = -E/t\n\nSolving for a:\n\na = -E / (I * w * t)\n\nFinally, substitute the expression for E to obtain:\n\na = -I * w² / (I * w * t)\n\na = -w/(t)\n\nSubstituting the values yields:\n\na = -(150 rev/min * 2*pi rad/rev * 1/60 min/sec)/(2.2 hr * 3600 sec/hr)\n\na = -(15.71 rad/s)/(7920 s)\n\na = -.001983 rad/s²\n\n(b) The rotational behavior of the fly wheel can be modeled as simply:\n\nR(t) = -1/2*a*t² + w*t + Ri\n\nwhere R(t) is measured in radians and Ri is the initial angular displacement. Assuming that Ri = 0 (we'll measure the number of rotations from its initial position), then the flywheel will revolve:\n\nR(t) = -1/2*a*t² + w*t (radians)\n\nR(t) = (-1/2*a*t² + w*t)/(2 * pi) (revolutions)\n\nSubtituting the values:\n\nR(2.2 hr) = R(7920 s) = (-.5*.001983*(7920 s)² + (15.71 rad/s)(7920 s))/(2*pi)\n\nR = 9900 revolutions\n\n(c) The tangential component of the linear acceleration of a point on a rotating body is given as:\n\nA-tangent = a * r\n\nwhere A = linear acceleration. The angular acceleration is constant, so for all w (including w=75 rev/min), the linear acceleration will also be constant. Thus:\n\nA-tangent = (-.001983 rad/s²)(.5 m)\n\nA-tangent = -9.915 x 10^(-4) m/s²\n\nA-tangent = -0.9915 mm/s²\n\n(d) The magnitude of the net linear acceleration on the particle is composed of a tangential and a normal component. The equation is of the form:\n\nAnet = [ (A-tangent)² + (A-normal)² ]^(1/2)\n\nOr:\n\nAnet = [ (a*r)² + (v-tangent²/r)² ]^(1/2)\n\nNote that the tangential velocity is given by:\n\nv-tangent = w * r\n\nThus, the magnitude of net linear acceleration is:\n\nAnet = [ (a * r)² + (w² * r)² ]^(1/2)\n\nSubstituting the known values yields (note that for this case, the angular velocity of 75 rev/min is equal to 7.854 rad/s):\n\nAnet = [ (-.001983 * .5)² + (7.85² * .5)² ]^(1/2)\n\nAnet = 30.81 m/s²\n\nFINAL ANSWERS:\n\n(a) a = -.001983 rad/s²\n(b) R = 9900 revolutions\n(c) A-tangent = -0.9915 mm/s²\n(d) Anet = 30.81 m/s²
783111
The stone has two componenst of velocity and displacement.  The x component of velocity stays the same, and the x displacement is x-velocity times time.  The y-velocity is g*t, acceleration times time, and the y displacement is ∫atdt = .5*a*t^2.  The total vector velocity and displacement is the vector sum of the x and y components √[x^2+y^2]
209347
"...Small stones to a few millimeters in size are slowed down so much by the atmosphere that they end up drifting slowly to the ground. \n\nPieces of space rock of a few millimeters to a few centimeters burn up completely in the atmosphere of the Earth before they can reach the ground, and can be seen at night as falling stars. If it is sufficiently dark and clear outside, then you can see a few falling stars every night. \n\nIf the rock is bigger, then part of it can survive the trip through the atmosphere and reach the ground as a meteorite. If the stone weighs less than a few tons, then it doesn't fall very fast anymore in the end, and it makes a small dent in the ground.   "\n\nhttp://www.phys.uu.nl/~strous/AA/en/antwoorden/meteorieten.html
505249
no.its the force produced due to the movement of other particles(electrons).so..