an object's change in position over time when compared with a reference point?
103569
Displacement.  If you mean its change in position *divided* by the time it took, that would be its average velocity.
314759
Yes, it would be more accurate, but the calculations are much harder. Whether it is worth it depends on what you have to do with the information. One thing that happens is 'frame dragging', where a rotating mass drags along space and time with the rotation. We now have a probe in orbit around the earth to measure exactly this phenomenon.
1118534
No.  The only way of confirming its accuracy is to date objects that we already know how old they are.  These fall into the scope of the last 5000 years, a period which they already say is less accurate than when measuring older objects.  Scientists always seem to be arguing about the age of things regardless of what the carbon dating says.
733431
Because there is no absolute reference frame for motion, you can only define uniform motion with respect to a specified coordinate frame.\n\nFor example, a ball falling in a vacuum is not in uniform motion with respect to an earth-fixed reference frame; the ball accelerates.\n\nWhen the forces on an obejct sum to zero, that object is in uniform motion with respect to a reference frame that is non-accelerated. That means that if you compare gravity acceleration between the a frame fixed to the object, and the frame you are wondering about, if the gravity is the same, the frames are not accelerated with respect to each other.\n\nOthers have suggested that uniform motion requires absence of forces. This is not quite true - it requires absence of *net* forces. For example, a car traveling a constant velocity on a highway is not in a force-free environment: you can identify forces of air friction, propulsion from the wheels, weight, and support from the pavement. However it is considered to be in uniform motion with respect to a reference frame fixed to the earth, because all of those forces sum to zero, and because the apparent gravity in a frame fixed to the car is the same as in a frame fixed to the earth.
783229
I left school a while ago, but isn't it (0,2).  2 on the y axis.
1030184
basicly, it tells you that you can´t get the position of an element without changing its velocity, and vice versa, you can´t know the velocity of an element without changing its position.\nThis applies to a variety of parameters couple.\n\nOne of the consequences in quantum mechanics is that further predictions of an element motions can result in having two positions  for one time. \nMeaning that the element is indeed in two places at the same time until you try to detect it in one of those places.
837255
its the point which you are rotating the object around
1173332
Regard Schrodinger's equation (Hy=C.dy/dt) H is the energy operator and you will have the answer it's the ability to change, in fact its the change it self, even if you go around and around again. Even if the object are instantly changing and in average aspect does not change.
451086
Exact position must be listed in terms of latitude and longitude degress.
53850
If you use a reference then the equation result can be updated correctly when the referenced value changes or the equation is moved to a new cell.
957617
2d point of no return
849322
I hope I can explain this....when you are aligning with three datums, your first datum (datum A) will contact your object at three points.  The second (datum b) at two and the third (datum c) at one.  If you rearrange the order of constraint on your object, you might actually move it.  It is based on degrees of freedom.\n\nAny object has six degrees of freedom....movement in three directions and rotation in three directions.  the first datums stops rotation in two direction and the second will stop rotation in the third....and it takes all three datums to stop translation.  So you can see that if you attach to the wrong datum first, you may be forcing the object to rotate or translate in a direction you don't really want it to....
796769
The point P moves in such a way that its Cartesian coordinates with respect to an origin O at time t is given by x= e^(-t) , y= 2te^(-t). the distance OP is denoted by r and the angle between OP and the x-axis by θ.\n(a)Using an analytical method, find the value of t when r^2 is stationary. Determine the nature of the stationary point.\n\nr² = Δx² + Δy²\n= (e^(-t) - 0)² + (2te^(-t) - 0)²\n= e^(-2t) + 4t²e^(-2t)\n= e^(-2t)(1 + 4t²)\nd(r²)/dt = -2e^(-2t)(1 + 4t²) +8te^(-2t)\n= 2e^(-2t)(1 - 4t + 4t²)\n= 0 for stationary points\nSo 1 - 4t + 4t² = 0 as e^(-2t) > 0 for all t\nie (1 - 2t)² = 0, ie t = ½, ½\nWhen t <½, d(r²)/dt > 0 and when t >½, d(r²)/dt > 0\nThus when t = ½ there is a stationary inflexion\n\n(b) Find, in terms of t, the rate of change of θ with respect to\n\ntanθ = Δy/Δx\n= 2te^(-t)/e^(-t)\n=2t\n\nsec²θ * dθ/dt = 2\nie\n\ndθ/dt = 2/sec²θ\n= 2/(1 + tan²θ)\n= 2/(1 + 4t²)\n\nCan you show the non analytical method for part (a) too!\n\nx = e^(-t) → -t = lnx ie t = -lnx\n\ny = 2te^(-t) \n= -2xlnx\n\nx: .. 0 ...... 0.25 ... 0.5 .... 0.75 . 1.0 .. 1.25 .. 1.5\ny: undef . -0.15 . -0.36 . -0.22 ... 0 .... 0.28 . 0.61\n... (→0)\n\nLim x → 0 (x lnx) = 0 and OP increases with x \n(limiting min value of OP = 0)\n\nMin value of xlnx occurs when lnx + 1 = 0 ie x = -1/e and at this point OP² will show its inflexion as it cannot be a minimum and nor can it be a maximum so all it can do is slow its rate of growth.
315076
no, it influences but not completely.  it stays to a certain extent but it still can be changed through time
731189
Yes. Shifting the binary point one position right doubles both the top and bottom.
44499
Try www.objectdesktop.com it's great !!!
186910
Each line is a change in elevation. There should be a key that shows the value of each line.