What happens to a moving force as a function of time and distance?
93773
well force is measured in kg*m/s^2 so, force will cause a function of time and distance to increase exponentially over the interval of the force applied.\n\nFor example, an object dropped from 100 meters will go 5 meters in the first second, 20 meters in the 2nd second, and  25 meters after 3 seconds, etc.
269311
assuming that force is applied in the same direction that displacement....\nwork = force * displacement * cos(angle between force and displacement)\nwork = (10N)*(5.0m)*cos(0)\nwork = 50 Joule
826484
Bernoulli Effects\n\nSee how to demonstrate it here --> http://www.physics.umn.edu/outreach/pforce/Bernoulli.html
680834
A force is a push or pull upon an object resulting from the object's interaction with another object.\n\nThe formula is F=ma, where F= force, m=mass, a=acceleration
709931
kaksi_guy: I think charlesellis753 is correct.  If we interpret the question as a fill in the blank, mass appears to be the best answer.\n\nIf the ______ of an object doubles, then the force required to accelerate it at the same rate will double.\n\nThat sentence is true is the blank is filled with 'mass'.
643953
depends on the direction of the force
1118162
If we imagine this all happening on in 1 dimension, a line, (i.e. no up/down or side to side movement possible), then the only way something can travel is toward or away from the observer.\nEven if this is not the exact case as in your example, this effect is easily done by merely turning one's head in order to align one's reference frame to the direction of the car's motion and thus would have the same effect.\n\nIn this case, our observer sitting at some location on the line and in his reference frame, everything else is moving either toward him or away from around him.\nIn his reference frame, he sees a car heading Straight towards him at relativistic speeds in the opposite direction he is facing.\n\nA length contraction occurs only in the direction of motion as observed by the observer.  In other words, if the car is moving toward/away from the observer, the car's length shortens, not its width or height.\nThe length contraction is in the same direction as the relativistic motion.  Since we defined this to be a single dimensional system with only one direction of travel possible, the approaching car would not get thinner or shorter (less tall), but rather, its length would contract.\n\nThe effects of the car appearing wider and taller as it got closer and closer would still apply as normal since the car is not traveling in either of those directions.\n\n\nIn your own frame of reference, taking a measurement and finding that the car has experienced a length contracting still makes that measurement every bit as real and valid as a measurement taken from the other car's frame of reference (v = 0) where there is no length contraction.
846277
It decreases to 1/4 of the original force.  This is because the force of gravity is inversely proportional to the distance between the two bodies.  Thus, increasing distance by a factor of n reduces force to 1/n^2.
888452
f=ma\n\na=change in velocity/change in time = (5.53-2.25)/1 \n\nm= 57\n\nf = 57 (5.53-2.25)\n\npunch the numbers in to your calculator, and there it is!
771568
Basically you have the right idea. Use momentum (mass x velocity) for the motion and the sudden change in speed and/or direction (acceleration) and you get force x time = kinetic energy trying to rip you apart or squash you like a bug.
1058239
Yeah, force is just energy applied on a particle. Force per time is power.
792607
0 km/hr^2 .... because your speed doesn't change with time due to constant as you mentioned
103569
Displacement.  If you mean its change in position *divided* by the time it took, that would be its average velocity.
859502
What you are likely looking is W=F*D \n\nWork = A Force * A Distance\n\nCheck out wikipedia for additional help:\nhttp://en.wikipedia.org/wiki/Mechanical_work
705496
assume that just for 1 second gravity goes off and comes back, what will happen??? is it possible??? no funky answers please. genuine replies are appreciated.
873423
velocity is given by :\nv(t) = -1 + e^(1-t)\n\nacceleration is given by :\na(t) = - e^(1-t)\n\nSo :\na) \nv(3) = e^(-2) -1\na(3) = - e^(-2)\n\nb) at time t = 3, acceleration is negative. The speed of the particle is decreasing.\n\nc) whenever the particle changes direction, it is velocity is momentarily zero.\n\n=> 0 = -1 + e^(1-t)\n=> 1 - t = 0\n=> t = 1\n\nNote that velocity is positive for all times < 1, and is constantly decreasing. For all times > 1, velocity is negative. Thus, the particle changes direction at time t = 1.\n\nd) at time t = 0, \nx(t) = - e\n\nat time t = 1,\nx(t) = -1 - 1 = -2\n\ntotal distance travelled thus far = (e - 2)\n\nat time t = 3,\nx(t) = - 3 - e^(-2)\n\ntotal distance by the particle over the interval 1<=t<=3 is :\n1 + e^(-2)\n \ntotal distance travelled is :\n(e - 2) + (1 + e^(-2))\n= e - 1 + e^(-2)
918805
Time And Relative Dimension In Space