Is this question a powerful, answer-repelling force field?
339746
t t   t   t r y  in g  t  an s w e r  f or c e   is t t o s  s t ro n    . . .
856622
Glad to see you're thinking, but this is an easily answered question.  \n\nTake for example a magnet.  Regardless of what is in front of your magnet it will still attract an object to a distance equal to its strength.\n\nex:\n\n.....(|)<<<< <[]<<<<<<<[]          \n\n......^..............^...... ......^\nMagnet......Metal..... ..Metal\n.................Object.... Object\n\nThe magnet is going to exert force on both of the objects without interference from first object.  \n\nBased on your definition of propagate...come to your own conclusion...that is the joy of thinking...we all have our own thoughts.
1160187
someone picking my answer as best (hint hint)
810779
Let the two +Q charges be placed left to right with the distance r separating them.  Let a third charge of unknown charge q be placed a distance x to the right of the first charge.  Now, look at each of the +Q charges and sum the electric force acting on them.  For example, the left hand +Q charge experiences a repellent force from the +Q on the right, pushing it towards the left.  To balance this out, there must be an attractive force pulling it to the right, so the q charge is negative.  However, this is just an observation.  In order to maintain the integrity of the algebra, we should continue to assume that q is positive, but recall the negative result llater when we check our answer..  The proportionality factor for electric field strength can be ignored, because it will cancel out, so you can treat Q/r^2 as the value under consideration.  In this case, we need Q/r^2 + q/x^2 = 0.  However, because the +Q charges are equal in magnitude, I will make the assumption that the distance x = r/2, so Q/r^2 = -4q/r^2 ==> Q = -4q ==> q = -Q/4.  Perform the same force balance on the right hand +Q charge to verify.\n\nIf the initial charges had been unequal, x would have remained a variable, and the complementary distance (r - x) would have been used for the second force balance in order to form a system of equations.
685243
Either the object is not immovable or the force is not irresistible.  Both things can not be true at the same time.
1052765
Consider the graph of effective potential (i.e. that including the angular momentum contribution) vs. radius 'r' for a test particle in orbit around a Schwarzschild hole. i.e. for timelike geodesics.   The slope of this graph at any value of r can be regarded as the force required to keep an object in orbit at that radius.  At r=3M, the slope can be zero (if the angular momentum's right), but for smaller r (i.e. approaching the Schwarzschild radius r=2M from larger values of r), this slope increases rather sharply (as r->0 the r^-3 term dominates).  Now, this is still FINITE, not infinite at r=2M (so the thread posted by the Canadian guy has incorrect info) but it can still be very large...  So, to keep your arm intact, you'd have to keep your whole arm more or less in orbit at a succession of raidi approaching r=2M.  \n\nThis then becomes a medical question: how strong is your arm?  If it's not strong enough, your hand breaks off (or something) and the lost bits freely fall into the hole.  If your arm is "strong enough", however... hmmm.  That's a good question!\n\n...Oh, and tell your ship's captain: "Don't forget to compensate for the decay of our orbit due to the gravitational radiation we're emitting!"
210510
No, because you might have other forces acting on the object such as magnetic force and gravity.
490956
gravity exists.  force fields do not.  \ngravity is an atractive force.  force-fields are described as a repelling force in most science fiction.
884867
is this a trick question ?
296792
Your experimental model is corrupt in terms of current known information about matter and energy in a space-time continuum.  No prediction or calculation can be applied.\nWith two point masses, the only force in the system would be do to gravity.  This is an assumption based upon multiple moving masses in what we would call the normal physical universe.\nIf the distance between their centers (d) was sufficiently small, the points would be touching one another and no force would be exerted on the masses.  If the distance was sufficiently large, then it is possible that the force would be so small that the masses would not move.  Only in a finite range of distances would any gravity possibly exist.\nOne can make no assumptions about G as a gravitational constant, since G assumes that one of the masses is the earth.  One would need additional information about the masses to solve the problem.
151370
is this a trick question?
641452
opposite force being applied
1137815
Got the answer yet? (f11)
730142
Possibly so. Possibly no. I'd say the Heisenberg uncertainty principle comes foremost into play here. Maybe also Schrödinger's cat. Is it dead or still alive? That would make a significant difference, you know. These questions can't just be taken so lightly. And how could you even consider Y!A as a forum to resolve such complex issues as this? More to the point, even assuming Planck scale dynamics can be fully understood with the mathematical tools we have at hand today, where exactly does that get us? We still don't know what topping to put on our pizza this weekend. Or whether a third eye would make a good chick magnet. Or even whether that darn cat is still alive or dead.... 'tis an enduring mystery. And we wouldn't want to take all the mysterious challenge out of life, now would we?....Meow.
393415
o this hap pend to me. just put it in the closest 1. done worry, it ll still get answered.
534010
Practically anything, as this event can not occur in our universe.  The existence of an irresistible force precludes the existence of an impenetrable barrier (or immovable object, as this question is normally phrased).  Conversely, if I know of the existence of an impenetrable barrier, I can state definitely that there is no such thing as an irresistible force.  The two items cannot exist in the same universe at the same time.
224380
I´m rather sure you could be rather disappointed by an answer as there would be several clear "No" contained in it.\nAre you prepared to receive such an answer ?