This was on my physics test, and I just wrote "Master Shake" as the answer (I got it right), but it makes me wonder: what would happen?
200882
My mother-in-law gets a black eye.
441014
The answer is OH MY GOD MY BRAIN JUST EXPLODED.
379573
haha too bad its on a comp...my sci teacher did this to us last year. i lied. i failed. LOL!
477753
1) all three can happen, but most 'common', or at the most angles, the answer is refracted.\n\n2) b.  The example I usually give is shaking a slinky back and forth.\n\n3) mohorviric discontinuity, or moho for short.  It's about 6 km for ocean crust, and 30 km for continental crust.\n\n4) d for sure!
292251
*******************************************************\n*It's not a paradox to begin w/!!!!  It's perspective.*\n*******************************************************\n\nThe question is "immovable object & irresistable force." It's a perspective thing. They do exist. The problem lies in the fact you're on the outside looking in on the system. \n\nWhat is the irresistable force? There IS only one. Gravity.\n\nWhat is your immovable object? Pick something. Anything.\n\nNow drop it. From your perspective the object falls to the ground. BUT what happens from the object's perspective? The ground jumps up and hits it, while the object remained still. It's your basic car window perspective question.\n\nIf using large enough objects, it can become a "matter shake."
101972
probably explode or though im not sure,it would be fun if it hit someone ,just kidding,he,he
285437
My response is to use the pythagoras theory to answer the question. After all, the question was set for you, so it wouldn't be right to answer it for you, now would it?
146634
This may actually have to do with Newtons laws of physics. This one would seem to be "Every action has an equal and opposite reaction"
159450
The ball would reach the other end and turn back and so forth i.e. it would oscillate. The whole motion is based on the gravitational force. Its a physics problem. check out a physics book chapter on gravitation.
201343
i will just answer your question with!!! what do you think would happen
618877
Fascinating question! The surface of the vibrating object would alternately cause reflected light to red shift and blue shift, but that's about as far as my thinking can go. Would the shifts cancel each other, rendering the object invisible?
471917
Well, in the first place you'd be dead...but let's ignore that.\n\nIt would take a good bit of energy to move the boulder, energy that our muscles coundn't produce. Inertia is the tendency for things to either keep moving or stay still.  So, it fits right in there...the boulder would "want" to stay in the same state, it would take energy to change it...we don't have the energy to move it all that well.
59521
yes try posting question under \n\nHome > Science & Mathematics > Physics \n\nYou will get an answer there.
512825
i wouldnt freak out about that..shit happens..i would honestly think it was a joke or somethin....
474418
As long as neither of the balls starts spinning (or changes the rate at which it's already spinning) after the collision (in which case some of their linear momentum would have been converted to angular momentum), the total linear momentum of the system before and after the collision must be the same.\n\nLet p11 and p21 be the momentum of balls 1 and 2 before the collision; let p12 and p22 be the momenum of balls 1 and 2 after ths collision; and let m be the mass of the balls (which we are told have the same mass).  The momenta are all vector quantities, which I'll write as order pairs of numbers (i.e.,  p = (x,y) = x*i + y*j, where i and j are the unit vectors in the +x and +y directions, respectively).\n\nLet the +x axis be pointed in the east direction, and the +y axis be in the north direction.\n\nThen, writing each \n\np11 = m*(2 m/s, 0 m/s)\np21 = m*(-1.2 m/s, 0 m/s)\np22 = m*(0 m/s, 1.43 m/s)\np12 = m*v12 (v12 is the vector velocity of ball 1 after the collision)\n\nBy conservation of momentum, we have that:\n\np11 + p21 = p12 + p22\n\nm*(2 m/s, 0 m/s) + m*(-1.2 m/s, 0 m/s) = m*v12 + m*(0 m/s, 1.43 m/s)\n\nv12 = ((2 - 1.2 - 0) m/s, (0 + 0 -1.43) m/s)\n\nv12 = (0.8 m/s, -1.43 m/s)\n\nThe speed of ball 1 after the collision is given by:\n\nsqrt((0.8 m/s)^2 + (-1.43 m/s)^2) = 1.639 m/s\n\nThe angle the velocity vector of ball 1 makes with respect to the +x axis after the collision is given by:\n\narctan(-1.43/0.8) = -1.061 radians = -60.776 degrees\n\n(It helps to draw a picture of this to visualize why this relationship holds.)\n\nThe first ball is traveling 60.776 degrees south of east after the collision.
39033
Yes. It will superheat and have a violent reaction when something touches it.
291726
Oh good grief.  If it's anything like what I've just done it'll include velocity, centrepetal forces, magnetic fields, friction, gravity, forces generally, working out the mass of something and the speed at which it will fall, balances with pivot points, static electricity, AC & DC electricity, transformers, vectors and so on and so on.  It's quite hard but it's maths based so if you like maths, you should be ok.  I wish you luck.