Momentum: shell fragments?
955270
I won't do it for you, but here is what you need to do:\n\nFind the height at which the explosion occurs = h.\n\nFind the horizontal component of momentum at the instant before the explosion (which is actually the entire momentum, since there is no velocity in the vertical direction at that instant!) = Mtot x Vhoriz =m1 (Mtot is the entire shell).......(1)\n\nNow, the total momentum the instant after the explosion is:\n\n(M1 x 0) + (M2 x V2) .......(2)(since the large part of the shell stops travelling horizontally, and is already not travelling vertically) Where V2 is the horizontal velocity of the small part of the shell.\n\nSince momentum is conserved, these two quantities are equal, so:\n\nMtot x Vhoriz = M2 x V2, so you can calculate V2, the new horizontal velocity of the small part of the shell.\n\nNow, simply calculate the time taken for the shell to travel vertically to the ground using s = 1/2at^2, and from V2, you can calculate the horizontal distance travelleed by the small part of the shell...presto!
55028
I can, the problem is that it's a bit long (but not that hard). the demo can be found in every physics book (e.g. Physics of solid state, C. Kittel)
240384
http://hyperphysics.phy-astr.gsu.edu/hbase/hframe.html
76628
By "shells," do you mean the energy levels of the electron orbitals?  If that's what you mean, all you do is look on the period table and count how many rows from the top the element is.  So H and He each have one electron shell, C, N, O, F, etc. all have two, Si, B, Na, Cl, etc. all have three, and so on.  Sorry about the scattered examples (I was listing them by memory), but hopefully you get the idea.
809913
1. Momentum=Massxvelocity=3600\n2.3600=.2KgxVelocity=18000\nThats all i know
458035
my jigsaw has a framulator. Do your own homework.\nMIKE
280156
solid\nliquid\ngas\nplasma
804774
A good source is Wikipedia, the free, online encyclopedia:\nhttp://en.wikipedia.org/wiki/Momentum\n\nHope this helps!
606473
Quite frankly, we don't know. Like listed above, there is a variety of smaller particles that we have been able to smash an atom into. However, each time we build a bigger smasher, we get new particles. The thought is that there is a fundamental particle that is no longer divisible, but I don't think science has commited to it yet.
190933
ARES\n\nhttp://aresgalaxy.sourceforge.net/
710256
You got the CG just right.  Your method might work with enough concentration., but I would look at making the angular momentum come to zero:\nL = Iω = (2.7*.325²  + 1.3*.675²) * 26rpm*2π/60 = 2.389 kg m²/s\n\nTo stop the movement the angular impulse must equal the - of the angular momentum:\n\nT * t = L→ T = L/t = 2.389/5.8 = .4118 Nm (ccw)
903139
A bullet of course.  A gun uses a small amount of kinetic energy for the trigger hammer to strike the expolsive cap in the shell but thereafter the explosive force is sent into the bullet, with and equal opposite reaction energy dissipated by providing recoil which the gun uses (if automatic) to load the next shell.\n\nI assume the trick in the question is to ignore the equal and opposite reaction rule but not all of the force thus generated is in the form of kinetic energy which relates to bodies in motion.
970362
Yes.  All p shells will have a total of 3 atomic orbitals.\n\nhttp://en.wikipedia.org/wiki/Electron_configuration#Worked_example
458509
Conserve both momentum and total energy (i.e., potential energy plus kinetic energy).\n\nMy coordinate system will have the pendulum lifted to the left and the block on the right moving left, so that the initial momentum is - M2 Vo2.\n\nIt will also have the block at height 0, so the initial potential is M1 g h where h is the initial height of the pendulum, which looks like L(1 - sin(10 degrees)).\n\nThe initial kinetic energy is of course (1/2) M2 (Vo2)^2.\n\nSo ...\n\ninitial momentum: - M2 Vo2\ninitial energy: M1 g h + (1/2) M2 (Vo2)^2\n\nAfter the collision the potential energy has decreased to zero, and if w is the combined velocity of the two masses then ...\n\nfinal momentum: (M1 + M2)w\nfinal energy: (1/2)(M1 + M2)w^2 + [the energy "loss" from the slight heating and deformation of the two masses sticking].
1069017
Hmm.... I can't seem to find one on the sites I've checked, but based on inference, I would guess it would be something like Argmt. or Agmt. These aren't official, mind you, but with a little hard work they could be ;) .
727943
Isn't it SDRadiant that you're supposed to use for W:ET? (GTKRadiant will kinda work if you try hard enough, but I think SD is a fork that's specific to Wolfenstein.)
1022160
I hope this will help. \nFor dynamics see reference 2