A 28 g rifle bullet traveling 230 m/s buries itself in a 3.6 kg pendulum hangingon a 2.8 m long string, which?
923444
Short Answer: \n\nvert comp = .161 meters\nhori comp = .936 meters                        \n\nLong Answer:\n\nMb=mass of bullet\nMp= mass of pendulum\nv=velocity of bullet\ng=9.81 m/s/s\nh=vertical displacement\n\n\nBy conversation of momentum:\n\nmomentum of pendulum with bullet=momentum of flying bullet\n\nsqrt(2*g*h)(Mb+Mp)=Mb*v= (28/1000)*230=6.44 \n\nsolving that you can find h=0.161 meters\n\n\nBy the distance formula:\n\nsqrt(2.8^2+0^2)=sqrt( (2.8-0.161)^2 + (horizontal position)^2)=2.8\n\nsolving that you can find horizontal position = 0.936 m
752440
The force applied to the beast (by the bullet) would have to be calculated by the angle/arc of it's descent, this new force differential may cause a new vector.  ie  If it was falling straight down and you fired straight up...no change.....if it were coming in at 45^ angle down, the bullets force will slow it slightly, and it's arc down will have increased due to the continuing force of gravity...with less momentum.  That is...also only if the head stays attached....if not then...the two bodies must be calculaated by the individual resistance/force of the bullet....    In the end...take one step to the right...after your shot...and watch out for falling bits and pieces.
234424
Depends on the gun and the bullet.  Different kinds of bullets have aerodynamic designs and powder charges that vary their speed when leaving the barrel.  For most handguns (say a Colt .38), that speed is around 300 meters per second.  On other kinds of wapons (say a Desert Eagle Cal. .50) The speed can go to around 400 meters per second with a heavier projectile, and so on.
1066435
it is seven letters long.\n"a string"
30590
nothing's wrong with the above and your conversion is correct.\n\ndoesn't seem like very much eh?  the reason is b/c of what i explained in the other question.. if you plot a graph of the force versus time, it will look like this:\n\n_____|___|___|___|___|___|___|___|_____\n\nthose sharp vertical lines represent the bullet impacts, and the rest of the time the force is zero.  if you average over this time span, the number turns out to be very low.  but the height of those peaks is very high because so much momentum is exchanged in a very short time!  so force is kind of misleading.  in physics the more meaningful way to talk about collisions is to talk about the momentum exchanged in a collision, which is called an "impulse" because it happens over such a short time.\n\nput another way, the bullet has a momentum of m * v, or .003 kg * 740 m/s = 2.22 kg*m/s.  think about what that means.  it's like getting whacked in the chest with a 2.22 kilogram (around 4.5 lbs) object moving at 1 meter per second - pretty heavy and pretty fast!  \n\nand that's just one bullet.  think about 8 of them hitting you every second.  furthermore, 3 grams seems like pretty low estimate for the mass of a bullet.   that's something like a .22 or a .25.  consider that a .44 magnum bullet, which is the sort of gun that actually can knock you off your feet, has a mass of 250 "grains" which is closer to 16 grams.  that's more like getting hit with a 10-kg object.  that's like someone winding up and throwing a 22-lb weight at you!\n\nso there you go.  the averaging and the short time interval for the impulse is what's misleading you.
862116
KE = 0.5*m*(v)^2 = 0.5*0.023*(848)^2 = 8269.696 J\n\nThis must = work to accelerate the bullet\n\nW = F*d = 8269.696\n\nF= 8269.696/0.803 = 10298.5 N
914639
T = mg.\n\nThe length of the cable is irrelevant in this example. The beam is hanging off the cable.\n\nThe tension in the cable therefore is m*g, which is 2450 N.
802288
the impuls (m*v) is conserved, so the impulse in one direction;\n0.150*20*sin(30deg)\nEquals the impulse in the other direction\n50*v_recoil\nthe recoil velocity is thus 0.03 m/s
109636
How long is a piece of string! comes to mind
713008
Bound's hubby here:\n\nYou have provided insufficient information. The amount of drop at 200 yards depends upon several factors beyond the bullet weight: the distance the rifle is sighted for; what type of bullet (service full metal jacket or factory hunting bullet) being fired; and the muzzle velocity of the cartridge fired!
864791
A.) 3.02891 seconds\nB.) 290.1 m/s\n\nTrust me... =)
387792
Literally a bullet with a hollow tip. They're able to pierce armor.
71702
If the only factor involved in this would be gravity, the bullet would come back down at the same speed as it went up. But since there are other factors (friction, air, etc), the bullet would come down with less speed than it went up.\n\nThe same would happen with the place where it was fired, it probably wouldnt land on the same spot, since other factors have an effect on the bullet trayectory.\n\nchech the link from digg and wait for the episode of Mythbusters where they solve this myth ...
451765
A shell.\n\nFrom a tracer bullet.\n\nWhich fell from a Royal Air Force Spitfire in the Second World War in a dogfight with a German aircraft.\n\nI dug it up in my garden sixty years after it had been fired by some brave airman in defence of his country.
916585
Conservation of momentum so mv is constant.\n\nTotal final mass = 5333+14.6 = 5347.6g\n\nTotal final momentum = 5347 x 1.7 gm/s = 9090.92 gm/s\n\nBullet's initial momentum = 9090 gm/s\n\nMomentum/mass = velocity\n\n9090.92/14.6 = 622.666m/s
867152
The period equation for a simple pendulum is:\nT=2pi*(I/mgL)^0.5\nI =rotational inertia\nm=mass\ng= gravity\nL=length\n\nBecause the rock attached to the string is small then I=mL²\ninserting this into the period equation above you get:\nT=2pi*(L/g)\n\nRearranging to solve for L:\nL= (g*T²)/(4*pi²)\nNow plug in the given values:\n(9.8m/s²*9.22²s²)/(4*3.14²) = 21.1m\n\nThis will only be correct if the rock is just above the floor of the tower.
861607
The frequency, f, of a vibrating string of m kg/m and subject to a tension of T Newtons is given by the relation\n\nf=sqrt(T/m)/2/L\n\nHence the frequency is inversely proportional to the length of the string, assuming the tension remains unchanged.\n\nThus the required length of the string is\n40*474/624=30.385 cm.