what direction does a bullet spin when it's shot out of a gun?
969234
Definitely one or the other :-)\n\nMost rifles have a clockwise twist. There are, however, osme barrels that utilize a counterclockwise twist. Why, I'm not sure - it must be the preference of the barrel manufacturer.\n\nAs a long-range shooter, I can tell you that barrel twist does affect the exterior ballistics of the bullet. There is a phenomenon called spin drift that gradually pushes a bullet off-course - in the case of a right-hand-twist barrel, it moves the bullet to the right slightly. Not a concern, though, until you start shooting at distances of greater than 800 yards or so, where it can push your bullet about 2 inches off course.
496751
Well, if you spin the barrel again then the odds of my brains being splattered on the opposite wall are at 1 in 3.  (2 bullets in 6 chambers)\n\nBut if you don't spin the barrel, I'm feeling a bit better because my chances of survival rise to 75%.  Why?  Well, since the bullets are in adjacent chambers, there's only one bullet that I have to worry about - the one that might (or might not) be in the next chamber.  However, since there are four empty chambers, and since only one of those empty chambers is followed by a chamber with a bullet in it, there is only a 1 in 4 chance that the wall gets a fresh coat of paint.\n\nI say, don't spin it - I'll press my luck.
669407
yeah right...i suppose green pigs swim upside down in you're world too eh? my apologies...a friend told me today that a dummy round can be inserted into the breach which has a laser pointer mounted in it to allow fine scope adjustment...i eat my words...
1052340
If you don't spin, then there are five left.  Now since you said that the two bullets are next to each other and you didn't get shot the first time, then they both count as the same bullet, so that there are four possibilities left.  So that means the you have a 1 in 4 chance of getting shot.  You can also think of it this way: say you  have this situation.\n\n1 2 3 4 5 6\nX X X B B X\n\nso that there is a bullet in positions 4 and 5.  Since you shot once and got a blank, then that means you either were in the 1, 2, 3, or 6 position.  Out of those, only one (position 3) has a bullet the next time.  So the chance of getting shot is 1 in 4.\n\nIf you do spin, then you have a 2 in 6 chance of getting shot, since then you can consider that a repeat of the previous round.\n\nSince 1/4 is smaller than 2/6, I would go with not spinning.\n\n********************************************************************\n\nThis is a different result than if we couldn't assume whether the bullets are together or not.  Not spinning in that case would lead to a 2 in 5 chance of getting shot; spinning would lead to a 2 in 6 chance of getting shot.  In that case, spinning would be better.\n\nThis is an interesting question.  I'll need to keep it in mind for my students one day!
914874
yes if it is spinning on its own axis as it moves.  This is called Spin Angular Momentum
1127089
If time was stopped, and forces like gravity didn't pull the bullet to the ground, it should be sitting there in mid air and you would be able to touch it.\n\nIt would probably be very hot from the friction from the air and firing it.
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 ...
851529
approximately 1107 feet i think\nof course this is neglecting air resistance and mass of the bullet along with loss of energy heating\n\nmy work:\nfind the time that bullet is in air:\nwe know that height is 5 feet and the initial velocity of the downward component is 0 because the gun was fired horizontally.\nusing second kinematics equation :\n intial velocity(time) + 1/2(acceleration)(time)^2 = distance\nwe fill in:\n0(t)+1/2(32.67ft/sec^2)(t^2) = 5 and solve for t which = .553283\nNow that we know the time bullet spends in the air we can find the distance by using time and the same equation but using all horizontal component.\nwe will in:\n2000(.553283)+1/2(0*)9.5532830^2 which equals =1106.57 or about 1107\n\n* = there is no acceleration since there is no gravitaional effect on horizontal component\n\ni made have made a mistake so please check
19099
The knob on the side is your elevation, "or Up & Down". The Knob on top is your windage. "Left & right.\nTake 3 very steady shots at a target, not too far away. Look at your target, mark the holes with a pen or cover with tape or something so you do not loose them. Then raise of lower you elevation. Then 3 more shots... then adjust, over & over again. You want to hope that the 3 shats are always together, as you make each adjustment. Once you get the elevation, Next work on you windage. Turning the knob left, moves the shots to the right, & vice versa. You really do not want to have your target too far away in the beginning, because you need to get a tight 3 shot pattern to be able to make adjustments. once you are on target at say 25 or 30 yards, then move the target farther away. In the beginning, you just want that tight 3 shot pattern to make accurate adjustments. Don't forget to hold your breath & squeeze slowly. You will get a stray shot from time to time, but this can be expected. Takes about an hour to do it right.
1111942
Not necessarily. If your theoretical gun shoots a bullet that reaches escape velocity [1] (the speed something needs to travel to beat Earth's gravity), it can overcome the Earth's gravity [2] and fly off into space.\n\nHowever, I know of no such gun. Only rocket engines [3] are able to do this, because they sustain and increase their speed by burning fuel. A bullet has nothing to keep its speed up once it leaves the gun.\n\nSo the answer is, "Yes, the bullet comes back down." The bullet is always beholden to gravity, even before it's fired. You can view gravity as an invisible string always pointing to the center of the earth, always being pulled. The instant the bullet leaves the barrel of the gun, it will be constantly slowing down due to gravity (and air resistance).\n\nAnother interesting idea is this: If you were to take a gun and fire it horizontally, and at the same instant drop a rock from the same height as the gun, which would hit the ground first? The bullet or the rock? \n\nThe answer is that they both would hit the ground at the same time. The bullet would just be several hundred feet away when it hits. Both are affected by gravity's pull, and approach the ground at the same speed.
646492
according to the tv show mythbusters bullets shot straight in the air will fall side ways and land into the ground about 2 inches,how ever the force of the falling bullet is not powerful enough to penetrate the skull. there have been stories of people being hit(and even killed) by bullets fired into the air,but they were fired at an angle less than 90 degrees and traveled over a mile away, the bullet was not free falling but instead coming straight at the person because of the trajectory it fired from.
789141
The proper name is the cylinder.  In shooting circles it may also be called a "wheel" ... for example, "He's a good shot with that wheel gun."
840106
With in the casing above the trigger is the fireing mechanism, This consists of 2 springs ( 2 double barrel, 1 for singal barrel) and 2 hammers (1 hammer if singal barrel) The hammers are alined with what we call fireing pins which are also alined with the primer of the cartridge once the gun is loaded.\nOunce the trigget is pulled then the spring forces the hammer to hit the fireingpin, which hits the primer sending a chane reaction through the cartridge for the conbustion to take place with the gunpowder, then the casing(wad) which the shot is positioned behing projects the buckshot up the barrels at roughley 450ft per sec . That is just a simple explanation of a tipical double barrel shotgun. How ever some,  more common in over&unders work on the theory of inertia, where the recoile from the the first shot activates the mechanism for the second shot, its also the same princable for semioutomatic shotguns.
730340
Projectile range would be the equivalent of horizontal distance.\nThe forces involved are gravity (acting on the projectile's mass in the vertical direction), and the "push" from the initial thrust.\n\nAs the angle of the firing position increases above the horizontal, more of the initial push is used to combat gravity.\n\nIf the angle is too low, the projectile will meet the ground before it's maximum range (not enough action against gravity).\n\nFor most projectiles, the optimum angle is between 42 and 45 degrees (depending on the strength of the initial impetus). This is because the trajectory approximates a parabola.\n\nWith the given choices, the answer will have to be 45 degrees (answer B)...
649550
It is the reaction of shooting force on gun's body. let m=bullet mass and M=gun mass. Then if bullet is shot from gun by velocity of v, the recoil velocity would be V=m*v/(m+M) and its shock is M*V=m*M*v/(m+M)\n\nIt does not depend of gun/bullet type.\n\nIf you want to measure it, you must first measure the mass of bullet and empty gun, then measure the bullet shooting speed. This measurement could be done using a common method of shooting to a tough wooden block, which is hung by four equal size lines from a support. You must measure the maximum height (vertically) of displacement of block after shooting (name it H). Then if the mass of block be MB, then the speed of bullet when going out of gun can be calculated as follows:\n\n(MB+m)*g*H=1/2*m*v^2 then v=sqrt[2*(MB+m)*g*H/m]\nwhere m=bullet mass and g=9.81N/m^2 (gravity factor)\n\nAll units are metric (mass=grams; H=meters; v=meters/second)\n\nGood luck!
1097929
I'm not sure who usually uses compressed nitrogen to shoot a bullet out of a gun.  Typical firearms use gunpowder, an explosive chemical mixture.  The bullet is propelled by the force of expanding gas in the explosion.\n\nPractically anything can be used to propel a bullet: explosive material, springs, leverage, compressed gases, etc.  The most powerful guns (those used on battleships and gun implacements) used plastic explosives to push their projectiles.  \n\nThere have been some experiments with using charged particles as a source of projectile propulsion.  The repulsive force between powerfully charged electromagnets can be INCREDIBLY powerful.  Theoretically, it should be possible, using current technology, to fire a small projectile at an appreciable fraction of the speed of light.  Obviously, this would take an enormous amount of money and some incredible engineering.
887139
Momentum is the direct result of the following Newtonian Law of Inertia\nAny object in motion remains in motion and any object at rest remains at rest (unless a net force acts upon the object)\nMomentum (p) is the mass (m) of the object times its velocity (v).\np = mv\n\nForce (F) and Newtons Second Law can be derive from a change in momentum.  That is (t is time)\nF = delta p/delta t \nF = delta (m*v) / dt\nwe usually say m is constant so\nF = m * v/dt\nF =  ma    (a is acceleration or a change in v/time)\n\nNewtons third law is also based on momentum.  "For every action there is an equal, yet opposite, reaction."\n\nSo...  When a gun is fired what we see is the bullet propelled forward and the gun (and gun holder) propelled back.  Both received the same Force but because the gun and person holding it are much more massive than the bullet, the bullet will travel much faster. \ne.g F1 is the force on the bullet which is the same as the force on the person F2\nF1 = F2\ndelta p1/dt = delta p2/dt\nm1v1/dt = m2v2/dt\nm1v1 =m2v2\nImagine the bullet is 1 gram and a 180 lbs (~8182 grams) person shot the bullet...  The avg v on the kick back will be about 1/8182 of the v of the bullet...  Lucky for the shooter.