What does Ballistic Coefficient mean??
145197
Ballistic Coefficient (BC) is basically a measure of how streamlined a bullet is; that is, how well it cuts through the air. Mathematically, it is the ratio of a bullet's sectional density to its coefficient of form. Ballistic Coefficient is essentially a measure of air drag. The higher the number the less drag, and the more efficiently the bullet cuts through the air. So for purposes of flying through the air efficiently, the bigger the BC number the better.\n\nBC is what determines trajectory and wind drift, other factors (velocity among them) being equal. BC changes with the shape of the bullet and the speed at which the bullet is traveling, while sectional density does not. Spitzer, which means pointed, is a more efficient shape than a round nose or a flat point. At the other end of the bullet, a boat tail (or tapered heel) reduces drag compared to a flat base. Both increase the BC of a bullet.\n\nFor example, a Hornady 100 grain round nose 6mm bullet has a BC of .216; a Hornady 100 grain spire point 6mm bullet has a BC of .357, and a Hornady 100 grain boat tail spire point 6mm bullet has a BC of .400. All three of these bullets have a sectional density (which is the ratio of a bullet's diameter to its weight) of .242, because they are all .243" in diameter and weigh 100 grains. But the more streamlined bullets have a higher ballistic coefficient. They are the ones to choose for long range shooting where a flatter trajectory is important.\n\nTo illustrate the practical difference between these three styles of bullets, let's use Hornady's trajectory figures for the 100 grain 6mm bullets above. Starting all three bullets at a muzzle velocity of 3100 fps from a scoped 6mm rifle zeroed at 300 yards, the trajectories are as follows.\n\n.243" 100 grain Round Nose (BC .216): -1.5" @ muzzle, +4.8" @ 100 yards, +6" @ 200 yards, 0 at 300 yards, -15.9" @ 400 yards, -46" @ 500 yards.\n\n.243" 100 grain Spire Point (BC .357): -1.5" @ muzzle, +3.8" @ 100 yards, +4.7" @ 200 yards, 0 @ 300 yards, -11.1" @ 400 yards, -30.5" @ 500 yards.\n\n.243" 100 grain Spire Point BT (BC .400): -1.5" @ muzzle, +3.6" @ 100 yards, +4.4" @ 200 yards, 0 @ 300 yards, -10.4" @ 400 yards, -28.6" @ 500 yards.
347176
Modren artillery shells travel at supersonic speeds.  If you heard it - it missed.\n\nBTW, in real life you only hear a rocket or mortar round for about 1/2 second.  Then it has either flown past you or you have had a very loud noise closer to you than you'd really prefer.
1073194
I assume, since this is in "physics", that you are referring to the frictional force exerted on an object by a fluid through which it travels.\n\nIf, by "definition", you were wondering what it is, then the above paragraph should provide a simple answer. For more information, check out:\nhttp://en.wikipedia.org/wiki/Drag_%28physics%29\n\nIf, on the other hand, you were wondering about the mathematical definition of drag, the equation is\n\nFd = -1/2 * (rho) * v^2 * A * Cd\n\nwhere Fd is the force of the drag, rho is the density of the fluid through which the object in question travels, v is the velocity of the object in question in relation to the fluid, A is roughly the area of the object as viewed from the line of motion (e.g. a bullet's area would be a circle), and Cd is the drag coefficient, which is dependent on the surface material, texture, and shape of the object.\n\nFor more information about this formula, see:\nhttp://en.wikipedia.org/wiki/Drag_equation
786520
The muzzle velocity of the paint ball gun is higher. \nhttp://en.wikipedia.org/wiki/Muzzle_velocity
826715
bullet drops 0.02 m due to gravity\n\nfree fall distance formula:\n\nd=v1xt + 0.5xgxt^2, v1=0 ( initial vert velocity is 0) and d=0.02 m\n                                   g=9.81\n\nt= (2x0.02/g)^0.5 = 0.0639 s\n\nv=25m/.0639= 391.5 m/s
78749
the escape velocity is around 11 km/sec, or 25000 mph.  this number is computed without considering the effect of air resistance.  \n\nthe mach number is defined as the ratio of an object's speed to the speed of sound.  however the speed of sound changes with altitude, and so therefore the mach number changes with altitude even if the velocity is the same!\n\nhowever if you are launching some kind of projectile from the earth's surface, the muzzle velocity will be the escape velocity and that has a well-defined mach number.  the speed of sound in the atmosphere at the surface of the earth is around 340 m/s.  11 km/s is then around mach 32.
1102976
a) It will decrease in speed as soon as the accelerative force of the explosive gas ceases to be greater than the drag of the gun barrel or the air or other media it passes through. This could mean that it begins to slow down in the barrel, or soon after leaving it. Certainly, after a few feet from the gun barrel one would expect the gases from the gun to be dissipated and the bullet to begin to be dragged by the air. However, the reduction in speed of a high velocity bullet could be negligible for seconds after firing.\n\nb) As soon as the bullet leaves the barrel it will begin to drop towards the earth. The formula for this is: \ndistance =0.5*acceleration*(time-squared)\n\nAcceleration at sea level is 9.81 metres per second per second. Thus it will take almost exactly 2 seconds for the bullet to drop the 20 metres from your firing platform. (square root of [20/(0.5*9.81)]  )\n\nc) See b above. The bullet is acted upon by gravity at all times. It is only prevented from being accelerated downwards by a force equal or greater to its weight pushing it up. In your example the gun barrel supports it until the bullet leaves the barrel, then begins to fall. Even if you pointed the gun straight up, the bullet will be slowed by gravity as soon as it leaves the barrel and gases behind, gradually slowing to a stop and falling back to land on your head.\n\nIf the bullet had enough horizontal velocity to make it to the horizon within two seconds, it would have acheived the required "escape velocity" to leave or orbit the earth, always falling but never landing. As the horizon at sea level is eight miles away this would need a speed of 28,800 miles per hour or 46,340 kilometers per hour. \n\nThe speed of sound is c.1,220 km/hour, so this would be 38 times the speed of sound!\nThe muzzle velocity of most military calibres is c.700metres/second or 2, 520 km/h or twice the speed of sound.\n\nThus in the two seconds it takes the bullet to fall to earth it would have carried some 1,400 metres.\n\nd) Very much so!
1161161
The law of conservation of momentum applies in all cases.\nFor a given system, the initial momentum must equal the final momentum.\n\nIn your example, we can define the "System" to be the combination fighter plane + bullet.  If we call this initial momentum P, then at any time after the initial conditions, the sum of the momentums of all parts of our system must add up to P.\nWhen the bullet is fired foward, the fighter plan slows down to compensate and conserve momentum.\n\nMomentum (P) = mass (m) * velocity (v)\nP = mv\n\nThe bullet's mass is very tiny compared to that of the fighter plane, so even a large increase in the bullet's velocity would only mean a small decrease in the resulting velocity of the plane.\n\nIf you call the bullet's mass, m_1 (with final velocity v_2) and the plane's mass m_2 (with final velocity V_2),\nwe know that,\nP = m_1*v_1 + m_2*v_2\nif P_1 goes up, then P_2 must go down by just that amount.\n(delta P_1) = - (delta P_2)\nm_1 * (delta v_1) = -m_1 * (delta v_2)\nsince we assume the mass stays constant,\n(delta v_2) = - (m_1 * delta v_1) / m_2
69092
7.62 mm refers to the diameter of the lands in the barrel. The actual bullet is normally .308 in (7.82 mm), although Soviet weapons commonly use a .310 in (7.87 mm) bullet, as do older British and Japanese cartridges.\nHope this can answer your question.
711377
Bound's hubby here: \n\nThe name or designation given to a cartridge identifies two pieces of critical information about the cartridge: the bore diameter of the projectile and the name of the cartridge. It is kinda like shopping for a suit: business versus leisure, wide lapel versus narrow lapel, or single breasted versus double breasted.\n\nThe number represents the diameter of the bullet. That is the easy part! Bullet diameters can be identified by decimal diameter (.22, .25, .27, .28, .30, .32, .33, .35, .40, etc.) or by the metric diameter (5.45mm, 6 mm, 7mm, 8mm, 9mm, 10mm, 11 mm, etc.). Sometimes, if the diameter is metric, it will suggest the cartridge is of European origin, but not always the case.\n\nThe name will indicate two things: the designer/originator of the cartridge, and whether the cartridge is a magnum or not. Each cartridge design is unique, and is intended to exploit the cases volume which can affect the powder burn rate. Generally, cartridges of the same bore diameter and different designer are not interchangeable ... to do so (if the cartridge fits the chamber of the rifle) could be courting disaster.\n\nGenerally, you want to select a rifle and cartridge adequate for the game you are hunting. You can use the game and cartridge selector (linked below) to help in your cartridge selection!\n\nGood luck!
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!
282533
Muzzle velocity is the speed of a projectile leaving the barrel of a gun.\n\nMeasured in feet per second. \n\nIt is used as a comparison between similar projectiles, such as ammunition for a gun.
8368
Assuming Federal's published muzzle velocity of 2600 FPS from a 24" barrel, and using Sierra's published ballistic coefficient of .477, you can expect a total drop of 483.7 inches, or a little over 40 feet. Notice that the 168 goes subsonic somewhere around 900 yards.\n\n168 Grain:\nRange (yds)Vel. (fps)Total Drop\n0 2600 0\n100 2421 -2.7\n200 2249 -11.4\n300 2085 -26.9\n400 1927 -50.1\n500 1777 -83.4\n600 1637 -127.8\n700 1505 -184.4\n800 1380 -253.4\n900 1280 -348.2\n1000 1184 -458.7\n\n175 grain:\nRange (yds)Vel. (fps)Total Drop\n0 2600 0.0\n100 2428 -2.6\n200 2262 -11.3\n300 2104 -26.7\n400 1951 -49.7\n500 1805 -82.3\n600 1669 -125.8\n700 1540 -181.5\n800 1418 -249.5\n900 1315 -340.3\n1000 1219 -448.7
1031011
In physics, thermal conductivity, k, is the intensive property of a material that indicates its ability to conduct heat.\n\nIt is defined as the quantity of heat, Q, transmitted in time t through a thickness L, in a direction normal to a surface of area A, due to a temperature difference ΔT, under steady state conditions and when the heat transfer is dependent only on the temperature gradient.\n\nthermal conductivity = heat flow rate × distance / (area × temperature difference)\n\n\nElectrical conductance is the reciprocal of electrical resistance. It is a measure of how easily electricity flows along a certain path through an object. The SI derived unit of conductance is the siemens (symbol S, equal to 1/Ω; alias the "mho"). Oliver Heaviside coined the term in September 1885.
1053008
This question has been answered pretty well by everyone, the only thing I want to add is contridictory to a statement by TJ N. \n\n\nthe caliber of a bullet is expressed in a fraction of an inch.  not in millimeters a .22 caliber bullet is 22/100ths of an inch in diameter, a .50 caliber bullet is 50/100ths of an inch (or if you reduced the fraction its a full 1/2 inch in diameter).  Sometimes ammunitions are interchangable with NATO ammuntions IE. Remington .223 = NATO 5.56 mm (its 223/1000ths of an inch measured metricly its 5.56 mm. \n\nthe problem with TJ Ns "years of shooting and hunting" experience is that a ".22 caliber bullet" - commonly a squirel/vermin hunting round would be 0.86 inch in diameter.... which wouldn't leave much squirel meat after you hit it with TJ's .22 caliber gun... Which leaves me with more questions about how much of a recoil his .22 caliber gun has? I've fired the .50 caliber Barret Sniper rifle before, as well as the M2 HB .50cal MG. (they both fire the same basic .50 caliber (50/100ths or 1/2 inch round) The M2 has to be mounted on some sort of mount (vehicle or sandbagged Tripod) to fire.  The Barret, I've heard of people firing it standing up.... but the only way I could get behind that thing was in the prone, and after about 15 rounds through that thing I had a bruise on my shoulder the size of Texas. I can't imagine what TJ's .22 caliber that's 86/100ths of an inch would do to me....\n\nnow the difference between the calibers... obviously is the size (diameter) of the bullet.  what's considered High caliber and low caliber?  I suppose thats a matter of opinion.  Some people say anything above .308 is large caliber, and I tend to agree with that but that doesn't really give an indication of damage caused by a bullet for example a .30 caliber (7.62 mm) fired from a M240B MG would tend to cause greater damage to soft tissue at long ranges(greater than 50 meters) than a .45 caliber fired from a M1911 handgun. \n\nAnd yes various rifles fire various caliber rounds.
869691
a)F = MV=  m1v1 +/-  m2v2  \nusing these we get,  95*4.1 + 85*5.5 (since same direction)=857kgm/s = (95+85)* V. so, V = 4.76m/s\nb) MV = M (6) = (9500+m)*6 = 9500*16 + m*0. \nso, (9500+m) = 9500*16/6= 25333.33 kg\nso, m = 25333.33-9500 =15833.33kg\nc)work done against friction= kinetic energy of bullet\nW = FScos(tetha)  since its horizontal, angle is zero, \nand cos0 = 1, so W = FS\nwe know, coeff of friction = tangent of angle between, the frictional force  and normal reaction...\nso, the \nF = normal reaction=(1.1 +0.015)* 9.81=10.938N\nfrictional force= 0.25 *10.938 = 2.7345N\ndistance is 9.5m...\nso W = 2.7345*9.5 = 25.97 J\nthe kinetic energy of bullet has been transferred to the block...\nso the kinetic energy of bullet is 25.97J\n1/2*(0.015)* v^2 = 25.97\nso, v^2 = 25.97*2/0.015=3463.74\nso, v = 58.85m/s\nhope this helps...:)
999288
well, others might have told you it is pressure, and actually even i would have gone for it!\nbut since you have mentioned in particular on a fluid, then this ratio can also be called as Bulk's modulus(though this applies even to solids)\nit can also be the coeff of viscosity taken velocity gradient times\nthere can be various other depictions also