Yes. The bullet is already moving at 900 fps before it is even fired. When the round leaves the barrel it will be traveling at 1,800 fps. Motion is all relative.
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The barrel of the gun has lands and grooves (grooves and ridges) cut in a spiral. The bullet molds to these and starts to spin as it moves down the barrel. The bullet just continues to spin after it leaves the barrel.Correct. The ridges are known as 'lands'. It is possible to calculate how fast a bullet will spin if you know the twist rate of the barrel and the velocity of the bullet. My AR15 has a twist rate of 1-in-8 ie for every eight inches the bullet travels down the barrel, the bullet is rotated once. It fires a .223 round at approx 2,800 feet per second so... The formula is (bullet velocity x 720)/twist rate so... (2,800x720)/8 is an incredible 252,000RPM!
Usually between 2500 and 3200 feet per second out of a 16" barrel, depending on the weight of the projectile.
A 9mm Parabellum cartridge loaded with a standard bullet and only a primer will normally not be able to drive the bullet out of the barrel. It will usually lodge in the barrel (this is known as a squib load). If a second normal cartridge is fired behind this, it stands a very good chance of blowing up the gun.
There is no single answer to this question. It will depend on the cartridge, and the characteristics of the powder and bullet being fired. For instance, the 40 gr 22 Long Rifle cartridge accelerates from the moment of firing, until it travels about 12 inches. The gas produced by firing reaches it's max expansion, and the bullet now begins to SLOW to some degree from friction with the barrel. Acceleration is NOT uniform with any firearm cartridge- if graphed it would be a curve with a very sharp spike.
Depends on the speed of the bullet, and the length of the barrel. In the case of a .22 rifle, firing a bullet at 1200 feet per second, from a 16 inch barrel, it will take 1/75th of a second for the bullet to leave the barrel.
it is about 25,000 persecond
An MP40's muzzle velocity (speed of a bullet upon exiting the barrel) is about 1,250 feet/second.
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Typically between 2600 and 2800 feet per second. Exact speed depends on WHICH bullet, the powder charge, and the length of barrel it is fired from.
A 150 grain bullet shot from a 308 will start dropping the fraction of a second it leaves the barrel.
3 per second = 3 Hz
The barrel of the gun has lands and grooves (grooves and ridges) cut in a spiral. The bullet molds to these and starts to spin as it moves down the barrel. The bullet just continues to spin after it leaves the barrel.Correct. The ridges are known as 'lands'. It is possible to calculate how fast a bullet will spin if you know the twist rate of the barrel and the velocity of the bullet. My AR15 has a twist rate of 1-in-8 ie for every eight inches the bullet travels down the barrel, the bullet is rotated once. It fires a .223 round at approx 2,800 feet per second so... The formula is (bullet velocity x 720)/twist rate so... (2,800x720)/8 is an incredible 252,000RPM!
Usually between 2500 and 3200 feet per second out of a 16" barrel, depending on the weight of the projectile.
For a 9mm Parabellum (aka 9mm Luger) the muzzle velocity of a 124 gr bullet will run ABOUT 1120-1170 feet per second, depending on exact load and barrel length.
Due to gravity, the bullet starts to drop the second it leaves the barrel. You can calculate the drop by factoring mass and velocity with gravity (9.8 m/s²).
A 9mm Parabellum cartridge loaded with a standard bullet and only a primer will normally not be able to drive the bullet out of the barrel. It will usually lodge in the barrel (this is known as a squib load). If a second normal cartridge is fired behind this, it stands a very good chance of blowing up the gun.