ITS OVER 9000mph!!!!!!!!!!!!!!
Depending on the rounds fired there would be a slight drop in speed momentarily due to the blowback of the gun. If firing say a standard 5.5 machine gun then the effect would be minor,if it were larger missiles etc the effects would be greater but still neglidgeable due to the huge mass and kinetic energy of the aircraft.
Bullet Joe Bush played in 25 games at pitcher for the Philadelphia Athletics in 1915, starting in none of them. He made 7 putouts, had 39 assists, and committed 3 errors, equivalent to .12 errors per game (estimate based on total games played in). He had 3 double plays.
Bullet Joe Bush played in 37 games at pitcher for the New York Yankees in 1923, starting in none of them. He made 15 putouts, had 74 assists, and committed 3 errors, equivalent to .081 errors per game (estimate based on total games played in). He had 3 double plays.
Bullet Joe Bush played in 39 games at pitcher for the New York Yankees in 1924, starting in none of them. He made 24 putouts, had 60 assists, and committed one error, equivalent to .026 errors per game (estimate based on total games played in). He had 4 double plays.
No, Molly Pitcher, also known as Mary Ludwig Hays McCauley, she died while working at home in Pennsylvania.
Momentum = mass x velocity A bullet has a high momentum because its velocity is really high.
If the gun is stationary before the shot, then the momentum of the gun and the momentum of the bullet are equal and opposite after the shot.
In the recoil? This follows from conservation of momentum. Actually, the momentum of the gun will be exactly opposite - or the negative - of the bullet's momentum. It can also be derived from Newton's Second and Third Laws.
Momentum before = momentum after. Since there was no movement before, momentum before = 0 If you think of the bullet as forward/positive momentum and the gun as backward/negative momentum then the momentum of the bullet plus the momentum of the gun =0 and therefore the momentum of the bullet = the momentum if the gun. momentum = mass x velocity P=m/v 20gx150m/s = 2000g (2kg) x velocity 3000 = 2000v 3000 / 2000 = v v = 1.5m/s
In an isolated system the total momentum of a system remains conserved. For example If you fire a bullet from Gun , bullet go forward with some linear momentum and in order to conserve the linear momentum the gun recoils
In an isolated system the total momentum of a system remains conserved. For example If you fire a bullet from Gun , bullet go forward with some linear momentum and in order to conserve the linear momentum the gun recoils
A bullet fired from a gun
The elephant
d) they have the same momentum,but the bullet has greater kinetic energy the the rifle
Because linear momentum is conserved. Before the shot, the momentum of (gun + bullet) is zero, so it has to be zero after the shot. The bullet gains forward momentum when fired, so the gun must gain reverse momentum in order to maintain the zero sum.
Momentum is the product of velocity and mass.
The idea is to use conservation of momentum. The momentum before the shot is fired is assumed to be zero, so write an equation that states that the momentum after the shot is zero, and solve it. The total momentum, of course, is the sum of the momentum of the two parts; for each part, the momentum is mass x velocity.