after a disastrous descent into enemy territory, our hero has found himself completely weaponless, yet only slightly disheveled. in a feat of inconceivable (perhaps even implausible) strength, he pries a perfectly intact chain gun from its mountings beneath his chopper's wreckage and cradles it in his arms. as enemy troops swarm in, he unleashes a 1626 spm (shots per minute) barrage that stops them dead in their tracks. if the muzzle velocity is 3,474 ft/s and each bullet weighs 0.23-lb, what average force f is required of our hero to hold the weapon in place? neglect, among other things, the mass of both the ammunition belt and the discarded casings. round your answer to the nearest lb.

Respuesta :

The speed of a bullet as it exits the barrel is known as the muzzle velocity of a firearm.

What is the muzzle velocity formula?

A bullet's kinetic energy and momentum can be calculated using the formulas E k = (1/2) m v 2 and momentum p = m v if its mass m and muzzle velocity v are known.

Black powder musket muzzle velocities ranging from about 120 m/s (390 ft/s) to 370 m/s (1,200 ft/s) to more than 1,200 m/s (3,900 ft/s) in contemporary rifles with high-velocity cartridges like the. 220 Swift and. 22 Long Rifle.

A 180-grain (12 g) bullet fired from a. 357 magnum handgun may produce 580 foot-pounds of force at the muzzle (790 J). Depending on how the cartridge was made, a 110-grain (7.1 g) bullet could only provide 400 foot-pounds force (540 J) of muzzle energy.

To learn more about muzzle velocity refer to

https://brainly.com/question/14243531

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