Question

Question #ffa40

Answer 1

Solution : The power of the machine gun is `P=25 kW`

Explanation : The power of the machine gun is the rate at which it delivers energy. The energy is delivered as the kinetic energy of the bullets. If `E_o` is the kinetic energy of each bullet and if `(dN)/(dt)` is the rate at which bullets are fired from the gun then the power of the machine gun is :
`P=E_o(dN)/(dt);`
`E_o = 1/2 mv^2 = 1/2 (5\times10^{-2} kg)(500 m/s)^2 = 6250` J
`(dN)/(dt)= 240` bullets/min `= 4` bullets/sec.

Therefore the power of the machine gun is :
`P=(6250 J)\times(4 s^{-1}) = 25,000` W `=25` kW