An attack helicopter is equipped with a 20- mm cannon that fires 88.7 g shells in the forward direction with a muzzle speed of 9
29 m/s. the fully loaded helicopter has a mass of 4180 kg. a burst of 89.9 shells is fired in a 4.84 s interval. what is the resulting average force on the helicopter? answer in units of n
The impulse imparted to the shells equals the change in the momentum: Fav*(Delta t)= Delta m*v. The mass change is Delta m= n*m= (89.9shells)*(88.7g)=7.97Kg So the average force is F=((v)*(Delta m))/t= ((929)*(7.97))/4.84=1529.78 N Since the velocity of the shells is much greater than the velocity of the helicopter, there is no need to use relative velocity.
The particles always move parallel and perpendicular to the waves. The waves which are in the water moves a circle. Both up and down and back and forth.
