Answer:
The answer of this is question is A.
Answer:
320 N/m
Explanation:
From Hooke's law, we deduce that
F=kx where F is applied force, k is spring constant and x is extension or compression of spring
Making k the subject of formula then

Conversion
1m equals to 100cm
Xm equals 25 cm
25/100=0.25 m
Substituting 80 N for F and 0.25m for x then

Therefore, the spring constant is equal to 320 N/m
C. Patient info, name of med, dosage & route, special instructions, prescriber’s DEA#, and number of refills
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.