Answer:
The average recoil force on the gun during that 0.40 s burst is 45 N.
Step-by-step explanation:
Mass of each bullet, m = 7.5 g = 0.0075 kg
Speed of the bullet, v = 300 m/s
Time, t = 0.4 s
The change in momentum of an object is equal to impulse delivered. So,
![F* t=mv\\\\F=(mv)/(t)](https://img.qammunity.org/2021/formulas/physics/college/aof4voth18yf71yfodcse0fjrtyn851w60.png)
For 8 shot burst, average recoil force on the gun is :
![F=(8mv)/(t)\\\\F=(7.5)/(1000)\cdot(300)/(0.4)\cdot8\\\\F=45\ N](https://img.qammunity.org/2021/formulas/physics/college/s4dnxhfj3azy71pdry3dopniv1ncb4wno7.png)
So, the average recoil force on the gun during that 0.40 s burst is 45 N.