Answer:
K =6.697 Kg/s²
Explanation:
Given:
delta m =41 g = 0.041 kg
delta x = 6cm = 0.06m
g = 9.8 m/s²
according to the given formula
K = delta m g /delta x
K = (0.041 kg × 9.8 m/s²) / 0.06m
K =6.697 Kg/s²
Answer:
The average recoil force on the gun during that 0.40 s burst is 45 N.
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\times t=mv\\\\F=\dfrac{mv}{t}](https://tex.z-dn.net/?f=F%5Ctimes%20t%3Dmv%5C%5C%5C%5CF%3D%5Cdfrac%7Bmv%7D%7Bt%7D)
For 8 shot burst, average recoil force on the gun is :
![F=\dfrac{8mv}{t}\\\\F=\dfrac{7.5}{1000}\cdot\dfrac{300}{0.4}\cdot8\\\\F=45\ N](https://tex.z-dn.net/?f=F%3D%5Cdfrac%7B8mv%7D%7Bt%7D%5C%5C%5C%5CF%3D%5Cdfrac%7B7.5%7D%7B1000%7D%5Ccdot%5Cdfrac%7B300%7D%7B0.4%7D%5Ccdot8%5C%5C%5C%5CF%3D45%5C%20N)
So, the average recoil force on the gun during that 0.40 s burst is 45 N.
Answer:
B it decreases
Explanation:
the movement of a positive test charge in the direction of an electric field would be like a mass falling downward within Earth's gravitational field. Both movements would be like going with nature and would occur without the need of work by an external force. This motion would result in the loss of potential energy
GPE I am assuming is gravitational potential energy. I'll denote it as U for simplicity.
U = mgy
U = (70kg)(9.81m/s^2)(1m) = 686.7J
U = 686.7J
Hope this helps!