Complete question:
The recoil of a shotgun can be significant. Suppose a 3.6-kg shotgun is held tightly by an arm and shoulder with a combined mass of 15.0 kg. When the gun fires a projectile with a mass of 0.040 kg and a speed of 380 m/s, what is the recoil velocity of the shotgun and arm–shoulder combination?
Answer:
The recoil velocity of the shotgun and arm–shoulder combination is 1.013 m/s
Explanation:
Given;
combined mass of the shotgun and arm–shoulder, m₁ = 15 kg
mass of the projectile, m₂ = 0.04 kg
speed of the projectile, u₂ = 380 m/s
let the recoil velocity of the shotgun and arm–shoulder combination = u₁
Apply the principle of conservation of linear momentum;
m₁u₁ + m₂u₂ = 0
m₁u₁ = - m₂u₂

Therefore, the recoil velocity of the shotgun and arm–shoulder combination is 1.013 m/s
= (3,760 joule/sec) / (4,000 joule/sec)
= 3,760 / 4,000 = 0.94 = 94%
Answer:
D)
Explanation:
The Period-Luminosity relationship tells us that luminosity increases with the period, and of course the more luminosity a star has the more far away they can be seen, so from this we know that:
A) False since lower luminosities can be observed when they are close.
B) False since longer periods means higher luminosities
C) False since lower luminosities can be observed when they are close.
D) True: Variable stars with shorter periods have lower luminosities, so they can only be observed when they are close.
Answer:
3.675 m
Explanation:

X-direction | Y-direction
| 
|
| 
Hope it helps