Answer:
The velocity is 
Explanation:
From the question we are told that
The mass of the bullet is 
The initial speed of the bullet is 
The mass of the target is 
The initial velocity of target is 
The final velocity of the bullet is is 
Generally according to the law of momentum conservation we have that

=> 
=> 
Okay so don't quote me on this but I believe the answer is A) I'm saying this because B and C make no sense. and you can't change the mass of something without changing it totally.
Answer:
(a) 42.28°
(b) 37.08°
Explanation:
From the principle of refraction of light, when light wave travels from one medium to another medium, we have:
= sinθ
/sinθ
In the given problem, we are given the refractive indices of light which are parallel and perpendicular to the axis of the optical lens as 1.4864 and 1.6584 respectively.
For critical angle θ
= θ
, θ
= 90°; 
(a) 
= sinθ
/sin90°
0.6728 = sinθ![_{c}θ[tex]_{c} = sin^(-1) 0.6728 = 42.28°(b) [tex]n_{a} = 1.6584](https://tex.z-dn.net/?f=_%7Bc%7D%3C%2Fp%3E%3Cp%3E%CE%B8%5Btex%5D_%7Bc%7D%20%3D%20sin%5E%28-1%29%200.6728%20%3D%2042.28%C2%B0%3C%2Fp%3E%3Cp%3E%3C%2Fp%3E%3Cp%3E%28b%29%20%5Btex%5Dn_%7Ba%7D%20%3D%201.6584)
= sinθ
/sin90°
0.60299 = sinθ[tex]_{c}
θ[tex]_{c} = sin^(-1) 0.60299 = 37.08°