By what i know i think that the answer would be A a homogeneous mixture.
Answer:
COMPLETE QUESTION
A spring stretches by 0.018 m when a 2.8-kg object is suspended from its end. How much mass should be attached to this spring so that its frequency of vibration is f = 3.0 Hz?
Explanation:
Given that,
Extension of spring
x = 0.0208m
Mass attached m = 3.39kg
Additional mass to have a frequency f
Let the additional mass be m
Using Hooke's law
F= kx
Where F = W = mg = 3.39 ×9.81
F = 33.26N
Then,
F = kx
k = F/x
k = 33.26/0.0208
k = 1598.84 N/m
The frequency is given as
f = ½π√k/m
Make m subject of formula
f² = ¼π² •(k/m
4π²f² = k/m
Then, m4π²f² = k
So, m = k/(4π²f²)
So, this is the general formula,
Then let use the frequency above
f = 3Hz
m = 1598.84/(4×π²×3²)
m = 4.5 kg
Answer:
(a) 
(b) 
(c)
(d)
Solution:
As per the question:
Refractive index of medium 1, 
Angle of refraction for medium 1, 
Angle of refraction for medium 2, 
Now,
(a) The expression for the refractive index of medium 2 is given by using Snell's law:

where
= Refractive Index of medium 2
Now,

(b) The refractive index of medium 2 can be calculated by using the expression in part (a) as:


(c) To calculate the velocity of light in medium 1:
We know that:
Thus for medium 1
(d) To calculate the velocity of light in medium 2:
For medium 2:
Scattering occurs when light changes direction after colliding with particles of matter.
Here is a helpful video https://www.khanacademy.org/_render