Answer: The light bends because light travels fast but it slows down in a denser medium. For example light refracts in water or it bends after passing through air. When light passes through air ( a less dense medium ) then through water ( a more dense medium ) the beam of light bends because light travels more slowly in a denser medium then it picks up its pace again once it passes. The density of the substance determines how much the light is refracted. I hope this makes sense and I hope this answered your question!! :) 
 
        
             
        
        
        
Answer:
 P = 1 (14,045 ± 0.03 )  k gm/s
Explanation:
In this exercise we are asked about the uncertainty of the momentum of the two carriages
             Δ (Pₓ / Py) =?
  Let's start by finding the momentum of each vehicle
car X
         Pₓ = m vₓ
         Pₓ = 2.34 2.5
         Pₓ = 5.85 kg m
car Y
         Py = 2,561 3.2
         Py = 8,195 kgm
How do we calculate the absolute uncertainty at the two moments?
           ΔPₓ = m Δv + v Δm
           ΔPₓ = 2.34 0.01 + 2.561 0.01
           ΔPₓ = 0.05 kg m
          Δ = m Δv + v Δm
 = m Δv + v Δm
          ΔP_{y} = 2,561 0.01+ 3.2 0.001
          ΔP_{y} = 0.03 kg m
now we have the uncertainty of each moment
           P = Pₓ /  
           ΔP = ΔPₓ/P_{y} + Pₓ ΔP_{y} / P_{y}²
           ΔP = 8,195 0.05 + 5.85 0.03 / 8,195²
           ΔP = 0.006 + 0.0026
           ΔP = 0.009 kg m
The result is
            P = 14,045 ± 0.039 = (14,045 ± 0.03 )  k gm/s
 
        
             
        
        
        
We will have the following:

So, the heat to add is 3611.52 Joules.
 
        
             
        
        
        
Answer: d= 0.57* l
Explanation:
We need to check that before ladder slips the length of ladder the painter can climb.
So we need to satisfy the equilibrium conditions.
So for ∑Fx=0, ∑Fy=0 and ∑M=0
We have,
At the base of ladder, two components N₁ acting vertical and f₁ acting horizontal
At the top of ladder, N₂ acting horizontal
And Between somewhere we have the weight of painter acting downward equal to= mg
So, we have N₁=mg
and also mg*d*cosФ= N₂*l*sin∅
So,
d= * tan∅
 * tan∅
Also, we have f₁=N₂
As f₁= чN₁
So f₁= 0.357 * 69.1 * 9.8
f₁= 241.75
Putting in d equation, we have
d=  * tan 58
 * tan 58
d= 0.57* l
So painter can be along the 57% of length before the ladder begins to slip