Answer:
shown in the attachment
Explanation:
The detailed step by step and necessary mathematical application is as shown in the attachment.
Answer:
a) 46.5º b) 64.4º
Explanation:
To solve this problem we will use the laws of geometric optics
a) For this part we will use the law of reflection that states that the reflected and incident angle are equal
θ = 43.5º
This angle measured from the surface is
θ_r = 90 -43.5
θ_s = 46.5º
b) In this part the law of refraction must be used
n₁ sin θ₁ = n₂. Sin θ₂
sin θ₂ = n₁ / n₂ sin θ₁
The index of air refraction is n₁ = 1
The angle is this equation is measured between the vertical line called normal, if the angles are measured with respect to the surface
θ_s = 90 - θ
θ_s = 90- 43.5
θ_s = 46.5º
sin θ₂ = 1 / 1.68 sin 46.5
sin θ₂ = 0.4318
θ₂ = 25.6º
The angle with respect to the surface is
θ₂_s = 90 - 25.6
θ₂_s = 64.4º
measured in the fourth quadrant
Answer:
Ratio table of ordered pairs represent proportional relationship .
<em>Hope </em><em>it</em><em> is</em><em> helpful</em><em> to</em><em> you</em>
Answer:
The coefficient of rolling friction will be "0.011".
Explanation:
The given values are:
Initial speed,

then,


Distance,
s = 18.2 m
The acceleration of a bicycle will be:
⇒ 
On substituting the given values, we get
⇒ 
⇒ 
⇒ 
⇒ 
As we know,
⇒ 
and,
⇒ 
⇒ 
On substituting the values, we get
⇒ 
⇒ 