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The refractive index for glycerine is 
, while for air it is 
.
When the light travels from a medium with greater refractive index to a medium with lower refractive index, there is a critical angle over which there is no refraction, but all the light is reflected. This critical angle is given by:
where n1 and n2 are the refractive indices of the two mediums. If we susbtitute the refractive index of glycerine and air in the formula, we find the critical angle for this case:
When the light travels from a medium with greater refractive index to a medium with lower refractive index, there is a critical angle over which there is no refraction, but all the light is reflected. This critical angle is given by:
where n1 and n2 are the refractive indices of the two mediums. If we susbtitute the refractive index of glycerine and air in the formula, we find the critical angle for this case:
Answer:
a = 1 m/s² and
Explanation:
The first two parts can be seen in attachment
We use Newton's second law on each axis
Y axis
Ty - W = 0
Ty = w
X axis
Tx = m a
With trigonometry we find the components of tension
Sin θ = Ty / T
Ty = T sin θ
Cos θ = Tx / T
Tx = T cos θ
We calculate the acceleration with kinematics
Vf = Vo + a t
a = (Vf -Vo) / t
a = (20 -10) / 10
a = 1 m/s²
We substitute in Newton's equations
T Sin θ = mg
T cos θ = ma
We divide the two equations
Tan θ = g / a
θ = tan⁻¹ (g / a)
θ = tan⁻¹ (9.8 / 1)
θ = 84º
We see that in the expression of the angle the mass does not appear therefore you should not change the angle
