Answer:
The radius is 
Explanation:
From the question we are told that
The distance beneath the liquid is 
The refractive index of the liquid is 
Now the critical value is mathematically represented as
![\theta = sin ^{-1} [\frac{1}{n_i} ]](https://tex.z-dn.net/?f=%5Ctheta%20%3D%20%20sin%20%5E%7B-1%7D%20%5B%5Cfrac%7B1%7D%7Bn_i%7D%20%5D)
substituting values
![\theta = sin ^{-1} [\frac{1}{131} ]](https://tex.z-dn.net/?f=%5Ctheta%20%3D%20%20sin%20%5E%7B-1%7D%20%5B%5Cfrac%7B1%7D%7B131%7D%20%5D)

Using SOHCAHTOA rule we have that

=> 
substituting values


The answer is,
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Answer:
The radius of the gold nucleus is 7.1x10⁻¹⁵m
Explanation:
The nearest distance is:
(eq. 1)
Where
z = atomic number of gold = 79
e = electron charge = 1.6x10⁻¹⁹C
k = electrostatic constant = 9x10⁹Nm²C²
energy of the particle = 32 MeV = 5.12x10⁻¹²J
At the potential energy is zero, all the energy will be kinetic energy:

Where
m = 4 mp = mass of proton

Replacing in equation 1

Answer:
Given:
m=1000kg
u= 16.7m/s
v=0m/s
F=8000N
Required:
s=?
Solution:
F=m × a
8000N=1000kg × a
a=8m/s^2
Since it decelerate a= -8m/s^2
v^2 = u^2 + 2as
s=v^2 - u^2 / 2a
s= 0 - (16.7m/s)^2 / 2 × -8m/s^2
s= -278.89/-16
s= 17.43m
The car travels approximately 17.43m before it stops
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