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
Answer:
5.01 J
Explanation:
Info given:
mass (m) = 0.0780kg
height (h) = 5.36m
velocity (v) = 4.84 m/s
gravity (g) = 9.81m/s^2
1. First, solve for Kinetic energy (KE)
KE = 1/2mv^2
1/2(0.0780kg)(4.84m/s)^2 = 0.91 J
so KE = 0.91 J
2. Next, solve for Potential energy (PE)
PE = mgh
(0.0780kg)(9.81m/s^2)(5.36m) = 4.10 J
so PE = 4.10 J
3. Mechanical Energy , E = KE + PE
Plug in values for KE and PE
KE + PE = 0.91J + 4.10 J = 5.01 J
Relative to the positive horizontal axis, rope 1 makes an angle of 90 + 20 = 110 degrees, while rope 2 makes an angle of 90 - 30 = 60 degrees.
By Newton's second law,
- the net horizontal force acting on the beam is
where 
are the magnitudes of the tensions in ropes 1 and 2, respectively;
- the net vertical force acting on the beam is
where 
and 
.
Eliminating 
, we have
Solve for .
D Because If Your Going To Have A Contest Its Ganna Have To Be The Same Objectives For Both Contenders
