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:
3.3m/s
Explanation:
You first get the total time (80 + 70 = 150s).
Then you would find the displacement of the truck. To do that you would do component method (vector addition), so since its a right triangle (North and East), displacement is 400^2 + 300^2 = d^2.
d= 500m.
So now that you have displacement and time, you can find the velocity:
v=d/t
v=500/150
v=3.3
Answer:
<em>3924 Pa</em>
<em></em>
Explanation:
Volume of cylinder = 2 L = 0.002 m^3 (1000 L = 1 m^3)
diameter of the inner cylinder = 8 cm = 0.08 m (100 cm = 1 m)
radius of the inner cylinder = diameter/2 = 0.08/2 = 0.04 m
area of the inner cylinder = 
where
= 3.142,
and r = radius = 0.04 m
area of inner cylinder = 3.142 x
= 0.005 m^2
<em>height h of the water in this cylinder = volume/area</em>
h = 0.002/0.005 = 0.4 m
<em>pressure at the bottom of the cylinder due to the height of water = pgh</em>
where
p = density of water = 1000 kg/m^3
g = acceleration due to gravity = 9.81 m/s^2
h = height of water within this cylinder = 0.4 m
pressure = 1000 x 9.81 x 0.4 = <em>3924 Pa</em>
Answer:
A capon (from Latin: caponem) is a cockerel (rooster) that has been castrated or neutered, either physically or chemically, to improve the quality of its flesh for food, and, in some countries like Spain, fattened by forced feeding.
Explanation:
Well, if you're using the law to work with periods of Earth satellites,
then the most convenient unit is going to be 'hours' for the largest
orbits, or 'minutes' for the LEOs.
But if you're using it to work with periods of planets, asteroids, or
comets, then you'd be working in days or years.