A laser positioned on a ship is used to communicate with a small two man research submarine resting on the bottom of a lake. The laser is positioned 12 m above the surface of the water, and it strikes the water 20 m from the side of the ship. The water is 76 m deep and has an index of refraction of 1.33. How far is the submarine from the side of the ship

84.1 m

Explanation:

Given :

The distance from the ship to submarine :

20 + y

Using Pythagoras :

Tan θ = opposite / Adjacent

Tan θ = 20 / 12

12 tan θ = 20

θ = tan^-1(20/12)

20

θ = 59.036°

The angle phi;

n1sinθ1 = n2sin θ

Sin 59.036 = 1.33 * sin phi

Sin phi = sinsin(59.04) ÷1.33

0.8574907 = 1.33 * sin phi

Sin phi = 0.8574907 / 1.33

Sin phi = 0.6447298

phi = sin(0.6447298

Phi = 40.15°

From Pythagoras :

y = 76tan40.15°

y = 76 * 0.8435707

y = 64.11

20 + y

20 + 64.11 = 84.11

a electromagnetic waves

a=d/t

a="triangle"v/t

explanation:

a = acceleration