The tip of a triangle is held 12.0 cm above the surface of a flat pool of water. A submerged swimmer in the pool sees the tip of the triangle at what distance above the water? Let the indices of refraction nwater = 1.33 and nair = 1.00.

9cm

Explanation:

Refractive index can also be defined as the ratio of the real depth to the apparent depth.

Given that the

Real depth = 12 m

Refractive index of water = 1.33

Refractive index of air = 1.00

nair/nwater = real depth/apparent depth

Substitute all the parameters into the formula

1.33/1 = 12/ apparent depth

Cross multiply

1.33 Apparent depth = 12

Apparent depth = 12/1.33

Apparent depth = 9.02 cm

Therefore,  A submerged swimmer in the pool sees the tip of the triangle at 9cm approximately distance above the water.

c, 1a hope this

f=ma a/f/m a= 480/60 a = 8m/s/s that is quick wow