A swimmer enters a gloomier world (in one sense) on diving to greater depths. Given that the mean molar absorption coeffi cient of seawater in the visible region is 6.2 × 10−5 dm3 mol−1 cm−1 , calculate the depth at which a diver will experience (a) half the surface intensity of light and (b) one-tenth that intensity.

a) x= 1,118 10⁴ m, b)   x=  3.7 104 m

Explanation:

To calculate the decrease in intensity with depth, write the equation

          I = I₀  

Where Io is the intensity at the surface and alf is the absorption coefficient

        Ln (I / I₀) = - α x

        x = -1 /α   ln (I / I₀)

Let's apply this expression to our case

a) intensity is half of the surface

         I / I₀ = 0.5

We calculate

         x = -1 /6.2 10⁻⁵  ln (0.5)

         x = 0.1118 10⁵ m = 1,118 10⁴ m

b) the intensity is one tenth

         I / Io = 0.1

         x = -1 / 6.2 10⁵  ln (0.1)

         x = 0.37138 10⁵ = 3.7 104 m

we're given that v₀ = 0 and a = 5.4 m/s². therefore, since v = at + v₀ for constant acceleration the equation for velocity is

after 28s,

m/s

which i would round to 150 m/s as my final answer