The harbour contains salt water while the river contains
fresh water. So assuming that the densities of fresh water and salt water are:
density (salt water) = 1029 kg / m^3
density (fresh water) = 1000 kg / m^3
The amount of water (in mass) displaced by the barge
should be equal in two waters.
mass displaced (salt water) = mass displaced (fresh
water)
Since mass is also the product of density and volume, therefore:
<span>[density * volume]_salt water = [density * volume]_fresh
water ---> 1</span>
First we calculate the amount of volume displaced in the harbour
(salt water):
V = 3.0 m * 20.0 m * 0.70 m
V = 42 m^3 of salt water
Plugging in the values into equation 1:
1029 kg / m^3 * 42 m^3 = 1000 kg/m^3 * Volume fresh water
Volume fresh water displaced = 43.218 m^3
Therefore the depth of the barge in the river is:
43.218 m^3 = 3.0 m * 20.0 m * h
<span>h = 0.72 m (ANSWER)</span>
