Question

Water has a density that is 13.6 times less than that of mercury. If an undersea vessel descends to 1.6 km, how much pressure does the water exert in atm?

Answer 1

Answer:

155.25 atm

Explanation:

Distance, h = 1.6 km = 1600 m

density of water = 13.6 times less than the density of mercury

density of mercury = 13.6 x 10^3 kg/m^3

So, density of water = 1 x 10^3 kg/m^3

Pressure exerted by the water

P = distance x density of water x acceleration due to gravity

P = 1600 x 1000 x 9.8

$P=1.568\times10^{8}Pa$

We know that

1 atm = 1.01 x 10^5 Pa

So, $P=1.568\times10^{8}Pa$ = $P=\frac{1.568\times10^{8}}{1.01\times10^{5}}=155.25 atm$