Explanation:
Relation between gauge pressure and density and height is as follows.
[tex]P_{gauge} = \rho \times g \times h[/tex]
The given data is as follows.
[tex]\rho[/tex] = 1025 [tex]kg/m^{3}[/tex]
g = 9.81 [tex]m/s^{2}[/tex]
h = 2.5 m
Therefore, substitute the given values into the above formula as follows.
[tex]P_{gauge} = \rho \times g \times h[/tex]
= [tex]1025 kg/m^{3} \times 9.81 m/s^{2} \times 2.5 m[/tex]
= [tex]25138.125 N/m^{2}[/tex]
or, = [tex]25.138125 kN/m^{2}[/tex]
= 25.138125 kPa
Thus, we can conclude that the gauge pressure exerted by sea-water is 25.138125 kPa.
