Question

What is the pressure exerted by sea water at a depth of 500 m? The density of sea is 1020 kgm$^{–3}$ and g = 9.8 ms$^{–2}$​

Answer 1

Answer:

Given:

Depth of sea (h) = 500 m.

Density of sea water (d) = 1020kgm$^{–3}$

Acceleration due to gravity g = 9.8 ms$^{–2}$.

To find:

Presurre exerted by sea water at a depth of 500 m.

Solution:

Pressure of sea water (P) = Depth of sea × density of sea water × acceleration due to gravity.

P = h × d × g

$= 500(m) \times 1020 \: (kgm {}^{ - 3} ) \times 9.8 \: (ms {}^{ - 2} )$

= 4998000 Pa.

Thus, pressure exerted by sea water at a depth of 500 m is 4998000 Pa.

Answer 2

Answer:

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