Question

At a depth of 500m in an ocean.what is the absolute pressure?Given that the density of sea water is 1.03×10^3kg/m^3 and g=10m/s​

Answer 1

Answer:-

Here

$\sf {\bull P_a=1.01\times 10^5Pa }\\ \sf\ {\; \atop \bull\rho=1.03\times 10^3kgm^{-3}}$

We know that

$\boxed{\sf Absolute\:Pressure(P)==P_a+\rho gh}$

$\\ \rm\longmapsto P=1.03\times 10^5Pa+1.03\times 10^3\times 10\times 500Pa$

$\\ \rm\longmapsto P=1.01\times 10^5Pa+51.5\times 10^5Pa$

$\\ \rm\longmapsto P=52.5\times 10^5Pa$

$\\ \rm\longmapsto P=52atm$

$\\ \underline{\boxed{\bf{\therefore Absolute\:Pressure=52atm.}}}$

Answer 2

Answer:

5251325 Pa = 5.25 × 10^6 Pa

Explanation:

Depth = 500m

Density of sea water = 1.03 × 10^3 kg/m^3

Acceleration due to gravity = 10 m/s^2

Atmospheric pressure = 101325 Pa

Absolute Pressure = p

$p = p \frac{}{a} + (rho)gh \\ = 101325 + (1.03 \times {10}^{3} \times 10 \times 500) \\ = 101325 \: + (515 \times {10}^{4} ) \\ = 101325 \: + \: 5150000 \\ = 5251325 = 5.25 \times {10}^{6} \: kg \: {m}^{ - 1} \: {s}^{ - 2}$