To find the depth of the submarine below the surface of the water, we can use the relationship between pressure, density, gravity, and depth in fluids. The formula for pressure in a fluid is given by:
[tex]\[
P = \rho g h
\][/tex]
where:
- [tex]\( P \)[/tex] is the pressure,
- [tex]\( \rho \)[/tex] is the density of the fluid (in this case, water),
- [tex]\( g \)[/tex] is the acceleration due to gravity,
- [tex]\( h \)[/tex] is the depth.
We can rearrange this formula to solve for the depth [tex]\( h \)[/tex]:
[tex]\[
h = \frac{P}{\rho g}
\][/tex]
Given values are:
- The density of water, [tex]\( \rho = 10^3 \, \text{kg/m}^3 \)[/tex],
- The water pressure on the submarine, [tex]\( P = 2 \times 10^7 \, \text{N/m}^2 \)[/tex],
- The acceleration due to gravity, [tex]\( g = 9.81 \, \text{m/s}^2 \)[/tex].
Plug these values into the formula:
[tex]\[
h = \frac{2 \times 10^7 \, \text{N/m}^2}{10^3 \, \text{kg/m}^3 \times 9.81 \, \text{m/s}^2}
\][/tex]
Now, perform the division:
[tex]\[
h = \frac{2 \times 10^7}{10^3 \times 9.81}
\][/tex]
[tex]\[
h = \frac{2 \times 10^7}{9.81 \times 10^3}
\][/tex]
[tex]\[
h = \frac{2 \times 10^7}{9.81 \times 10^3} = \frac{2 \times 10^7}{9.81 \times 10^3} \approx 2038.74 \, \text{meters}
\][/tex]
Therefore, the depth of the submarine below the surface of the water is approximately 2038.74 meters.
