## Pressure Difference

At 10,994 meters below sea level, the Mariana Trench is the deepest known point in the Earth’s ocean. If a submarine is exploring the ocean depths 2,500 m above the Trench’s floor, what is the difference in pressure between the submarine and sea level? Assume the density of salt water is 1,026 kg/m^3 and is constant throughout the ocean’s depths.

##
__
__**Hint**

**Hint**

The pressure difference between two different points:

$$$P_2-P_1=-\gamma(z_2-z_1)=-\gamma h=-\rho gh$$$

where
$$\gamma $$
is the liquid’s specific weight,
$$\rho$$
is the liquid’s density, and
$$g$$
is the acceleration due to gravity

##
__
__**Hint 2**

**Hint 2**

$$P_2$$
,
$$P_1$$
,
$$z_1$$
,
$$z_2$$
, and
$$h$$
are defined as:

The pressure difference between two different points:

$$$P_2-P_1=-\gamma(z_2-z_1)=-\gamma h=-\rho gh$$$

where
$$\gamma $$
is the liquid’s specific weight,
$$\rho$$
is the liquid’s density,
$$g$$
is the acceleration due to gravity, and
$$P_2$$
,
$$P_1$$
,
$$z_1$$
,
$$z_2$$
, and
$$h$$
are defined as:

Based on the problem statement, the submarine is at
$$P_1$$
and sea level is at
$$P_2$$
. It also gives
$$z_1$$
and
$$z_2$$
for the fact about the Mariana Trench’s depth and the submarine’s location. Thus,

$$$P_2-P_1=\rho gh=\rho g(z_2-z_1)=(1,026\frac{kg}{m^3})(9.8\frac{m}{s^2})(10,994m-2,500m)$$$

$$$\Delta P=10,054.8\frac{N}{m^3}(8,494m)=85,405,471\frac{N}{m^2}=85\:MPa$$$

85 MPa