Given:
Depth of the bottom = 11.0 km
Let's calculate the pressure due to the ocean at a depth of 10.6 km.
Apply the formula:
![P=P_o+\rho gh](https://img.qammunity.org/2023/formulas/physics/college/2cf4tvvfmxih5gb67lg8tn3na732bppapx.png)
Where:
P is the pressure
ρ is the density of sea water = 1027 kg/m³
g is the acceleration due to gravity = 9.8 m/s²
h = 10.6 km = 19.6 x 10³ m
Po = 1 atm
Thus, we have:
![\begin{gathered} P=1\text{ atm+\lparen1027}*9.8*10.6*10^3) \\ \\ P=1\text{ atm+\lparen106684.76 *10}^3\text{\rparen} \end{gathered}](https://img.qammunity.org/2023/formulas/physics/college/l27tmes9x3dbt746gmsohki0yg7zhhf92n.png)
Where:
1 atm = 101325 Pa
We have:
![\begin{gathered} P=1\text{ atm + }(106684.76*10^3)/(101325) \\ \\ P=1+1052.90 \\ \\ P=1053.90\text{ atm} \end{gathered}](https://img.qammunity.org/2023/formulas/physics/college/e1etdskausvq1kvmep4rb3rxgeq08k9pgy.png)
Therefore, the pressure due to the ocean at the depth of 10.6 km is 1053.90 atm.
ANSWER:
1053.90 atm