Respuesta :
Given:
The height (depth) of the hatch from the seawater surface is: h = 25 m
The density of seawater is: ρ = 1.03 × 10³ kg/m³
The diameter of the hatch is: d = 0.25 m
The pressure inside the submarine is: Ps = 1 Atm = 101325 Pa
To find:
The magnitude of the force pressing the hatch from the outside by the seawater and the magnitude of force to open the hatch from the inside.
Explanation:
The net pressure on the hatch area is given as:
[tex]P_{net}=\frac{F_{net}}{A}..........(1)[/tex]Here, Pnet is the net pressure on the hatch of area A, and Fnet is the net force acting on the hatch from the outside.
The net pressure Pnet is given as:
[tex]P_{net}=P_{atm}+P_w-P_s[/tex]Here, Patm is the atmospheric pressure above the seawater surface, Pw is the pressure of the water and Ps is the pressure inside the submarine.
The atmospheric pressure Patm of 1 Atm will contribute to the net pressure on the hatch from the outside. The atmospheric pressure Patm will have no effect as it will cancel out with the pressure Ps of equal magnitude as it is in an opposite direction from inside the submarine.
Substituting the values of the atmospheric pressure Patm and the pressure inside the submarine Ps in the above equation, we get:
[tex]\begin{gathered} P_{net}=1\text{ Atm}+P_w-1\text{ Atm} \\ \\ P_{net}=P_w..........(2) \end{gathered}[/tex]Thus the net pressure on the hatch will be the pressure of the seawater only.
Now, the net pressure Pnet can be calculated as:
[tex]P_{net}=h\rho g[/tex]Here, g is the acceleration due to the gravity having a value of 9.8 m/s².
Substituting the values in the above equation, we get:
[tex]\begin{gathered} P_{net}=25\text{ m}\times1.03\times10^3\text{ kg/m}^3\times9.8\text{ m/s}^2 \\ \\ P_{net}=252350\text{ N/m}^2\text{..........\lparen3\rparen} \end{gathered}[/tex]The area of the hatch can e calculated as:
[tex]\begin{gathered} A=\pi r^2 \\ \\ A=\pi\times(\frac{0.25}{2})^2 \\ \\ A=0.0491\text{ m}^2..........(4) \end{gathered}[/tex]Substituting equation (3) and equation (4) in equation (1) and rearranging it, we get:
[tex]\begin{gathered} 252350\text{ N/m}^2=\frac{F_{net}}{0.0491\text{ m}^2} \\ \\ F=252350\text{ N/m}^2\times0.0491\text{ m}^2 \\ \\ F=12390.39\text{ N} \end{gathered}[/tex]A force of 12390.39 Newtons acts on the hatch by the seawater above it. To open the hatch from the inside, an equal magnitude of force is needed to apply from inside the submarine. Thus, a force of 12390 Newtons is required to open the hatch from the inside.
Final answer:
The magnitude of force pressing the hatch from the outside by the seawater is 12390.39 Newtons.
The magnitude of force required to open the hatch from inside the submarine is 12390.39 Newtons.
