Question

A submarine is stranded on the bottom of the ocean with its hatch 25 m below the surface. In this problem, assume the density of sea water is 1.03 x 10^3 kg/m^3. A) calculate the magnitude of the force in newtons pressing on the hatch from outside by the sea water given it is circular and 0.25m in diameter. The air pressure inside the submarine is 1.00atm (101,325 Pa). B) calculate the magnitude of the force, in newtons needed to open the hatch from the inside

Answer 1

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:

$P_{net}=\frac{F_{net}}{A}..........(1)$

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:

$P_{net}=P_{atm}+P_w-P_s$

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:

$\begin{gathered} P_{net}=1\text{ Atm}+P_w-1\text{ Atm} \\ \\ P_{net}=P_w..........(2) \end{gathered}$

Thus the net pressure on the hatch will be the pressure of the seawater only.

Now, the net pressure Pnet can be calculated as:

$P_{net}=h\rho g$

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:

$\begin{gathered} P_{net}=25\text{ m}\times1.03\times10^3\text{ kg/m}^3\times9.8\text{ m/s}^2 \\ \\ P_{net}=252350\text{ N/m}^2\text{..........\lparen3\rparen} \end{gathered}$

The area of the hatch can e calculated as:

$\begin{gathered} A=\pi r^2 \\ \\ A=\pi\times(\frac{0.25}{2})^2 \\ \\ A=0.0491\text{ m}^2..........(4) \end{gathered}$

Substituting equation (3) and equation (4) in equation (1) and rearranging it, we get:

$\begin{gathered} 252350\text{ N/m}^2=\frac{F_{net}}{0.0491\text{ m}^2} \\ \\ F=252350\text{ N/m}^2\times0.0491\text{ m}^2 \\ \\ F=12390.39\text{ N} \end{gathered}$

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.

Answer 2

(A) The magnitude of the force in newtons pressing on the hatch from outside by the seawater given it is circular and 0.25 m in diameter is 1.24 × 10 ⁴ N.

(B) The force applied needed to open the hatch from the inside is 1.24 × 10 ⁴ N.

What is force?

In the body, force is the result of either a push or a pull. The three main categories of forces are friction, nuclear, and gravitational. For instance, when a hand strikes a wall, the hand puts force on the wall and the wall also exerts a force on the hand. Newton was given a variety of laws to help him understand force.

Given :

The density of seawater, ρ = 1.03 x 10³ kg / m³ ,

The height of the hatch of the submarine, h= 25 m,

The gravitational acceleration, g = 9.8 m / s² ,

The radius of the hatch, r = 0.25 m

Calculate the area of the hatch by the formula given below,

A = π × r² = π  × (0.125 m)²   [r = d / 2]

A = 4.9 x 10 ⁻² m²

(A) Calculate the force of the water by the following formula,

$P = F / A$

F = P  × A            (P = ρ × g × h)

F = ρ × g × h × A

Substitute the values,

F = 1.03 x 10³ kg / m³  × 9.8 m / s²  ×  25 m × 4.9 x 10 ⁻² m²  

F = 12365.15 N or 1.24 × 10 ⁴ N

(B) The force needed to open the hatch  = The Force applied by water on the hatch.

Therefore, the magnitude of the force in newtons pressing on the hatch from outside by the seawater given it is circular and 0.25 m in diameter is 1.24 × 10 ⁴ N, and the force applied needed to open the hatch from the inside is 1.24 × 10 ⁴ N.

To know more about Force:

brainly.com/question/13191643

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