Answer:
F=126339.5N
Explanation:
To compute the force required to escape, a free-body diagram for the hatch must be drawn. We will equate the downward and upward forces, thus applying the following equation:
Fw=W+Fi+F
where
Fw= force or weight exerted by the water column above the submarine.
To calculate Fw, we can use:
Fw=h. γ. A
h=height
γ=
specific weight of seawater = 10074N / m ^ 3
A=Area
Fw=28x10074x0.7=197467N
w represents the hatch weight = 200N
Fi denotes the internal pressure force in the submarine, which is 1 atm = 101325Pa. We can calculate this force using:
Fi=PA=101325x0.7=70927.5N
Finally, the force needed to open the hatch is determined by the original equation:
Fw=W+Fi+F
F=Fw-W+Fi
F=197467N-200N-70927.5N
F=126339.5N
The structure with the least oxygen is the vena cava. I hope this information is useful!
Torque = Force * Distance perpendicular
Torque = 15 N * 2 m = 30 Nm
The density of mercury in its liquid form is

We understand that the equation determining the pressure at the base of a fluid column can be expressed through Stevin's law

where

represents the density of the liquid
g signifies the acceleration due to gravity
h indicates the height of the fluid column
Given that the pressure at the lower section of the beaker is

, we can manipulate the preceding formula to calculate the height of the mercury column