Under the sea

In Jules Verne's famous novel "Twenty Thousand Leagues Under the Sea", three heroes – professor Aronnax with his servant Conseil and the harpooner Ned Land – experience a journey by submarine Nautilus under the command of captain Nemo.
Suppose that the average density of sea water during the entire journey was ϱm = 1.028 g/cm³.
a) What distance d did Verne have in mind, when the original French "lieues" corresponded instead of miles to the so-called "walking hour", i.e. a distance of 4 km?
b) The submarine Nautilus had according to the description in the book a volume of V = 1 500 m³. What must be the mass m of the submarine fully submerged under the surface so that it neither sinks to the bottom nor rises to the surface?
c) What was the mass m1 of the submarine with empty balance tanks, if after surfacing 1/10 of the volume of the submarine was above the surface?
d) In one part of the book, the Nautilus dived to a depth of h = 16 000 m. By how much greater pressure than on the surface would the manometers of the submarine measure at this? Could it dive so deeply?
e) Shortly after, the Nautilus surfaced from a depth of h1 = 13 000 m to the surface in a time of t = 4 minutes. At what average speed v would it have to move? Is this plausible?

Final Answer:

d =  80000 km
m =  1542 t
m₁ =  1387.8 t
P =  161354.88 kPa
v =  54.1667 m/s

Step-by-step explanation:

$d=20000\cdot 4=80000\text{km}$

$$\begin{gathered} V=1500{\text{m}}^3 \\ \rho =1.028{\text{t/m}}^3 \\ m=\rho \cdot V=1.028\cdot 1500=1542\text{t} \end{gathered}$$

$$\begin{gathered} r_1=1-\frac{1}{10}=\frac{1\cdot 10}{10}-\frac{1}{10}=\frac{10}{10}-\frac{1}{10}=\frac{10-1}{10}=\frac{9}{10}=0.9 \\ {m}_1=r_1\cdot m=0.9\cdot 1542=1387.8\text{t} \end{gathered}$$

$$\begin{gathered} h_2=16000\text{m} \\ g=9.81{\text{m/s}}^2 \\ {\rho }_3=1000\cdot \rho =1000\cdot 1.028=1028{\text{kg/m}}^3 \\ P=h_2\cdot g\cdot {\rho }_3/1000=16000\cdot 9.81\cdot 1028/1000=161354.88\text{kPa} \end{gathered}$$

$$\begin{gathered} h_1=13000\text{m} \\ t=4\text{min}\to \text{s}=4\cdot 60\text{s}=240\text{s} \\ v=h_1/t=13000/240=54.1667\text{m/s} \end{gathered}$$

Try another example