Answer
(a) $F=3.4780\times10^6N$
(b) Weight of jetliner $W=1.1772\times10^6N$
Work Step by Step
The density of sea water is given as $\rho=1025kg/m^3$
depth of water is given as $h=11000m$
so pressure it this depth $P=P_{a}+\rho g h$
$P_{a}$=atmospheric pressure $=1.01325\times10^5Pa= 1.01325\times10^5N/m^2$
$g=9.81m/s^2$
$\rho=1025kg/m^3$
$h=11000m$
$P=1.01325\times10^5N/m^2+1025kg/m^3\times9.81m/s^2\times 11000m$
$P=1.01325\times10^5N/m^2 +1.1060775\times10^8N/m^2$
$P=1.1070907\times10^8N/m^2$
(a)
given observation window radius $r=0.10m$
area of the window $A=\pi r^2=\pi (0.10m)^2$
window $A=3.14159\times10^{-2}m^2$
since $P=\frac{F}{A}$
so force on window $F=PA$
$F=1.1070907\times10^8N/m^2\times3.14159\times10^{-2}m^2$
$F=3.4780\times10^6N$
(b)
Mass of the jetliner $M=1.2\times10^5kg$
acceleration due to gravity $g=9.81m/s^2$
so weight of the jetliner $W=Mg$
$W=1.2\times10^5kg\times9.81m/s^2$
$W=11.772\times10^5N$
$W=1.1772\times10^6N$
we can say that force is almost double that of the weight of jetliner.
