Asked by need help!

The Mariana trench is located in the Pacific Ocean at a depth of about 11 000 m below the surface of the water. The density of seawater is 1025 kg/m3. (a) If an underwater vehicle were to explore such a depth, what force would the water exert on the vehicle's observation window (radius = 0.109 m)? (b) For comparison, determine the weight of a jetliner whose mass is 2.98 x 105 kg.

Answered by bobpursley
What is the weight of water that height?

I will be happy to critique your thinking.
Answered by need help!
I know that the density of the water is 1025kg/m^3. And I know that to get the pressure you divide the force by the are. I also have the formula P2=P1+pgh. I just don't know where to go from here. Please help!
Answered by bobpursley
the weight of a column of water of is density*g*area*height

then pressure is weight/area= density*g*height.
Answered by bobpursley
Now, the force on the window is Net force, which is weight of water+ atmospheric pressure above- air pressure inside the vehicle. I would assume that the air pressure inside is approximately equal to atmospheric pressure, leaving the weight of the water as the force.
Answered by need help!
how would you know the are of the column of water? would you use the radius of the window or something else?
Answered by Taylor
a)P=P of atm+pvh
P= 1.01*10^5+1025*9.81*11000
P=1.117*10^13

A of circle = pi r^2
A=pi*.109^2
A=.03732

P=F/A
1.117*10^13=F/.03732
F=1.117*10^13*.03732
F=4.1686*10^11N!!!
Answered by Tony
ur all wrong.
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