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Certain underwater vehicles are capable of exploring some of the deepest regions of the ocean. consider the density of seawater to be 1025 kg/m3. if such a vehicle were submerged to a depth of 8513 m below the surface, what force would be exerted on the vehicle's observation window (radius = 9.3 cm)?

Respuesta :

skyluke89
The pressure exerted by the water at a depth of h=8513 m below the surface is given by Stevin's law:
[tex]p=p_a + \rho g h[/tex]
where
[tex]p_a =1.013\cdot 10^5 Pa[/tex] is the atmospheric pressure
[tex]\rho=1025 kg/m^3[/tex] is the water density
[tex]g=9.81 m/s^2[/tex] is the gravitational acceleration
[tex]h=8513 m[/tex] is the depth

If we plug the numbers into the formula, we find the pressure exerted on the window of the vehicle:
[tex]p=1.013 \cdot 10^5 Pa + (1025 kg/m^3)(9.81 m/s^2)(8513 m)=8.57 \cdot 10^7 Pa[/tex]

The area of the window is
[tex]A=\pi r^2 = \pi (0.093 m)^2 = 0.027 m^2[/tex]

And so the force exerted on the window is
[tex]F=pA=(8.57 \cdot 10^7 Pa)(0.027 m^2)=2.31 \cdot 10^6 N[/tex]

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