In the days before scuba gear, some divers descended to underwater depths in diving bells, which are basically just upside-down containers whose open ends face down. The bell allows the person inside to breathe the air trapped inside it, observe underwater objects and marine life, and work under the water. Required:
a. If the bell is submerged to a depth of 30m below sea level, what is the water pressure at the air-water interface inside the bell?
b. If the air pressure inside the bell before submersion into the water was 1 atm (101.825 kPa), what air pressure does the person experience at that depth?

"" would be the pressure outside the bell .as well as the air pressure inside it. A further explanation is below.

According to the question,

Depth,

Pressure,

Now,

→ The water pressure outside the bell will be:

=

By putting the values, we get

=

=

And inside the air pressure will be same as the water pressure.

Thus the response above is right.

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a)  P = 4.03 10⁵ Pa, b)  P = 4.03 10⁵ Pa

Explanation:

a) The pressure as a function of the depth of a fluid is

          P = + ρ g y

where Patm is the atmospheric pressure, the sea density of about 1025 kg / m³

let's calculate

         P = 1.01825 10⁵ + 1025 9.8 30

         P = 4.03 10⁵ Pa

b) When the hood is submerged, the water exerts a perpendicular force on the entire surface, in the equilibrium position, the air is compressed by this force until the pressure it exerts is equal to the external pressure (open at the lower), therefore the air pressure is

        P = 4.03 105 Pa

find the sum of the two horizontal vectors and the sum of the two vertical vectors

isoceles right angled triangle