A research submarine has a 10-cm-diameter window that is 8.4 cm thick. The manufacturer says the window can withstand forces up to 1.2×106 N . What is the submarine's maximum safe depth in salt water?

The pressure inside the submarine is maintained at 1.0 atm .

Express your answer using two significant figures.

Pnet = Po + dgh

Density of saltwater = 1030 kg/m^3.

Disregard the thickness. Assuming it's a circular window, then the area is pi(r^2).

d = 20 cm = 0.2 m
r = d/2 = 0.1 m

A = pi(r^2)
A = 3.14159265(.1^2)
A = 0.0314159265 m^2

p = F/A
p = (1.1 x 10^6) / (0.0314159265)
p = 35,014,087.5 Pa

1 atm = 101,325 Pa

P = Po + dgh
h = (P - Po) / dg
h = (35,014,087.5 - 101,325) / (1030 x 9.81)
h = 3 455.23812 m
h = 3.5 km

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ariston ariston · Answer 2 · 2019-03-05T03:44:26-05:00

Answer: 15 km

Explanation:

The maximum pressure that the submarine can withheld would be equal to the maximum pressure that the window can withstand.

The maximum pressure that window can sustain is given by:

\(Pressure\frac{force}{Area]\)

Force = 1.2×10⁶ N

Area of the window = \(\pi r^{2}= 7.85\times 10^{-5} m^2\)

∵radius = (0.1 m)/2 = 0.05 m

⇒P = 1.2×10⁶ N ÷ 7.85 ×10⁻³ m² = 1.52 ×10⁸ N/m²

The pressure of the salt water at depth h is given by

P = ρ g h

where, ρ = 1027 kg/m³ is the density of the salt water, g is the acceleration due to gravity and h is the depth.

⇒1.52×10⁸ Pa = (1027 kg/m³)(9.8 m/s²)(h)

⇒ h ≈ 15 km