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A boat has a porthole window, of area 0.00849 m^2, 6.25 m below the surface. The density of sea water is 1027 kg/m^3. The air inside the boat is at 1 atm. What is the net force on the window?

User Clarkitect
by
5.4k points

2 Answers

11 votes

Answer: 534.6 N

Step-by-step explanation:

This answer works for Acellus!

User Haris Ur Rehman
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5.2k points
7 votes

Answer:


325.65\ \text{N} outward.

Step-by-step explanation:


\rho = Density of water =
1027\ \text{kg/m}^3

g = Acceleration due to gravity =
9.81\ \text{m/s}^2

h = Height of the water = 6.25 m

A = Area of the window =
0.00849\ \text{m}^2


P_i = Internal pressure of the boat =
1\ \text{atm}=101325\ \text{Pa}

Pressure by the water on the window


P=\rho gh\\\Rightarrow P=1027* 9.81* 6.25\\\Rightarrow P=62967.9375\ \text{Pa}

Net pressure on the window


P_n=P-P_i=62967.9375-101325=-38357.0625\ \text{Pa}

Force is given by


F=P_nA\\\Rightarrow F=-38357.0625* 0.00849\\\Rightarrow F=-325.651460625\ \text{N}

Net force on the window is
325.65\ \text{N} outward.

User Ihor Pomaranskyy
by
5.4k points