We MUST Maintain Continuity - The Continuity Equation

Q10,13
P35,36,71

Question 10. For the barge to float with its top lower, the bottom of the barge must be at a greater depth. This means the buoyancy force must be greater. For this to happen and the barge still be in equilibrium, the mass of the boat must be greater. Therefore sand must be added to the barge.

Q 13. The buoyancy force depends only on g, the fluid density, and the submerged volume. The buoyancy force must therefore be the same whether it is submerged completely just below the surface, or deep below, since the submerged volume is the same in both cases.

Problems:
35. We are given the volume of the room, and are told that the air in the room is replaced in 10 minutes. This gives us the volumetric flow rate into/out of the room: Q = 0.345 m^3/s.

This is equal to the volumetric flow rate in, with A₁v₁ = Q.

Solving, v₁ = 3.80 m/s

36. The volume of the pool when it is filled is pi*(3.6 m)^2*(1.5 m) = 61.07 m^3

5/8 inch = .015875 m

Q = A₁v₁ = volume/delta_t

so delta_t = volume/(A₁v₁) = (61.07 m^3/(pi*(.015875 m)^2*.28 m/s) = 76.5 hours

71. Gauge pressure of -80 mm Hg means the atmospheric pressure is 680 mm Hg.

Using the conversion factor that 760 mm Hg = 1.013 x 10⁵ Pa

The pressure in the lungs is 9.064 x 10⁴ Pa

You should now imagine that the lungs are a closed container of gas on top of a column of water (like the problems we did on manometer day.)

Comparing the pressure at the bottom of the water column with the atmosphere:

P_lungs + rho_water*g*H = Patm

H = 1.088 m