Exercise 2-5 Solution

ð  Problem 1

This problem involves finding the mass given the volume.

mass = density x volume

mass = 3x10¹¹ kg

 

ð  Problem 5

 

The specific gravity of the fluid is found by taking the ratio of the densityh of the fluid to that of water - using the same volume for both liquids

 

SG_fluid = r_fluid/r_water = (m/V)_fluid/(m/V)_water = 0.8477

 

 

ð  Problem 7

 

a. P = W_leg/A = 7x10⁷ N/m²

 

b. P = W_elephant/A = 2x10⁵ N/m²

 

 

ð  Problem 14

 

a. To find absolute pressure, we use equation 10-3c. The total force is the absolute pressure times the area of the bottom of the pool.

 

P = P_o + rgh = 1.21 x 10⁵ N/m²

 

F = PA = 2.3x10⁷N

 

b. The pressure against the pool's side near the bottom is the same as the pressure at the bottom.

 

P = P_o + rgh = 1.21 x 10⁵ N/m²

 

 

ð  Problem 22

 

We know tht the difference in the actual mass and the apparent mass is just the mass of the water displaced by the rock.  The mass of the water displaced is the volume times its density; the folume of the rock is its jmass divided by its density. Combine these relationships to get an expression for the density of the rock.

 

density of the rock = 2.99 x 10³ kg/m³

 

 

ð  Problem 35

 

This is an example of the use of equation 10-4b, the equation of continuity for an incompressible fluid.  We compare the blood flow in the two areas.

 

(Av)_aorta = (Av)_arteries  simplifying gives

 

v_arteries = (A_aorta/A_arteries)v_aorta   =  0.9 m/s

 

 

ð  Problem 38

 

Use Torricelli's theorem to solve for v = 9.5 m/s

 

ð  Problem 45

 

We estimate the pressure of the air inside the hurricane by using Bernoulli's equation. We make the assumption that the pressure outside the hurricane is ari pressure, the speed of the wind outside the hurrican is 0. We also assume that the 2 pressure measurements are made at the same height.

 

Pressure = 0.96 atm