Libro Fisica Universitaria 12a Zemansky FLUID MECHANICS

Libro Fisica Universitaria 12a Zemansky

FLUID MECHANICS

14.1. IDENTIFY: Use Eq.(14.1) to calculate the mass and then use w mg = to calculate the weight.

SET UP: ρ = m V/ so m V = ρ From Table 14.1, 3 3 ρ = × 7.8 10 kg/m .

EXECUTE: For a cylinder of length L and radius R, 2 2 43 V RL ( ) (0.01425 m) (0.858 m) 5.474 10 m . π π − = = =×

Then 3 3 43 m Vρ (7.8 10 kg/m )(5.474 10 m ) 4.27 kg, − == × × = and 2 w mg == = (4.27 kg)(9.80 m/s ) 41.8 N (about

9.4 lbs). A cart is not needed.

EVALUATE: The rod is less than 1m long and less than 3 cm in diameter, so a weight of around 10 lbs seems

reasonable.

14.2. IDENTIFY: Convert gallons to kg. The mass m of a volume V of gasoline is m V = ρ .

SET UP: 3 3 1 gal 3.788 L 3.788 10 m − = =× . 3 1 m of gasoline has a mass of 737 kg.

EXECUTE:

3

3 3

1 gal 1 m 45.0 mi/gal (45.0 mi/gal) 16.1 mi/kg 3.788 10 m 737 kg −

⎛ ⎞⎛ ⎞ = = ⎜ ⎟⎜ ⎟ ⎝ ⎠ × ⎝ ⎠

EVALUATE: 1 gallon of gasoline has a mass of 2.79 kg. The car goes fewer miles on 1 kg than on 1 gal, since

1 kg of gasoline is less gasoline than 1 gal of gasoline.

14.3. IDENTIFY: ρ = m V/

SET UP: The density of gold is 3 3 19.3 10 kg/m × .

EXECUTE: 3 3 3 63 V (5.0 10 m)(15.0 10 m)(30.0 10 m) 2.25 10 m −− − − =× × × = × .

3 3

6 3

0.0158 kg 7.02 10 kg/m 2.25 10 m

m

V ρ − == = × × . The metal is not pure gold.

EVALUATE: The average density is only 36% that of gold, so at most 36% of the mass is gold.

14.4. IDENTIFY: Find the mass of gold that has a value of 6 $1.00 10 × . Then use the density of gold to find the volume

of this mass of gold.

SET UP: For gold, 3 3 ρ = × 19.3 10 kg/m . The volume V of a cube is related to the length L of one side by 3 V L = .

EXECUTE:

3

6 1 troy ounce 31.1035 10 kg ($1.00 10 ) 72.9 kg $426.60 1 troy ounce m

− ⎛ ⎞⎛ ⎞ × =× = ⎜ ⎟⎜ ⎟ ⎝ ⎠⎝ ⎠

. m

V ρ = so

3 3

3 3

72.9 kg 3.78 10 m

19.3 10 kg/m

m V

ρ

− == = × × . 1/3 L V== = 0.156 m 15.6 cm .

EVALUATE: The cube of gold would weigh about 160 lbs.

14.5. IDENTIFY: Apply / ρ = m V to relate the densities and volumes for the two spheres.

SET UP: For a sphere, 4 3

3 V r = π . For lead, 3 3

l ρ = × 11.3 10 kg/m and for aluminum, 3 3

a ρ = × 2.7 10 kg/m .

EXECUTE: 4 3 m 3 = = ρV πr ρ . Same mass means 3 3

aa 11 r ρ = r ρ .

1 3 1 3 3

a 1

3

1 a

11.3 10 1.6

2.7 10

r ρ

r ρ

⎛ ⎞ ⎛ ⎞ × == = ⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠ × .

EVALUATE: The aluminum sphere is larger,

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