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YF_ISM_14

YF_ISM_14 - FLUID MECHANICS 14 14.1 IDENTIFY Use Eq(14.1 to...

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14-1 F LUID M ECHANICS 14.1. IDENTIFY: Use Eq.(14.1) to calculate the mass and then use wm g = to calculate the weight. SET UP: / mV ρ = so = From Table 14.1, 33 7.8 10 kg/m . EXECUTE: For a cylinder of length L and radius R , 22 4 3 ( ) (0.01425 m) (0.858 m) 5.474 10 m . VR L ππ == = × Then 4 3 (7.8 10 kg/m )(5.474 10 m ) 4.27 kg, == × × = and 2 (4.27 kg)(9.80 m/s ) 41.8 N g = (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 = . SET UP: 1 gal 3.788 L 3.788 10 m × . 3 1 m of gasoline has a mass of 737 kg. EXECUTE: 3 1 gal 1 m 45.0 mi/gal (45.0 mi/gal) 16.1 mi/kg 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: / = SET UP: The density of gold is 19.3 10 kg/m × . EXECUTE: 333 6 3 (5.0 10 m)(15.0 10 m)(30.0 10 m) 2.25 10 m V −−− × × = × . 63 0.0158 kg 7.02 10 kg/m m V = × × . 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, . The volume V of a cube is related to the length L of one side by 3 VL = . EXECUTE: 3 6 1 troy ounce 31.1035 10 kg (\$1.00 10 ) 72.9 kg \$426.60 1 troy ounce m × = . m V = so 72.9 kg 3.78 10 m m V × . 1/3 0.156 m 15.6 cm LV = . EVALUATE: The cube of gold would weigh about 160 lbs. 14.5. IDENTIFY: Apply / = to relate the densities and volumes for the two spheres. SET UP: For a sphere, 3 4 3 Vr π = . For lead, l 11.3 10 kg/m and for aluminum, a 2.7 10 kg/m . EXECUTE: 3 4 3 m ρ V π r ρ . Same mass means aa 11 r ρ r ρ = . 13 3 a1 3 1a 11.3 10 1.6 2.7 10 r ρ r ρ × = × . EVALUATE: The aluminum sphere is larger, since its density is less. 14.6. IDENTIFY: Average density is / = . SET UP: For a sphere, 3 4 3 = . The sun has mass 30 sun 1.99 10 kg M and radius 8 6.96 10 m × . EXECUTE: (a) 30 30 sun 83 2 73 4 sun 3 1.409 10 kg/m (6.96 10 m) 1.412 10 m M V π ×× = (b) 30 30 16 3 43 1 3 3 4 3 5.94 10 kg/m (2.00 10 m) 3.351 10 m π = × 14

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14-2 Chapter 14 EVALUATE: For comparison, the average density of the earth is 33 5.5 10 kg/m . × A neutron star is extremely dense. 14.7. IDENTIFY: wm g = and mV ρ = . Find the volume V of the pipe. SET UP: For a hollow cylinder with inner radius 1 R , outer radius 2 R , and length L the volume is 22 21 () . VR R L π = 2 1 1.25 10 m R and 2 2 1.75 10 m R EXECUTE: 4 3 ([0.0175 m] [0.0125 m] )(1.50 m) 7.07 10 m V = . 4 3 (8.9 10 kg/m )(7.07 10 m ) 6.29 kg == × × = . 61.6 N g == . EVALUATE: The pipe weights about 14 pounds. 14.8. IDENTIFY: The gauge pressure 0 pp at depth h is 0 g h = . SET UP: Ocean water is seawater and has a density of 1.03 10 kg/m × . EXECUTE: 2 7 0 (1.03 10 kg/m )(9.80 m/s )(3200 m) 3.23 10 Pa = × . 7 0 5 1 atm (3.23 10 Pa) 319 atm 1.013 10 Pa ⎛⎞ = ⎜⎟ × ⎝⎠ . EVALUATE: The gauge pressure is about 320 times the atmospheric pressure at the surface.
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YF_ISM_14 - FLUID MECHANICS 14 14.1 IDENTIFY Use Eq(14.1 to...

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