Physics Solution Manual for 1100 and 2101

# 8 77 c chapter 12 answers to focus on concepts

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Unformatted text preview: W = mg (Equation 4.5). 97. REASONING AND SOLUTION The figure at the right shows the two forces that initially act on the box. Since the box is not accelerated, the two forces must have zero resultant: FB0 − Wbox = 0 . Therefore, FB0 = Wbox B0 (1) W From Archimedes' principle, the buoyant force on the box is equal to the weight of the water that is displaced by the box: FB0 = Wdisp F box (2) Combining (1) and (2) we have Wbox = Wdisp, or mbox g = ρwatergVdisp. Therefore, mbox = ρwaterVdisp Since the box floats with one-third of its height beneath the water, Vdisp = (1/3)Vbox, or Vdisp = (1/3)L3. Therefore, Chapter 11 Problems mbox = ρ water L3 (3) 3 The figure at the right shows the three forces that act on the box after water is poured into the box. The box begins to sink when Wbox + Wwater > FB 621 F B (4) The box just begins to sink when the equality is satisfied. From Archimedes' principle, the buoyant force on the system is equal to the weight of the water that is displaced by the system: FB = Wdisplaced. W box W water The equality in (4) can be written as mboxg + mwaterg = mdisplacedg (5) When the box begins to sink, the volume of the water displaced is equal to the volume of the box; Equation (5) then becomes mbox + ρwaterVwater = ρwaterVbox. The volume of water in the box at this instant is Vwater = L2h, where h is the depth of the water in the box. Thus, the equation above becomes mbox + ρwaterL2h = ρwaterL3 Using Equation (3) for the mass of the box, we obtain ρ water L3 3 + ρ water L2 h = ρ water L3 Solving for h gives 2 3 2 3 h = L = (0.30 m) = 0.20 m 98. REASONING AND SOLUTION The mercury, being more dense, will flow from the right container into the left container until the pressure is equalized. Then the pressure at the bottom of the left container will be P = ρwghw + ρmghmL and the pressure at the bottom of the right container will be P = ρmghmR. Equating gives ρwghw + ρmg(hmL − hmR) = 0 (1) 622 FLUIDS Both liquids are incompressible and immiscible so hw = 1.00 m and hmL + hmR = 1.00 m Using these in (1) and solving for hmL gives, hmL = (1/2)(1.00 − ρw/ρm) = 0.46 m. So the fluid level in the left container is 1.00 m + 0.46 m = 1.46 m from the bottom. 99. SSM WWW container is REASONING AND SOLUTION The pressure at the bottom of the P = Patm + ρwghw + ρmghm We want P = 2Patm, and we know h = hw + hm = 1.00 m. Using the above and rearranging gives hm = Patm − ρ w gh ( ρm − ρw ) g = ( )( ) 1.01 × 105 Pa − 1.00 × 103 kg/m3 9.80 m/s 2 (1.00 m ) (13.6 × 10 3 kg/m − 1.00 × 10 kg/m 3 3 3 )(9.80 m/s2 ) = 0.74 m 100. REASONING AND SOLUTION We refer to Table 11.1 for values of 3 3 ρsoda = ρwater = 1.000 × 10 kg/m and ρAl = 2700 kg/m3 for the density of aluminum. The mass of aluminum required to make the can is mAl = mtotal − msoda where the mass of the soda is 3 3 msoda = ρsodaVsoda = (1.000 × 10 kg/m )(3.54 × 10 –4 Therefore, the mass of the aluminum used to make the can is mAl = 0.416 kg − 0.354 kg = 0.062 kg Using the definition of density, ρ = m/V, we have VAl = mAl ρ Al = 0.062 kg = 2.3 ×10 –5 m3 2700 kg/m3 3 m ) = 0.354 kg Chapter 11 Problems 623 101. REASONING The number of gallons per minute of water that the fountain uses is the volume flow rate Q of the water. According to Equation 11.10, the volume flow rate is Q = Av , where A is the cross-sectional area of the pipe and v is the speed at which the water leaves the pipe. The area is given. To find a value for the speed, we will use Bernoulli’s equation. This equation states that the pressure P, the fluid speed v, and the elevation y at any two points (1 and 2) in an ideal fluid of density ρ are related according to 2 2 P + ρ v1 + ρ gy1 = P2 + ρ v2 + ρ gy2 (Equation 11.11). Point 1 is where the water leaves 1 1 2 1 2 the pipe, so we are seeking v1. We must now select point 2. Note that the problem specifies the maximum height to which the water rises. The maximum height occurs when the water comes to a momentary halt at the top of its path in the air. Therefore, we choose this point as point 2, so that we can take advantage of the fact that v2 = 0 m/s. We know that P = P2 = Patmospheric and v2 = 0 m/s. 1 SOLUTION Substituting this information into Bernoulli’s equation, we obtain 2 Patmospheric + ρ v1 + ρ gy1 = Patmospheric + ρ ( 0 m/s ) + ρ gy2 1 2 2 1 2 Solving for v1 gives v1 = 2 g ( y2 − y1 ) , which we now substitute for the speed v in the expression Q = Av for the volume flow rate: Q = Av1 = A 2 g ( y2 − y1 ) ( = 5.00 ×10−4 m 2 ) 2 (9.80 m/s2 ) (5.00 m − 0 m ) = 4.95 ×10−3 m3 / s To convert this result to gal / min, we use the facts that 1 gal = 3.79 × 10−3 m3 and 1 min = 60 s: m3 1 gal Q = 4.95 ×10−3 3.79 × 10−3 m3 s 60 s = 78.4 gal / min 1 min 102. REASONING AND SOLUTION The upward buoyant force must equal the weight of the shell if it is floating, ρwgV = W. The submerged volume of the shell is V = (4/3)π R23, where R2 is its outer radius. Now R23 = (3/4)m/(ρwπ ) gives R2 = 3 3m 4π ρ w = 3 ( 3 (1.00 kg ) 4π 1.00 × 10 kg/m...
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