Physics Solution Manual for 1100 and 2101

095 10 6 m 3 08467 vcopper vcopper g solving for

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Unformatted text preview: s volume is beneath the water. Thus, ( ) ρduckVduck g = ρ water 1 Vduck g 4 14 244 4 3 1442443 Weight of duck Magnitude of buoyant force Solving this equation for the average density of the duck (and taking the density of water from Table 11.1) gives ρduck = 1 4 ρ water = 1 4 (1.00 ×103 kg/m3 ) = 250 kg/m3 44. REASONING The density ρ of an object is equal to its mass m divided by its volume V, or ρ = m/V (Equation 11.1). The volume of a sphere is V = 4 π r 3 , where r is the radius. 3 According to the discussion of Archimedes’ principle in Section 11.6, any object that is solid throughout will float in a liquid if the density of the object is less than or equal to the density of the liquid. If not, the object will sink. SOLUTION a. The average density of the sun is ρ= m m 1.99 ×1030 kg = = V 4 π r 3 4 π 6.96 ×108 m 3 3 ( ) 3 = 1.41× 103 kg/m3 b. Since the average density of the solid object (1.41 × 103 kg/m3) is greater than that of 3 3 water (1.00 × 10 kg/m , see Table 11.1), the object will sink . c. The average density of Saturn is 588 FLUIDS ρ= m m 5.7 ×1026 kg = = 4 V 3 π r 3 4 π 6.0 ×107 m 3 ( ) 3 = 0.63 ×103 kg/m3 3 3 The average density of this solid object (0.63 × 10 kg/m ) is less than that of water (1.00 × 3 3 10 kg/m ), so the object will float . 45. SSM WWW REASONING According to Equation 11.1, the density of the life jacket is its mass divided by its volume. The volume is given. To obtain the mass, we note that the person wearing the life jacket is floating, so that the upward-acting buoyant force balances the downward-acting weight of the person and life jacket. The magnitude of the buoyant force is the weight of the displaced water, according to Archimedes’ principle. We can express each of the weights as mg (Equation 4.5) and then relate the mass of the displaced water to the density of water and the displaced volume by using Equation 11.1. SOLUTION According to Equation 11.1, the density of the life jacket is mJ ρJ = (1) VJ Since the person wearing the life jacket is floating, the upward-acting buoyant force FB balances the downward-acting weight WP of the person and the weight WJ of the life jacket. The buoyant force has a magnitude that equals the weight WH 2 O of the displaced water, as stated by Archimedes’ principle. Thus, we have FB = WH 2 O = WP + WJ (2) In Equation (2), we can use Equation 4.5 to express each weight as mass m times the magnitude g of the acceleration due to gravity. Then, the mass of the water can be expressed as mH O = ρ H OVH O (Equation 11.1). With these substitutions, Equation (2) becomes 2 2 2 mH O g = mP g + mJ g ( ρ H OVH O ) g = mP g + mJ g or 2 2 2 Solving this result for mJ shows that mJ = ρ H OVH 2 2 O − mP Chapter 11 Problems 589 Substituting this result into Equation (1) and noting that the volume of the displaced water is VH O = 3.1× 10−2 m3 + 6.2 × 10−2 m3 gives 2 ρJ = ρ H OVH 2 2 O − mP (1.00 ×103 kg/m3 ) (3.1×10−2 m3 + 6.2 ×10−2 m3 ) − 81 kg = 390 kg/m3 = 3.1×10−2 m3 VJ 46. REASONING The mass m of the shipping container is related to its weight by W = mg (Equation 4.5). We neglect the mass of the balloon and the air contained in it. When it just begins to rise off the ocean floor, the shipping container’s weight W is balanced by the upward buoyant force FB on the shipping container and the balloon: mg = FB (1) Both the shipping container and the balloon are fully submerged in the water. Therefore, Archimedes’ principle holds that the magnitude FB of the buoyant force is equal to the weight Wwater = ρ water gV of the water displaced by the total volume V = Vcontainer + Vballoon of the container and the balloon: FB = Wwater = ρ water g (Vcontainer + Vballoon ) (11.6) The volume of the container is the product of its length l, width w, and height h: Vcontainer = lwh (2) The volume of the spherical balloon depends on its radius r via Vballoon = 4 π r 3 3 (3) SOLUTION Substituting Equation 11.6 into Equation (1), we obtain m g = ρ water g (Vcontainer + Vballoon ) or m = ρ water (Vcontainer + Vballoon ) (4) Replacing the volumes of the container and the balloon in Equation (4) with Equations (2) and (3) yields the mass m of the shipping container: ( m = ρ water lwh + 4 π r 3 3 ( ) ) 3 = 1025 kg/m 3 ( 6.1 m )( 2.4 m )( 2.6 m ) + 4 π (1.5 m ) = 5.4 × 104 kg 3 590 FLUIDS 47. REASONING When an object is completely submerged within a fluid, its apparent weight in the fluid is equal to its true weight mg minus the upward-acting buoyant force. According to Archimedes’ principle, the magnitude of the buoyant force is equal to the weight of the fluid displaced by the object. The weight of the displaced fluid depends on the volume of the object. We will apply this principle twice, once for the object submerged in each fluid, to find the volume of the object. SOLUTION The apparent weights of the object in ethyl alcohol and in water are: Ethyl alcohol 1 24 mg ρalcohol gV 15.2 N = { − 14 244 43 4 3 Water 1 24 mg ρ water gV 13.7 N = { − 14244 43 4 3 Weight in alcohol Weight in water True weight True weight (1) M...
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This note was uploaded on 04/30/2013 for the course PHYS 1100 and 2 taught by Professor Chastain during the Spring '13 term at LSU.

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