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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 upwardacting buoyant force balances
the downwardacting 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 upwardacting buoyant force FB
balances the downwardacting 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 upwardacting 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.
 Spring '13
 CHASTAIN
 Physics, The Lottery

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