Physics Solution Manual for 1100 and 2101

# 9b solving this equation for fthread and using the

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Unformatted text preview: mass of 3 3 M = ρV = (2200 kg/m )(0.025 m ) = 55 kg 566 FLUIDS The mass of concrete removed to make the hole is then 55 kg − 33 kg = 22 kg. This corresponds to a volume V = (22 kg)/(2200 kg/m3) = 0.010 m3. Since the hole is spherical V = (4/3)π r3 so r= 7. 3 3V = 4π 3 ( 3 0.010 m3 4π ) = 0.13 m REASONING According to the definition of density ρ given in Equation 11.1, the mass m of a substance is m = ρV, where V is the volume. We will use this equation and the fact that the mass of the water and the gold are equal to find our answer. To convert from a volume in cubic meters to a volume in gallons, we refer to the inside of the front cover of the text to find that 1 gal = 3.785 × 10 –3 3 m. REASONING Using Equation 11.1, we find that ρ Water VWater = ρ Gold VGold Using the fact that 1 gal = 3.785 × 10 Table 11.1, we find VWater = –3 or 3 m V Water = ρ Gold VGold ρ Water and densities for gold and water from ρ GoldVGold ρ Water 19 bb b c 300 kg / m h0.15 mg0.050 mg0.050 m g 1 gal F = G × 10 H 3.785 1000 c kg / m h 3 −3 3 8. m 3 I= J K 1.9 gal REASONING The total mass mT of the rock is the sum of the mass mG of the gold and the mass mQ of the quartz: mT = mG + mQ. Thus, the mass of the gold is mG = mT − mQ (1) The total volume VT of the rock is the sum of the volume VG of the gold and volume VQ of the quartz: VT = VG + VQ (2) The volume of a substance is related to the mass and the density of the substance according to the definition of density: ρ = m/V (Equation 11.1). These relations will enable us to find the mass of the gold in the rock. Chapter 11 Problems 567 SOLUTION Substituting mQ = ρQVQ from Equation 11.1 into Equation (1) gives mG = mT − mQ = mT − ρQVQ (3) Solving Equation (2) for the volume VQ of the quartz and substituting the result into Equation (3) yields mG = mT − ρQVQ = mT − ρQ(VT − VG) (4) Substituting VG = mG/ρG from Equation 11.1 into Equation (4) gives m mG = mT − ρ Q (VT − VG ) = mT − ρ Q VT − G ρG (5) Solving Equation (5) for the mass of the gold, and using the densities for gold and quartz given in Table 11.1, gives VT − mG = 9. mT ρQ 1 1 − ρG ρQ = 12.0 kg ( 4.00 ×10−3 m3 ) − 2660 kg/m3 1 1 − 3 19 300 kg/m 2660 kg/m3 = 1.6 kg SSM WWW REASONING The period T of a satellite is the time for it to make one complete revolution around the planet. The period is the circumference of the circular orbit (2π R) divided by the speed v of the satellite, so that T = (2π R)/v (see Equation 5.1). In Section 5.5 we saw that the centripetal force required to keep a satellite moving in a circular orbit is provided by the gravitational force. This relationship tells us that the speed of the satellite must be v = GM / R (Equation 5.5), where G is the universal gravitational constant and M is the mass of the planet. By combining this expression for the speed with that for the period, and using the definition of density, we can obtain the period of the satellite. SOLUTION The period of the satellite is T= 2π R 2π R R3 = = 2π v GM GM R 568 FLUIDS According to Equation 11.1, the mass of the planet is equal to its density ρ times its volume V. Since the planet is spherical, V = 4 π R 3 . Thus, M = ρ V = ρ 3 ( 4 π R3 ) . Substituting this 3 expression for M into that for the period T gives T = 2π R3 R3 3π = 2π = GM Gρ G ρ 4 π R3 3 ( ) The density of iron is ρ = 7860 kg/m (see Table 11.1), so the period of the satellite is 3 3π = Gρ T= ( 6.67 ×10 3π −11 )( N ⋅ m 2 /kg 2 7860 kg/m3 ) = 4240 s 10. REASONING The total mass of the solution is the sum of the masses of its constituents. Therefore, ρsVs = ρwVw + ρgVg (1) where the subscripts s, w, and g refer to the solution, the water, and the ethylene glycol, respectively. The volume of the water can be written as Vw = Vs − Vg . Making this substitution for Vw , Equation (1) above can be rearranged to give Vg Vs = ρs − ρ w ρg − ρ w (2) Equation (2) can be used to calculate the relative volume of ethylene glycol in the solution. SOLUTION The density of ethylene glycol is given in the problem. The density of water is given in Table 11.1 as 1.000 ×103 kg/m3 . The specific gravity of the solution is given as 1.0730. Therefore, the density of the solution is ρs = (specific gravity of solution) × ρ w = (1.0730)(1.000 ×103 kg/m3 ) = 1.0730 ×103 kg/m3 Substituting the values for the densities into Equation (2), we obtain Vg Vs = ρs − ρ w 1.0730 × 103 kg/m3 − 1.000 ×103 kg/m3 = = 0.63 ρg − ρ w 1116 kg/m3 − 1.000 × 103 kg/m3 Chapter 11 Problems 569 Therefore, the volume percentage of ethylene glycol is 63% . 11. SSM REASONING Since the inside of the box is completely evacuated; there is no air to exert an upward force on the lid from the inside. Furthermore, since the weight of the lid is negligible, there is only one force that acts on the lid; the downward force caused by the air pressure on the outside of the lid. In order to pull the lid off the box, one must supply a force that is at least equal in magnitude and opposite in direction to the force exerte...
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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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