Physics Solution Manual for 1100 and 2101

Negative root x 2 d 2123 m 246 m p according to a

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Unformatted text preview: that the speed of sound in the bar is v= L 0.83 m = = 4.4 × 103 m/s −4 t 1.9 × 10 s where L is the length of the rod and t is the time required for the wave to travel the length of the rod. The mass density of the bar is, from Equation 11.1, ρ = m / V = m /( LA) , where m and A are, respectively, the mass and the cross-sectional area of the rod. The density of the rod is, therefore, m 2.1 kg ρ= = = 1.9 × 104 kg/m3 LA (0.83 m)(1.3 ×10 –4 m 2 ) Using these values, we find that the bulk modulus for the rod is ( )( Y = ρ v 2 = 1.9 ×104 kg/m3 4.4 ×103 m/s ) 2 = 3.7 ×1011 N/m2 Comparing this value to those given in Table 10.1, we conclude that the bar is most likely made of tungsten . ______________________________________________________________________________ 860 WAVES AND SOUND 46. REASONING Let vP represent the speed of the primary wave and vS the speed of the secondary wave. The travel times for the primary and secondary waves are tP and tS, respectively. If x is the distance from the earthquake to the seismograph, then tP = x/ vP and tS = x/ vS. The difference in the arrival times is tS − tP = 1 x x 1 − = x − v vS vP S vP We can use this equation to find the distance from the earthquake to the seismograph. SOLUTION Solving the equation above for x gives tS − tP 78 s = = 8.0 × 105 m 1 1 1 1 − − vS vP 4.5 × 103 m/s 8.0 × 103 m/s ______________________________________________________________________________ x= 47. REASONING Since the sound wave travels twice as far in neon as in krypton in the same time, the speed of sound in neon must be twice that in krypton: vneon = 2vkrypton (1) Furthermore, the speed of sound in an ideal gas is given by v = γ kT , according to m Equation 16.5. In this expression γ is the ratio of the specific heat capacities at constant pressure and constant volume and is the same for either gas (see Section 15.6), k is Boltzmann’s constant, T is the Kelvin temperature, and m is the mass of an atom. This expression for the speed can be used for both gases in Equation (1) and the result solved for the temperature of the neon. SOLUTION Using Equation 16.5 in Equation (1), we have γ kTneon m 1 24 4 neon 3 =2 Speed in neon γ kTkrypton mkrypton 14243 Speed in krypton Squaring this result and solving for the temperature of the neon give γ kTneon mneon γ kTkrypton = 4 mkrypton or m Tneon = 4 neon m krypton Tkrypton Chapter 16 Problems 861 We note here that the mass m of an atom is proportional to its atomic mass in atomic mass units (u). As a result, the temperature of the neon is m 20.2 u Tneon = 4 neon Tkrypton = 4 ( 293 K ) = 283 K mkrypton 83.8 u ______________________________________________________________________________ 48. REASONING The speed vtruck of the truck is equal to the distance x it travels divided by the time ttruck it takes to travel that distance: vtruck = x (1) ttruck Since sound also moves at a constant speed, the distance x it travels in reaching the truck is the product of the speed v of sound and the time tsound for it to travel that distance, or x = v tsound. The time for the sound to reach the truck is one-half the round-trip time tRT, so tsound = 1 tRT . Thus, the distance traveled by the sound can be written as 2 x=v ( 1 tRT ) 2 Substituting this expression for x into Equation (1) gives vtruck = x ttruck = v ( 1 tRT ) 2 ttruck (2) Since the air is assumed to be an ideal gas, the speed of sound is related to the Kelvin temperature T and the average mass m of an air molecule by Equation 16.5: v= γ kT m where γ is the ratio of the specific heat capacity of air at constant pressure to that at constant volume, and k is Boltzmann’s constant. Substituting this expression for v into Equation (2) gives γ kT 1 t m 2 RT vtruck = ttruck ( ) SOLUTION The temperature of the air must be its Kelvin temperature, which is related to the Celsius temperature Tc by T = Tc + 273.15 (Equation 12.1): T = 56 °C + 273.15 = 329 K. 862 WAVES AND SOUND The average mass m (in kg) of an air molecule is the same as that determined in Example 4, −26 namely, m = 4.80 × 10 kg. Thus, the speed of the truck is γ kT ( 7 5 ) (1.38 ×10−23 J/K )( 329 K ) 1 ( 0.120 s ) 2 4.80 ×10−26 kg m vtruck = = = 5.46 m/s ttruck 4.00 s ______________________________________________________________________________ 1t 2 RT 49. REASONING AND SOLUTION The speed of sound in an ideal gas is v= γ kT m so that the mass of a gas molecule is m= γ kT v 2 = (1.67) (1.38 ×10−23 J/K ) ( 3.00 × 102 K ) ( 363 m/s ) 2 = 5.25 × 10−26 kg = 5.25 × 10−23 g We now need to determine what fraction of the gas is argon and what fraction is neon. First find the mass of each molecule. mar = 39.9 g/mol = 6.63 × 10−23 g 23 6.022 × 10 /mol mne = 20.2 g/mol = 3.35 × 10−23 g 23 6.022 × 10 /mol Let q be the fraction of gas that is in the form of argon and p be the fraction that is neon. We know that q + p = 1. Also, qmar + pmne = m. Substituting p = 1 – q in this equation gives qmar + (1 – q)mne = m and we can now solve for the f...
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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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