Physics Solution Manual for 1100 and 2101

# 0 10 4 wm 2 2 10 db log 2 10 db log 89 db

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Unformatted text preview: raction q= m − mne mar − mne Suppressing the units and algebraically canceling the factor of 10–23, we find q= 5.25 − 3.35 = 0.57 = 57% argon 6.63 − 3.35 and p = 1 – q = 1 – 0.57 = 0.43 = 43% neon Chapter 16 Problems 863 ______________________________________________________________________________ 50. REASONING Both the sound reflected from the ceiling and from the floor of the cavern travel the same distance through the ground to reach the microphones. But the sound reflected from the floor travels from the ceiling to the floor and back again, a distance 2h farther than the sound reflected from the ceiling, where h is the height of the cavern. Therefore, the time ∆t = 0.0437 s − 0.0245 s = 0.0192 s that elapses between the arrival of the first and second reflected sounds is the time it takes the sound that penetrates the cavern to travel the distance 2h in the cavern. If the speed of sound in the air of the cavern is v, then Equation 2.1 gives 2h v= (2.1) ∆t We are assuming that air behaves like an ideal gas, so the speed v of sound is given by γ kT (Equation 16.5), where γ = cP cV is the ratio of the specific heat capacities of v= m air at constant pressure and constant volume, k is Boltzmann’s constant, m is the mass of a single molecule of air, and T is the Kelvin temperature of the air. Since we do not have the values of γ or m, we will use the fact that, at T0 = 273.15 K + 20.0 °C the speed of sound in air is v0 = 343 m/s (see Table 16.1) to determine, with Equation 16.5, the speed v of sound when the temperature is T = 273.15 K + 9.0 °C. SOLUTION Solving Equation 2.1 for the height h of the cavern yields h= v∆t 2 (1) Equation 16.5 gives both the speed v of sound in air at T = 273.15 K + 9.0 °C and the speed v0 of sound in air at T0 = 273.15 K + 20.0 °C: v= γ kT v0 = and m γ kT0 (2) m Taking the ratio of Equations (2), we obtain γkT v = v0 m γ k T0 = T T0 m Substituting Equation (3) into Equation (1) yields or v = v0 T T0 (3) 864 WAVES AND SOUND T v0 ∆t T0 v ∆t T v∆t h= = =0 2 2 2 T0 = 51. ( 343 m/s )( 0.0192 s ) 2 (4) 273.15 K + 9 oC = 3.23 m 273.15 K + 20.0 o C SSM WWW REASONING We must determine the time t for the warning to travel the vertical distance h = 10.0 m from the prankster to the ears of the man when he is just under the window. The desired distance above the man's ears is the distance that the balloon would travel in this time and can be found with the aid of the equations of kinematics. SOLUTION Since sound travels with constant speed vs , the distance h and the time t are related by h = vst . Therefore, the time t required for the warning to reach the ground is t= h 10.0 m = = 0.0292 s vs 343 m/s We now proceed to find the distance that the balloon travels in this time. To this end, we must find the balloon's final speed vy, after falling from rest for 10.0 m. Since the balloon is 2 2 dropped from rest, we use Equation 3.6b ( v y = v0 y + 2a y y ) with v0y = 0 m/s: 2 v y = v0 y + 2a y y = ( 0 m/s )2 +2(9.80 m/s2 )(10.0 m) = 14.0 m/s Using this result, we can find the balloon's speed 0.0292 seconds before it hits the man by solving Equation 3.3b ( v y = v0 y + a y t ) for v0 y : v0 y = v y − a y t = (14.0 m/s) − (9.80 m/s 2 )(0.0292 s) = 13.7 m/s Finally, we can find the desired distance y above the man's head from Equation 3.5b: 1 2 1 2 y = v0 y t + a y t 2 = (13.7 m/s)(0.0292 s) + (9.80 m/s 2 )(0.0292 s) 2 = 0.404 m ______________________________________________________________________________ 52. REASONING AND SOLUTION The energy carried by the sound into the ear is Energy = IAt = (3.2 × 10–5 W/m2)(2.1 × 10–3 m2)(3600 s) = 2.4 × 10 –4 J ______________________________________________________________________________ Chapter 16 Problems 53. 865 SSM WWW REASONING AND SOLUTION Since the sound radiates uniformly in all directions, at a distance r from the source, the energy of the sound wave is distributed P over the area of a sphere of radius r. Therefore, according to I = (Equation 16.9) 4π r 2 with r = 3.8 m, the power radiated from the source is P = 4π Ir 2 = 4π (3.6 ×10 –2 W/m 2 )(3.8 m) 2 = 6.5 W ______________________________________________________________________________ 54. REASONING a. The source emits sound uniformly in all directions, so the sound intensity I at any 2 distance r is given by Equation 16.9 as I = P / 4 π r , where P is the sound power emitted ( ) by the source. Since patches 1 and 2 are at the same distance from the source of sound, the sound intensity at each location is the same, so I1 = I2. Patch 3 is farther from the sound source, so the intensity I3 is smaller for points on that patch. Therefore, patches 1 and 2 have equal intensities, each of which is greater than the intensity at patch 3. b. According to Equation 16.8, the sound intensity I is defined as the sound power P that passes perpendicularly through a surface divided by the area A of that surface, I = P/A. The area of the surface is, then, A = P/I. Since...
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