Physics Solution Manual for 1100 and 2101

Therefore the doppler shifted frequency that you

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Unformatted text preview: ources can be found from P (Equation 16.9), where P is the power output of a source and r is the distance I= 4π r 2 between the source and point A. The sound intensities are equal at point A, so we have that I1 = I 2 = P 1 4π r12 = P2 4π r2 2 (1) Additionally, we know that the power P1 emitted by source 1 is four times greater than the power P2 emitted by source 2: P = 4 P2 1 (2) SOLUTION In Equation (1), r1 = x, and r2 = d – x when A is taken as the point of equal sound intensity. Alternatively, we could say that x is the position of point B relative to the origin (source 1), so that the distance r2 between source 2 and point B would be r2 = x – d. It Chapter 16 Problems 869 does not make a difference which we choose, since r2 is squared in Equation (1), and r22 = ( d − x ) = ( x − d ) . In fact, because Equation (1) is quadratic, either choice will yield two values of x, one for position A and one for position B. Substituting Equation (2), r1 = x, and r2 = d − x into Equation (1) yields 2 2 4 P2 4π x 2 = P2 4π ( d − x ) 4 1 = 2 x ( d − x )2 or 2 (2) Cross-multiplying Equation (2) and taking the square root of both sides, we obtain x2 = 4 (d − x) 2 or x = ±2 ( d − x ) (3) Solving Equation (3) separately for the positive root and the negative root gives Positive root x = +2 ( d − x ) = 2 d − 2 x x = −2 ( d − x ) = −2d + 2 x 3 x = 2d − x = −2 d x = 2 d 3 = 2(123 m) 3 = 82.0 m 61. Negative root x = 2 d = 2(123 m) = 246 m P , according to A Equation 16.8. The area is given directly, but the power is not. Therefore, we need to recast this expression in terms of the data given in the problem. Power is the change in energy per unit time, according to Equation 6.10b. In this case the energy is the heat Q that causes the Q temperature of the lasagna to increase. Thus, the power is P = , where t denotes the time. t As a result, Equation 16.8 for the intensity becomes SSM REASONING Intensity I is power P divided by the area A, or I = I= PQ = A tA (1) According to Equation 12.4, the heat that must be supplied to increase the temperature of a substance of mass m by an amount ∆T is Q = cm∆T, where c is the specific heat capacity. Substituting this expression into Equation (1) gives I= Q cm∆T = tA tA (2) 870 WAVES AND SOUND SOLUTION Equation (2) reveals that the intensity of the microwaves is I= cm∆T 3200 J/ ( kg ⋅ C° ) ( 0.35 kg ) ( 72 C° ) = = 7.6 × 103 W/m 2 −2 2 tA ( 480 s ) 2.2 × 10 m ( ) ______________________________________________________________________________ 62. REASONING Since the sound is emitted from the rocket uniformly in all directions, the energy carried by the sound wave spreads out uniformly over concentric spheres of increasing radii r1, r2, r3, . . . as the wave propagates. Let r1 represent the radius when the measured intensity at the ground is I (position 1 for the rocket) and r2 represent the radius when the measured intensity at the ground is 1 3 I (position 2 of the rocket). The time required for the energy to spread out over the sphere of radius r1 is t1 = r1 / vs , where vs is the speed of sound. Similarly, the time required for the energy to spread out over a sphere of radius r2 is t2 = r2 / vs . As the sound wave emitted at position 1 spreads out uniformly, the rocket continues to accelerate upward to position 2 with acceleration ay for a time t12. Therefore, the time that elapses between the two intensity measurements is t = t2 − t1 + t12 (1) From Equation 3.3b, the time t12 is t12 = ( v2 − v1 ) ay (2) where v1 and v2 are the speeds of the rocket at positions 1 and 2, respectively. Combining Equations (1) and (2), we obtain t = t2 − t1 + ( v2 − v1 ) ay (3) 2 The respective values of v1 and v2 can be found from Equation 3.6b ( v 2 = v0 y + 2a y y ) with y v0 y = 0 m/s, and y = r . Once values for v1 and v2 are known, Equation (3) can be solved directly. SOLUTION From the data given in the problem statement, r1 = 562 m . We can find r2 using the following reasoning: from the definition of intensity, I1 = P /(4π r12 ) and I 2 = P /(4π r22 ) . Since I1 = 3I2 , we have Chapter 16 Problems P /(4π r12 ) = 3P /(4π r22 ) or 871 r2 = r1 3 = 973 m The times t1 and t2 are t1 = r1 562 m = = 1.64 s vs 343 m/s t2 = r2 and vs = 973 m = 2.84 s 343 m/s Then, taking up as the positive direction, we have from Equation 3.6b, v1 = 2a y r1 = 2(58.0 m/s 2 )(562 m) = 255 m/s and v2 = 2a y r2 = 2(58.0 m/s 2 )(973 m) = 336 m/s Substituting these values into Equation (3), we have t = 2.84 s − 1.64 s + ( 336 m/s − 255 m/s ) = 2.6 s 58.0 m/s 2 ______________________________________________________________________________ 63. REASONING AND SOLUTION The sound intensity level β in decibels (dB) is related to the sound intensity I according to Equation 16.10, β = (10 dB) log ( I / I 0 ) , where I0 is the reference intensity. If β1 and β2 represent two different sound level intensities, then, I2 I I2 / I0 I = 10 dB ) log 2 − (10 dB ) log 1 = (10 dB ) log I...
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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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