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Physics Solution Manual for 1100 and 2101

# 8 103 w 3 2 a3 62 10 m 2 i 3 029 wm these answers

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Unformatted text preview: ( ) xruler = vair ( 1 t ) = ( 343 m/s ) ( 1 ) 3.29 × 10 s = 5.64 m 2 2 −2 Thus, the ruler displays a distance of xruler = 5.64 m. As expected, the reading on the ruler’s display is less than the actual distance of 25.0 m. ______________________________________________________________________________ 41. REASONING AND SOLUTION a. In order to determine the order of arrival of the three waves, we need to know the speeds of each wave. The speeds for air, water and the metal are va = 343 m/s, vw = 1482 m/s, vm = 5040 m/s The order of arrival is metal wave first, water wave second, air wave third . b. Calculate the length of time each wave takes to travel 125 m. tm = (125 m)/(5040 m/s) = 0.025 s tw = (125 m)/(1482 m/s) = 0.084 s ta = (125 m)/(343 m/s)= 0.364 s Therefore, the delay times are ∆t12 = tw – tm = 0.084 s – 0.025 s = 0.059 s ∆t13 = ta – tm = 0.364 s – 0.025 s = 0.339 s ______________________________________________________________________________ 42. REASONING Two facts allow us to solve this problem. First, the sound reaches the microphones at different times, because the distances between the microphones and the source of sound are different. Since sound travels at a speed v = 343 m/s and the sound arrives at microphone 2 later by an interval of ∆t = 1.46 × 10−3 s, it follows that L2 − L1 = v∆t (1) Second, the microphones and the source of sound are located at the corners of a right triangle. Therefore, the Pythagorean theorem applies: 2 L2 = D 2 + L1 2 (2) Chapter 16 Problems 857 Since v, ∆t, and D are known, these two equations may be solved simultaneously for the distances L1 and L2. SOLUTION Solving Equation (1) for L2 gives L2 = L1 + v∆t Substituting this result into Equation (2) gives 2 ( L1 + v∆t )2 = D 2 + L1 2 2 L1 + 2 L1v∆t + ( v∆t ) = D 2 + L1 2 2 L1v∆t + ( v∆t ) = D 2 2 Solving for L1, we find that ( 2 (1.50 m ) − ( 343 m/s ) 1.46 ×10−3 s D 2 − ( v∆t ) L1 = = 2v∆t 2 ( 343 m/s ) 1.46 × 10−3 s 2 2 ( ) ) 2 = 2.00 m Solving Equation (2) for L2 and substituting the result for L1 reveal that 2 L2 = D 2 + L1 = (1.50 m )2 + ( 2.00 m )2 = 2.50 m ______________________________________________________________________________ 43. SSM REASONING The sound will spread out uniformly in all directions. For the purposes of counting the echoes, we will consider only the sound that travels in a straight line parallel to the ground and reflects from the vertical walls of the cliff. Let the distance between the hunter and the closer cliff be x1 and the distance from the hunter to the further cliff near wall far wall be x2 . The first echo arrives at the location of the hunter after x1 traveling a total distance 2 x1 in a time t1, so that, if vs is x2 the speed of sound, t1 = 2 x1 / vs . Similarly, the second echo arrives after reflection from the far wall and in an point of amount of time t2 after the firing of the gun. The firing quantity t2 is related to the distance x2 and the speed of sound vs according to t2 = 2 x2 / vs . The time difference between the first and second echo is, therefore 858 WAVES AND SOUND ∆ t = t2 – t1 = 2 ( x2 – x1) vs (1) The third echo arrives in a time t3 after the second echo. It arises from the sound of the second echo that is reflected from the closer cliff wall. Thus, t3 = 2 x1 / vs , or, solving for x1, we have vt x1 = s 3 (2) 2 Combining Equations (1) and (2), we obtain ∆t = t2 – t1 = Solving for x2 , we have vst 3 2 x2 – vs 2 v x2 = s (∆ t + t3 ) 2 (3) The distance between the cliffs can be found from d = x1 + x 2 , where x1 and x2 can be determined from Equations (2) and (3), respectively. SOLUTION According to Equation (2), the distance x1 is x1 = (343 m/s)(1.1 s) = 190 m 2 According to Equation (3), the distance x2 is x2 = (343 m/s ) 2 (1.6 s + 1.1 s ) = 460 m Therefore, the distance between the cliffs is d = x1 + x 2 = 190 m + 460 m = 650 m ______________________________________________________________________________ 44. REASONING AND SOLUTION The speed of sound in an ideal gas is given by text Equation 16.5: γ kT v= m Chapter 16 Problems 859 where γ = cp/cv, k is Boltzmann's constant, T is the Kelvin temperature of the gas, and m is the mass of a single gas molecule. If mTOTAL is the mass of the gas sample and N is the total number of molecules in the sample, then the above equation can be written as γ kT v= (mTOTAL / N ) = γ N kT (1) mTOTAL For an ideal gas, PV = NkT, so that Equation (1) becomes, v= γ PV mTOTAL = (1.67)(3.5×105 Pa)(2.5 m3 ) = 8.0 ×102 m/s 2.3 kg ______________________________________________________________________________ 45. SSM REASONING Equation 16.7 relates the Young's modulus Y, the mass density ρ, and the speed of sound v in a long slender solid bar. According to Equation 16.7, the Young's modulus is given by Y = ρ v 2 .The data given in the problem can be used to compute values for both ρ and v. SOLUTION Using the values of the data given in the problem statement, we find...
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