Physics Solution Manual for 1100 and 2101

Since we do not have the values of or m we will use

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Unformatted text preview: 3kT = = = 431 m/s m 6.63 ×10−26 kg b. The speed of sound in argon is, according to Equation 16.5, v= γ kT (1.67 ) (1.38 ×10−23 J/K ) ( 298 K ) = = 322 m/s m 6.63 × 10−26 kg ______________________________________________________________________________ 36. REASONING The speed of sound in an ideal gas depends on its temperature T through γ kT Equation 16.5 as v = , where γ = 1.40 for air, k is Boltzmann’s constant, and m is the m mass of a molecule in the air. Thus, the speed of sound depends on the air temperature, with a lower temperature giving rise to a smaller speed of sound. The speed of sound is greater in the afternoon at the warmer temperature and is the speed that the car must exceed. SOLUTION The speed of sound in air is v = γ kT , where the temperature T must be m expressed on the Kelvin scale (T = Tc + 273, Equation 12.1). Taking the ratio of the speed of sound at 43 °C to that at 0 °C, we have 854 WAVES AND SOUND v43 °C v0 °C γ k ( 43 + 273) = 316 K m = 273 K γ k ( 0 + 273) m The speed of sound at 43 °C is v43 °C = v0 °C 316 K 316 K = ( 331 m/s ) = 356 m/s 273 K 273 K Therefore, the speed of your car must exceed vcar = 356 m/s . ______________________________________________________________________________ 37. REASONING AND SOLUTION The wheel must rotate f = (2200 Hz)/20 = 110 Hz. The angular speed ω of the wheel is at a frequency of ω = 2π f = 2π (110 Hz) = 690 rad/s ______________________________________________________________________________ 38. REASONING Under the assumptions given in the problem, both the bullet and the sound d of the rifle discharge travel at constant speeds. Therefore, we can use v = (Equation 2.1, t where v is speed, d is distance, and t is elapsed time) to determine the distance that the bullet travels, as well as the time it takes the rifle report to reach the observer. The speed of the bullet is given, and we know that the speed of sound in air is 343 m/s, because the air temperature is 20 °C (See Table 16.1). SOLUTION Applying Equation 2.1, we can express the distance db traveled by the bullet as d b = vbt (1) where vb = 840 m/s and t is the time needed for the report of the rifle to reach the observer. The observer is a distance d = 25 m from the marksman, so we can use Equation 2.1 with the speed of sound v = 343 m/s to determine t: t= d v (2) Substituting Equation (2) into Equation (1), we find the distance traveled by the bullet before the observer hears the report: d v d ( 840 m/s ) ( 25 m ) d b = vb t = vb = b = = 61 m v 343 m/s v Chapter 16 Problems 855 39. REASONING AND SOLUTION The sound wave will travel at constant speed through the water (speed = vwater) and the copper block (speed = vcopper). Thus, the times it takes for the sound wave to travel downward through the water a distance dwater and downward through the copper a distance dcopper are ∆twater = d water vwater ∆tcopper = and d copper vcopper The time required for the incident sound wave to reach the bottom of the copper block just before reflection is equal to the time required for the reflected sound wave to reach the surface of the water. Using values for the speed of sound in water and copper from Table 16.1 in the text, we find that the total time interval between when the sound enters and leaves the water is, then, d copper d ∆ttotal = 2(∆twater + ∆tcopper ) = 2 water + vwater vcopper 0.15 m 0.45 m –4 ∆t total = 2 + = 6.7 × 10 s 1482 m/s 5010 m/s ______________________________________________________________________________ 40. REASONING Generally, the speed of sound in a liquid like water is greater than in a gas. And, in fact, according to Table 16.1, the speed of sound in water is greater than the speed of sound in air. Since the speed of sound in water is greater than in air, an underwater ultrasonic pulse returns to the ruler in a shorter time than a pulse in air. The ruler has been designed for use in air, not in water, so this quicker return time fools the ruler into believing that the object is much closer than it actually is. Therefore, the reading on the ruler is less than the actual distance. SOLUTION Let x be the actual distance from the ruler to the object. The time it takes for the ultrasonic pulse to reach the object and return to the ruler, a distance of 2x, is equal to the distance divided by the speed of sound in water vwater: t = 2x/vwater. The speed of sound in water is given by Equation 16.6 as vwater = Bad / ρ , where Bad is the adiabatic bulk modulus and ρ is the density of water. Thus, the time it takes for the pulse to return is t= 2x 2x = = vwater Bad ρ 2 ( 25.0 m ) 2.37 × 10 Pa 3 1025 kg/m 9 = 3.29 × 10−2 s 856 WAVES AND SOUND The ruler measures this value for the time and computes the distance to the object by using the speed of sound in air, 343 m/s. The distance xruler displayed by the ruler is equal to the speed of sound in air multiplied by the time the object: 1 2 t it takes for the pulse to go from the ruler to...
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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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