Physics Solution Manual for 1100 and 2101

Therefore for both 0 and 35 we have s s0 1 2 1 2

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Unformatted text preview: uation 8.2). The time Δt it takes the light to travel from the rotating mirror to ω= ∆t 2d the fixed mirror and back is given by c = (Equation 2.1), where c = 3.00×108 m/s is the ∆t speed of light in a vacuum, and d = 35 km is the distance between the rotating mirror and the fixed mirror. Together, Equations 8.2 and 2.1 will allow us to determine the minimum angular speed ω of the rotating mirror. The angles between the rays of light shown in Figure 24.12 are exaggerated. In reality, the diameter of the rotating mirror is so much smaller than the distance d to the fixed mirror that these two rays may be considered to be parallel. SOLUTION Solving c = 2d (Equation 2.1) for Δt yields ∆t 2d ∆t = c Substituting Equation (1) into ω = (1) ∆θ (Equation 8.2), we obtain ∆t ∆θ ∆θ c ( ∆θ ) ( 3.00 ×108 m/s ) ( 0.125 rev ) = = = = 540 rev/s ∆t 2d 2d 2 ( 35 × 103 m ) c ______________________________________________________________________________ = ω 17. REASONING Since the speed at which an electromagnetic wave travels is known, the round-trip travel time t of a single pulse from the lidar gun can be used to determine the distance d of the speeding vehicle from the gun as d = ct/2. Here, we have divided by a factor of two in order to account for the fact that the time given is that for the electromagnetic wave to travel out to the vehicle and return. Thus, in effect, the two pulses are used to measure the distances of the vehicle from the gun at two different instants. The difference in the distances is the distance that the speeder travels in the interval between the pulses. We can determine the vehicle’s speed by dividing the travel distance by the time interval of 0.450 s. Chapter 24 Problems 1287 SOLUTION Applying the expression d = ct/2 for each pulse, we obtain the distance D traveled by the speeding vehicle between the two pulses as D = d 2 − d1 = c ( 1 t2 ) − c ( 1 t1 ) 2 2 Dividing this distance by the interval tpulses between the pulses gives the speed v of the vehicle as (1 ) (1 ) −7 t −t c 2 t2 − c t1 D 1 3.00 × 108 m/s 1.27 × 10 v= 2 = = 1 c 2 1 = 2 t 0.450 s 2 tpulses tpulses pulses ( ) s = 42.3 m/s 18. REASONING Let R denote the average rate at which the laptop downloads information, measured in bits per second (bps). This average rate is equal to the number N of bits downloaded in a time t divided by the time: R = N/t. Therefore, the number N of bits that the laptop downloads is given by N =Rt (1) We note that 1 Mbps (megabit per second) is equal to 106 bps. The time t is the time it takes the wireless signal to travel the distance d = 8.1 m between the router and the laptop. This time is determined by Equation 2.1 as t= d c (2) where c = 3.00×108 m/s is the speed of light in a vacuum. SOLUTION Substituting Equation (2) into Equation (1), we obtain d = R t R N= c (3) To convert the download rate R into bits per second, we use the equivalence 1 Mbps = 106 bps, and rewrite “bps” as “bits/s” ( R = 260 Mbps 6 10 bps 260 260 ) 1 Mbps = ×106 bps = ×106 bit/s Therefore, from Equation (3), the average number N of bits downloaded is ( ) 8.1 m d N= R = 260 ×106 bits/s 7.0 bits = 8 c 3.00 × 10 m/s ______________________________________________________________________________ 1288 ELECTROMAGNETIC WAVES 19. REASONING The distance between Polaris and earth is equal to the speed of the light multiplied by the time it takes for the light to make the journey. The time is given. Since light is an electromagnetic wave, and all electromagnetic waves travel through a vacuum at the speed of light c, the speed of the light is also known. SOLUTION The distance s between Polaris and earth is s = ct , where t is the time for the light to travel this distance. Using the fact that = 3.156 ×107 s (see the table of 1 yr conversion factors at the front of the book), we find that 3.156 ×107 s yr = 6.4 × 1018 m 1 yr ______________________________________________________________________________ = ct s= ( 3.00 × 108 m/s ) ( 680 ) 20. REASONING Because the flash from the gunshot travels at the speed c = 3.00×108 m/s of light in a vacuum, it can make N round trips between the two mirrors in the time Δts that it takes the sound of the gunshot to make one round trip, returning as an echo. Therefore, in terms of the time Δtf for one round-trip of the light flash, the number N of round trips of the flash is given by ∆t N= s (1) ∆tf The time Δt it takes either the sound or the flash to travel the round-trip distance d between d the gun and the cliff at a constant speed v is given by ∆t = (Equation 2.1). The sound of v the gunshot travels at a speed v = 343 m/s, so Equation 2.1 yields both d = ∆ts v Sound of echo d = and ∆tf c Light flash SOLUTION Substituting Equations (2) into Equation (1) yields ∆ts = = N ∆tf d v d c c 3.00 ×108 m/s = = = 8.75 ×105 343 m/s v (2) Chapter 24 Problems 21. REASONING AND SOLUTION The time t that it takes for the telephone call to go from one city to the other is equal to the distance s...
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