Unit 16 Solutions

# Unit 16 Solutions - Problems[3.69 3.72 3.73 69 Strategy...

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1021 Problems: [3.69], 3.72, 3.73 69. Strategy Consider the relative motion of the ship and the water. Solution The relative speeds are: upstream ship water up s w downstream ship water d s w w up up s w up s w d d s w d s w up d Find the speed of the current, . ( ) , so (1). ( ) (2). vv v v v v v v v v v xx x v t v vt v v x v t v v v tt =− = = =+ = = + ΔΔ Δ= Δ = − Δ = − Δ= Δ = + Δ = + Subtract (1) from (2). ww du p p 11 2 0 8 k m 1 1 2 0.42 km h . 2 2 19.2 h 20.8 h x ⎛⎞ Δ ⎜⎟ −= = = = ⎝⎠ 72. Strategy The minimum air velocity is in the same direction as the airplane’s. Solution 210 m s east 160 m s east 50 m s east 73. Strategy The upstream component of the velocity of the boat must be equal in magnitude to that of the current. Solution Let + y -direction be toward the opposite shore. 1 1.8 (4.0 km h)sin 1.8 km h, so sin 27 . 4.0 θθ == = ° The direction of the velocity of the boat relative to the water is 27° upstream. θ Opposite shore Upstream 4.0 km/h 1.8 km/h y Conceptual Questions 26.5, 26.8 5. The astronaut would measure about 52 beats per minute. He is in his own rest frame, so he measures the proper time between heartbeats—the same as if he were taking his pulse back on Earth. 8. The answer depends upon the orientation of the rod. The rod will be length contracted with respect to the Earth observer if its axis is directed along the direction of motion. If the axis of the rod is perpendicular to the direction of motion, the length of the rod will appear unchanged. Problems: 26.1, [26.3] 1. Strategy Compute the times required for the optical signal and the train to reach the station and find the difference. Solution The time required for the optical signal to reach the station, as measured by an observer at rest relative to the station (the stationmaster) is

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Unit 16 Solutions 1022 5 1.0 km 3.3 s. 3.00 10 km s d c == μ × Measured by the same observer, the time needed for the train to arrive is 5 1.0 km 5.6 s. 0.60 0.60 3.00 10 km s dd vc = μ ⋅× The difference in the arrival times is 5.56 s 3.33 s 2.2 s . μ −μ 3. (a) Strategy The velocity of the spaceship relative to the Earth is SE 0.13 . = The velocity of the radio transmission relative to the spaceship is TS . = − This velocity is negative since the radio transmission is moving opposite the direction of the spaceship. Galilean velocity addition gives the velocity of the radio transmission relative to the Earth Solution Compute the velocity of the transmission relative to the Earth. TE TS SE 0.13 0.87 vv v c c c =+= + = . The speed is TE 0.87 0.87 . c =− = (b) Strategy According to Einstein’s Postulates, all electromagnetic waves, including radio waves, move at speed c relative to all inertial frames. Solution The Earth can be taken as an inertial frame, so the speed of the radio transmission relative to the Earth is . c Problems: 26.5, 26.9, [26.11], 26.17, 26.19, [26.21] 5. Strategy The proper time interval 0 t Δ is the time interval measured by the Rolex. The time interval measured at mission control is Δ t = 12.0 h. The time-dilation equation can be used to find 0 . t Δ Solution 28 2 0 2 (2.0 10 m s) 1 (12.0 h) 1 8.9 h (3.00 10 m s) tv tt c γ Δ× Δ= = Δ − = = × 9. (a) Strategy
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Unit 16 Solutions - Problems[3.69 3.72 3.73 69 Strategy...

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