s16 - P16.1 5 Replace x by x vt= x 4. t to get P16.5 y= 6 (...

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P16.1 Replace x by 4.5 xv tx t −=− to get () 2 6 4. 53 y xt = ⎡⎤ −+ ⎣⎦ P16.5 The distance the waves have traveled is ( ) ( )( ) 7.80 km s 4.50 km s 17.3 s dt t == + where t is the travel time for the faster wave. Then, ( ) ( ) ( ) 7.80 4.50 km s 4.50 km s 17.3 s t −= or ( ) 23.6 s t and the distance is ( ) 7.80 km s 23.6 s1 8 4 km d . P16.9 ( ) ( ) 0.020 0 ms i n2.11 3.62 yx =− t in SI units 2.00 cm A = 2.11 rad m k = 2 2.98 m k π λ 3.62 rad s ω = 0.576 Hz 2 f 23 . 62 1.72 m s 22 . 11 vf k = = P16.13 0.250si n0.300 40. 0 m t Compare this with the general expression ( ) sin y Ak (a) 0.250 m A = (b) 40.0 rad s = (c) 0.300 rad m k = (d) 20.9 m k = (e) 20.9 m1 3 3 m s λλ ππ ⎛⎞ = = ⎜⎟ ⎝⎠ (f) The wave moves to the right, in direction x + . P16.17 0.120 i n4 8 t =+ 473
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474 Wave Motion (a) dy v dt = : () ( ) 0.120 4 cos 4 8 xx π t ⎛⎞ =+ ⎜⎟ ⎝⎠ 0.200 s, 1.60 m1 . 51 m s v =− dv a = : ( ) 2 0.120 m4 s i n4 8 ax t + m0 a = (b) 2 8 k λ == : 16.0 m = 2 4 T ωπ : 0.500 s T = 32.0 m s v T = P16.23 2 3 1350 kg m s 520 m s 5.00 10 kg m T v μ = × P16.26 T v = 22 2 2 2 2 34 8920 kg m 7.50 10 m2 0 0 m s 631 N Tv A v r v T T μρ ρ = = P16.32 Refer to the diagrams. From the free-body diagram of point A : and 1 0s i n y FT M θ =⇒ = g 1 0c o s x F TT = ⇒=
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This note was uploaded on 08/31/2010 for the course PHYSICS 7E taught by Professor Staff during the Fall '08 term at UC Irvine.

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s16 - P16.1 5 Replace x by x vt= x 4. t to get P16.5 y= 6 (...

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