4.175. u=_- vvo(V1 + (v/v 0 ) 2 —1) (^) 2vvvo = 0.5 m/s.
4.176. 0.) = aAv (1/1 + (Av/v 0 ) 2 — 1) = 34 s- 1.
4.177. v= vo /V1 + 2wt I v =1.35 kHz.
4.178. (a) 'v = vo/(1 —1^2 ) = 5 kHz; (b) r =11x1 + ii 2 = 0.32 km.
4.179. Decreases by 2u/(v + u) = 2.0%.
4.180. v = 2v ou/(v + u) = 0.60 Hz.
4.181. 1n( 2 ( rlir 'irl = 6 .10)) -3 m- 1.
4.182. (a) L' = L — 20Tx log e = 50 dB; (b) x = 0.30 km.
4.183. (a) L = Lo + 20 log (ro/r) = 36 dB; (b) r > 0.63 km.
4.184. [i = ln(rB /rA) (rB — rA)w1 = 0.12 s- 1.
4.185. (a) Let us consider the motion of a plane element of the
medium of thickness dx and unit area of cross-section. In accordance
with Newton's second law p dx = —dp, where dp is the pressure
increment over the length dx. Recalling the wave equation =
= v 2 (a 2 V0x 2 ), we can write the foregoing equation as
pv 2 -adx = — dp.
Integrating this equation, we get
Ap = — pv^2 4, + const.
In the absence of a deformation (a wave) the 'surplus pressure is
Op = 0. Hence, coast = 0.
4.186. (D) = 3111 2 (Ap);„/2pvk = 11 mW.
4.187. (a) (Ap),,., = VpvP/27tr^2 = 5 Pa,^ (Ap)„,1 p = 5.10-5;
(b) a = (Ap),,/22tvpv = 311,m, alt = 5.10 -6.
4.188. P = 43tr^2 e^2 Yrio • 10L = 1.4 W, where L is expressed
in bels.
4.189. AX = (1/1/Z —1) c/v —50 in.
4.190. t = 2 (1/ — lre;) //c In (ei/ez)•
4.191. j/jdis = a/2nveco = 2..
4.192. H = Ve^0 /11^0 [kErn] cos (ckt), where c is the velocity of
the wave in vacuum.
4.193. (a) H = ezEm eo/R0 cos kx = — 0. 3 0ez;
(b) H = ezEm lico/tto cos (ckto — kx) = 0.18ez. Here ez is the unit vector
of the z axis, H is expressed in A/m.
4.194. 8, = 2irv1 2 E,,/c = 13 mV.
4.196. (S) = 1 /,keoc^2 EPe).
4.197. (a) jai, = 1/ 2sovEm = 0.20 mA/m^2 ;^ (b) (8) =
(^112 6 0) cE;r = i 3.3 pW/m 2.
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joyce
(Joyce)
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