12.66.Set Up:Take the tension in the wire to be the weight of the suspended mass; neglect the small variation intension along the wire that is due to the mass of the wire. The time is the distance (14.0 m) divided by the wavespeed. Let mbe the mass of the wire and let Mbe the mass of the suspended object.Solve:12.67.Set Up:The heart wall first acts as the listener and then as the source. The positive direction is from listenerto source. The heart wall is moving toward the receiver so the Doppler effect increases the frequency and the finalfrequency received,is greater than the source frequency,Solve:Heart wall receives the sound:gives Heart wall emits the sound:andReflect:and so the approximation we made is very accurate. Within thisapproximation, the frequency difference between the reflected and transmitted waves is directly proportional to thespeed of the heart wall.12.68.Set Up:gives Iif is specified. Then relates I,rand P.
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Decibel, 0.800 kg, 1 7.50 kg, 5 0.500 m, 5 1.26 km