Question

A dolphin (Dolphin A) is swimming under the ocean at a speed of 4.87 m/s, emitting sonar sound wave at a frequency of 1222 Hz. The speed of sound waves under the ocean is 5518.8 km/h. Another dolphin (Dolphin B), located away is such that the two dolphins move towards each other. If Dolphin B is moving at 7.31 m/s, calculate the frequency that would be detected by a receiver attached to Dolphin B as the two fish swim towards each other. Answer: Sound waves are used in speed camera. Select one: O a. True O b. False A dung beetle's image located parallel to the central axis of a spherical mirror is exactly at 13.86 cm. If the focal length of the spherical mirror is 20.46 cm. Calculate the distance (in cm) of the beetle from the spherical mirror. Answer:

Answer 1

So as two dolphins move toward each other that's why resulting frequancy heard by receiver attached to dolphin B increases due to doppler effect

and given by

$f'=\frac{(v+v_r)}{(v-v_s)}f$

here f= 1222Hz

v= 5518.8 km/h = 1533m/s

and vr = 7.31 m/s
vs = 4.87 m/s

so we get

f' = 1231.74 Hz

2) Speed camera measures speed of vehicle using doppler effect so sound waves must be used here

true

3) So,

q= 13.86 cm

f= 20.46cm

so,

f 9 D

so

$\frac{1}{20.46}=\frac{1}{13.86}-\frac{1}{p}$

so we get object distance from spherical mirror

p = 42.97 cm