A company designed and sells an ultrasonic receiver, which detects sounds unable to be heard by the human ear. The receiver can
detect mechanical and electrical sounds such as leaking gases, air, corona, and motor friction noises. It can also be used to hear bats, insects, and even beading water. The receiver is 2121 in. deep. The focus is 7 in.7 in. from the vertex. Find the diameter of the outside edge of the receiver.
1 answer:
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Apparent frequency heard by the staff: 389 Hz
Explanation:
The phenomenon described in this situation is called Doppler effect.
Doppler effect occurs when there is a source emitting a wave in relative motion with respect an observer. In such situation, the frequency of the wave as perceived by the observer ("apparent frequency") is shifted from the real frequency of the sound ("proper frequency"). In particular:
- The observer perceives a higher frequency if the source is moving towards them
- The observer perceives a lower frequency if the source is moving away from them
The formula to calculate the apparent frequency in the Doppler effect is
where
f is the proper frequency
f' is the apparent frequency
v is the speed of the wave
is the velocity of the observer (positive if they are moving towards the source, negative if moving away)
is the velocity of the source (positive if it is moving away, negative if moving towards the observer)
First of all, in this problem we have to calculate the proper frequency of the sound wave emitted from the ambulance; we have:
v = 343 m/s (speed of sound wave)
(wavelength)
So the proper frequency is
Now we can calculate the apparent frequency heard by the staff at the hospital when the ambulance moves away; we have:
(velocity of the ambulance)
(velocity of the staff)
Substituting,
Learn more about frequency and wavelength:
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Answer:
The graphs are attached
Explanation:
We are told that he starts with a constant speed of 25 m/s for a distance of 100 m.
At constant velocity, v = distance/time
time(t) = distance(d)/velocity(v)
t1 = 100/25
t1 = 4 s
Now, we are told that he applies his brakes and accelerates uniformly to a stop just as he reaches a wall 50m away.
It means, he decelerate and final velocity is zero.
Thus;
v² = u² + 2as
0² = 25² + 2a(50)
25² = - 100a
625 = - 100a
a = - 625/100
a = - 6.25 m/s²
v = u + at
0 = 25 + (-6.25t)
25 = 6.25t
t = 25/6.25
t = 4 s
With the values gotten, kindly find attached the distance-time and velocity-time graphs.
