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Zinaida [17]
3 years ago
14

If the bat emits a sound at a frequency of 80.2 kHzkHz and hears it reflected at a frequency of 83.5 kHzkHz while traveling at a

speed of 3.9 m/sm/s , calculate the speed of the insect.
Physics
1 answer:
solong [7]3 years ago
8 0

To solve this problem we will apply the concepts related to the Doppler Effect. The Doppler effect is the change in the perceived frequency of any wave movement when the emitter, or focus of waves, and the receiver, or observer, move relative to each other.

Mathematically it can be described as

v' = v (\frac{f' ( v - v_{bat}) - f ( v + v_{bat})}{f'(v-v_{bat}) + f ( v + v_{bat})})

Where,

v = Speed of sound

f' = Reflected frequency

f = Emitted Frequency

v_{bat} = Speed of the object

Assuming that the speed of the medium is the speed of sound of 343m / s, we will then have that:

v' = 343 (\frac{(83.5) (343 -3.9) - 80.2(343+3.9)}{83.5(343-3.9) + 80.2(343+4.3)})

v' = 3.0134m/s

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How do I solve such problem???
pashok25 [27]

As far as I'm concerned, this is a bogus question, or at least a severely corrupted one.

The three numbers given can NOT all be true on Earth.

-- It rolled off the table at 7.6 m/s .  By golly, there you are!  Its initial horizontal velocity is 7.6 m/s, and it has no vertical velocity until it leaves the table.

-- There are no horizontal forces that we're aware of acting on the object.  So it maintains the same horizontal velocity for the rest of the story.  It's 10.5m away from the table in (10.5 m) / (7.8 m/s) = 1.35 second .

-- Vertically, it's just an object dropped from 17.6m off the floor.  Shockingly, the distance it falls in time 'T' is (1/2 g) T².  In 1.35 second, that's 8.88 meters ! . . . only about halfway to the floor !

-- In order to fall 17.6 m to the floor, it would need 1.89 seconds.  In <u>that</u> length of time, however, it would travel (7.8 m/s) x (1.89 s) = 14.78 m away from the base of the table.

So you see, either . . .

-- the table is NOT 17.6m tall, or

-- the object does NOT roll off of the table at 7.8 m/s, or

-- it does NOT land 10.5 m away from the base of the table.

OR . . .

-- the table is not on Earth, and gravity is not 9.8 m/s² !

We often see questions posted on Brainly with not enough given information, OR with some information given that's not needed because it's not involved the answer.  

THIS one is different, and it's unusual.  In this one, we have<em> too much</em> given information, we can't ignore any of it because it's all related, but it's inconsistent and it CAN't all be true.

(Unless the whole story takes place on a mystery planet that is not Earth.  Which I'm not going to take the time and effort right now to figure out what the acceleration of gravity has to be in order to make all of the given information compatible.)

7 0
3 years ago
is the following sentence true or false? the faster the particles of a substance are moving, the more energy they have.
svetlana [45]
I think true. I'm pretty sure, but check w/ others too.
8 0
3 years ago
Read 2 more answers
A cat dozes on a stationary merry-go-round, at a radius of 4.6 m from the center of the ride. then the operator turns on the rid
ioda
<span>Radius = 4.6 m
 Time for one complete rotation t = 5.5 s.
 Distance = 2 x 3.14 x R = 2 x 3.14 x 4.6 m = 28.888.
 Velocity V = distance / time = 28.888 / 5.5 s = 5.25 m/s
 Force exerted by cat Fc = mV^2 / R = (mx 5.25^2) / 4.6 m
  Force of the cat Fc = 6m, m being the mass.
 Normal force = Us x m x g = Us x m x 9.81 = Us9.81m
 equating the both forces => Us9.81m = 6m => Us = 6 / 9.81 => Us = 0.6116 So coefficient of static friction = 0.6116</span>
4 0
3 years ago
A typical wall outlet voltage is 120 volts in the United States. Personal MP3 players require much smaller voltages, typically 2
butalik [34]

Answer:        

Number of turns in secondary will be 7

Explanation:

We have given primary voltage V_p=120volt

Number of turns in the primary is N_p=3575

Secondary voltage is given V_s=235mV=0.235volt

We have to find the number of turns in secondary

We know that \frac{N_p}{N_s}=\frac{V_p}{V_s}

So \frac{3575}{N_s}=\frac{120}{0.235}

N_s=6.60

As the number of turns can not be in decimal so number of turns will be 7

6 0
3 years ago
A circular coil of radius r = 5 cm and resistance R = 0.2 is placed in a uniform magnetic field perpendicular to the plane of th
Yuri [45]

Answer:

the question is incomplete, the complete question is

"A circular coil of radius r = 5 cm and resistance R = 0.2 ? is placed in a uniform magnetic field perpendicular to the plane of the coil. The magnitude of the field changes with time according to B = 0.5 e^-t T. What is the magnitude of the current induced in the coil at the time t = 2 s?"

2.6mA

Explanation:

we need to determine the emf induced in the coil and y applying ohm's law we determine the current induced.

using the formula be low,

E=-\frac{d}{dt}(BACOS\alpha )\\

where B is the magnitude of the field and A is the area of the circular coil.

First, let determine the area using \pi r^{2} \\ where r is the radius of 5cm or 0.05m

A=\pi *(0.05)^{2}\\ A=0.00785m^{2}\\

since we no that the angle is at 0^{0}

we determine the magnitude of the magnetic filed

B=0.5e^{-t} \\t=2s

E=-(0.5e^{-2} * 0.00785)

E=-0.000532v\\

the Magnitude of the voltage is 0.000532V

Next we determine the current using ohm's law

V=IR\\R=0.2\\I=\frac{0.000532}{0.2} \\I=0.0026A

I=2.6mA

6 0
3 years ago
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