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Vadim26 [7]
3 years ago
12

Where is the frequency of ultrasound in relation to the range of human ability to hear

Physics
2 answers:
Arte-miy333 [17]3 years ago
8 0

The frequency of "ultra"sound is <em>much much much higher</em> than any frequency that any human being can hear.

Ultrasound frequencies in diagnostic radiology range from 2 MHz to approximately 15 MHz (although even higher frequencies may be used in some situations) !  That's a range from <em>100 times</em> to <em>750 times</em> the average limit of human hearing (20,000 Hz) !  (See more numbers below.)

Even when ultrasound blasts through your canal, shakes your tympanum, oscillates your ossicles, and attacks your hair cells, your brain never has a clue.

= = = = = = = = = = = = = = =

Looking at it another way:

-- The exact upper and lower limits of human hearing are different for just about every individual on Earth.  As a broad average for the whole human race, we commonly consider the range to be from 20 Hz to 20,000 Hz .

That's a range of almost exactly <u><em>10 octaves</em></u>.

-- Ultrasound operating at 2 MHz is <u><em>another 6.6 octaves</em></u> above the upper audible frequency limit.

--  Ultrasound operating at 15 MHz is <u><em>another 9.5 octaves</em></u> above the upper audible frequency limit.  

kykrilka [37]3 years ago
4 0

Answer:

ultra sounds have frequency higher than the upper audible limit of human hearing, for healthy, young adults.

Explanation:

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galina1969 [7]

Answer:

Second option 6.3 N at 162° counterclockwise from  

F1->

Explanation:

Observe the attached image. We must calculate the sum of all the forces in the direction x and in the direction y and equal the sum of the forces to 0.

For the address x we have:

-F_3sin(b) + F_1 = 0

For the address and we have:

-F_3cos(b) + F_2 = 0

The forces F_1 and F_2 are known

F_1 = 5.7\ N\\\\F_2 = 1.9\ N

We have 2 unknowns (F_3 and b) and we have 2 equations.

Now we clear F_3 from the second equation and introduce it into the first equation.

F_3 = \frac{F_2}{cos (b)}

Then

-\frac{F_2}{cos (b)}sin(b)+F_1 = 0\\\\F_1 = \frac{F_2}{cos (b)}sin(b)\\\\F_1 = F_2tan(b)\\\\tan(b) = \frac{F_1}{F_2}\\\\tan(b) = \frac{5.7}{1.9}\\\\tan^{-1}(\frac{5.7}{1.9}) = b\\\\b= 72\°\\\\m = b +90\\\\\m= 162\°

Then we find the value of F_3

F_3 = \frac{F_1}{sin(b)}\\\\F_3 =\frac{5.7}{sin(72\°)}\\\\F_3 = 6.01 N

Finally the answer is 6.3 N at 162° counterclockwise from  

F1->

7 0
3 years ago
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tekilochka [14]

Answer:

changing the direction in which a force is exerted

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Answer:

No

Explanation:

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The graph below shows the displacement of an object as a constant
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3 years ago
In 2 1/2 hours an airplane travels 1150 km against the wind. It takes 50 min to travel 450 km with the wind. Find the speed of t
Tamiku [17]

Answer:

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Explanation:

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Applying the value of velocity of wind to the first equation

V_a-40=460\\\Rightarrow V_a =500\ km/h

∴ Velocity of airplane in still air is 500 km/h and Velocity of wind is 40 km/h

5 0
3 years ago
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