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3 years ago

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 much much much higher 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 100 times to 750 times 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 10 octaves.

-- Ultrasound operating at 2 MHz is another 6.6 octaves above the upper audible frequency limit.

-- Ultrasound operating at 15 MHz is another 9.5 octaves 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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You are driving your 1700 kg car at 21 m/s down a hill with a 5.0∘ slope when a deer suddenly jumps out onto the roadway. You sl

podryga [215]

Answer:

d = 27.7 m

Explanation:

Here the car is driving on the inclined plane

So here we can say that work done by the gravity and work done by friction is equal to change in kinetic energy of the system

So here we can write it as

$mgsin\theta \times d - F_f \times d = \frac{1}{2}mv_f^2 - \frac{1}{2}mv_i^2$

now we have

m = 1700 kg

$v_f = 0$

$v_i = 21 m/s$

$F_f = 1.5 \times 10^4 N$

$(1700)(9.81)sin5 (d) - (1.5\times 10^4)d = 0 - \frac{1}{2}(1700)(21^2)$

$1453.5 d - (1.5\times 10^4)d = -374850$

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3 years ago

A bucket that has a mass of 20 kg when filled with sand needs to be lifted to the top of a 15 meter tall building. You have a ro

prohojiy [21]

Answer:

work done lifting the bucket (sand and rope) to the top of the building,

W=67.46 Nm

Explanation:

in this question we have given

mass of bucket=20kg

mass of rope= $.2\frac{kg}{m}$

height of building= 15 meter

We have to find the work done lifting the bucket (sand and rope) to the building =work done in lifting the rope + work done in lifting the sand

work done in lifting the rope is given as, $W_{1}=Force \times displacement$

= $\int\limits^{15}_0 {.2x} \, dx$ ..............(1)

= $.1\times 15^2$

=22.5 Nm

work done in lifting the sand is given as, $W_{2}=Force \times displacement$

$W_{2}=\int\limits^{15}_0 F \, dx$ .................(2)

Here,

F=mx+c

here,

c=20-18

c=2

m= $\frac{20-18}{15-0}$

m=.133

Therefore,

$F=.133x+2$

Put value of F in equation 2

$W_{2}=\int\limits^{15}_0 (.133x+2) \, dx$

$W_{2}=.133 \times 112.5+2\times15\\W_{2}=14.96+30\\W_{2}=44.96 Nm$

Therefore,

work done lifting the bucket (sand and rope) to the top of the building, $W=W_{1}+W_{2}$

W=22.5 Nm+44.96 Nm

W=67.46 Nm

4 0

3 years ago

An object is moving initially with a velocity of 4.7 m/s . After 3.9 s the object's velocity is -2.1 m/s . What is the object's

IgorC [24]

Answer: The acceleration of the object is 0.67m/s^2 west.

Explanation: Here we are given the initial velocity and final velocity as well as the time taken. Acceleration is the change in velocity per unit time, thus the equation becomes.

a=dv/t

a=vf-vi/t

a=-2.1-4.7/3.9

a= 0.66m/s^2 west

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3 years ago

A bucket of mass m is hanging from the free end of a rope whose other end is wrapped around a drum (radius R, mass M) that can r

mr Goodwill [35]

Answer:

Explanation:

Let T be the tension .

Applying newton's second law on the downward movement of the bucket

mg - T = ma

On the drum , a torque of TR will be acting which will create an angular acceleration of α in it . If I be the moment of inertia of the drum

TR = Iα

TR = Ia/ R

T = Ia/ R²

Replacing this value of T in the other equation

mg - T = ma

mg - Ia/ R² = ma

mg = Ia/ R² +ma

a ( I/ R² +m)= mg

a = mg / ( I/ R² +m)

mg - T = ma

mg - ma = T

mg - m x mg / ( I/ R² +m) = T

mg - m²g / ( I/ R² +m ) = T

mg - mg / ( 1 + I / m R² ) = T

b ) T = Ia/ R²

I = TR² / a

c ) Moment of inertia of hollow cylinder

I = 1/2 M ( R² - R² / 4 )

= 3/4 x 1/2 MR²

= 3/8 MR²

I / R² = 3/8 M

a = mg / ( I/ R² +m)

a = mg / ( 3/8 M + m )

T = Ia/ R²

= 3/8 MR² x mg / ( 3/8 M + m ) x 1 /R²

= $\frac{3mMg}{(3M +8m)}$

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4 years ago

After the box comes to rest at position x1, a person starts pushing the box, giving it a speed v1. when the box reaches position

KiRa [710]

As we know by work energy theorem

total work done = change in kinetic energy

so here we can say that wok done on the box will be equal to the change in kinetic energy of the system

$W_p = KE_f - KE_i$

initial the box is at rest at position x = x1

so initial kinetic energy will be ZERO

at final position x = x2 final kinetic energy is given as

$KE_f = \frac{1}{2}mv_1^2$

now work done is given as

$W_p = \frac{1}{2}mv_1^2 - 0$

so we can say

$W_p = \frac{1}{2}mv_1^2$

so above is the work done on the box to slide it from x1 to x2

3 0

3 years ago