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3 years ago

Describe how ultrasound and infrasound are used in specific industrial applications and provide detailed examples.

1 answer:

4 0

In scientific terms, ultrasound is a sound pressure, cyclic in nature, that has a greater frequency than the limit at the top of human hearing capabilities. What this means is that an ultrasonic sound can’t be heard by the human ear because their frequency is too high for our ears to pick up. In healthy young adults, this upper hearing capability is an average of 20 kilohertz. Ultrasound has many applications in several fields. Perhaps the best known application for ultrasound is sonography. This is where medical staff use the high pitched noise to produce a picture of a fetus while in the mother’s womb. Another use however, doesn’t directly concern humans at all. Bats use the high pitched noises to see in the dark and get an accurate reading on their preys internal structure. A popular belief is that an ultrasonic sound has the ability to turn the locking mechanism in a door lock, as demonstrated on some spy movies. On the opposite side of this are infrasonic sounds. These are noises with a frequency less than the lowest level of human hearing capabilities is 20 hertz. It is possible for humans to perceive infrasonic sounds, but only if the air pressure is sufficient. Although the war is the main tool for hearing these low sounds, it is possible for other parts of the body to “feel them”. Infrasound can be used to send signals in the army to special machines that can pick them up. These can be used to transmit vital data. Animals are able to pick up some low infrasonic noises which warn them of natural disasters before they happen, generally earthquakes and tsunamis.

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3. The propeller of a World War II fighter plane is 2.30 m in diameter. (a) What is its angular velocity in radians per second i

Darya [45]

Answer:

(a) Angular velocity will be 125.6 rad/sec

(b) Linear velocity will be 144.44 m /sec

Explanation:

We have given diameter d = 2.30 m

So radius r = $\frac{d}{2}=\frac{2.30}{2}=1.15m$

(a) Speed is given as 1200 rev/min

We know that angular velocity is given by $\omega =\frac{2\pi N}{60}=\frac{2\times 3.14\times 1200}{60}=125.6rad/sec$

(b) Linear speed is given by $v=\omega r=125.6\times 1.15=144.44m/sec$

We know that $g=9.81m/sec^2$

So $18141.66m/sec^2=\frac{18141.664}{9.81}=1849.3031g$

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2 years ago

If a rock is dropped from the top of a tower at the front of it and takes 3.6 seconds to hit the ground. Calculate the final vel

expeople1 [14]

Answer:

35.28m/s; 63.50m

Explanation:

<u>Given the following data;</u>

Time, t = 3.6 secs

Since it's a free fall, acceleration due to gravity = 9.8m/s²

Initial velocity, u = 0

To find the final velocity, we would use the first equation of motion;

$V = u + at$

Substituting into the equation, we have;

$V = 0 + 9.8 * 3.6$

V = 35.28m/s

Therefore, the final velocity of the penny is 35.28m/s.

To find the height, we would use the second equation of motion;

$S = ut + \frac {1}{2}at^{2}$

Substituting the values into the equation;

$S = 0(3.6) + \frac {1}{2}*9.8*(3.6)^{2}$

$S = 0 + 4.9*12.86$

$S = 0.5 *36$

S = 63.50m

Therefore, the height of the tower is 63.50m.

6 0

3 years ago

Question 14 of 25

natali 33 [55]

the answer is sueist

Explanation:

7 0

3 years ago

Read 2 more answers

What does hydraulic mean?

Harrizon [31]

Answer:

denoting, relating to, or operated by a liquid moving in a confined space under pressure.

4 0

2 years ago

When you irradiate a metal with light of wavelength 433 nm in an investigation of the photoelectric effect, you discover that a

sergij07 [2.7K]

Answer:

The right solution is:

(a) 2.87 eV

(b) 1.4375 eV

Explanation:

Given:

Wavelength,

= 433 nm

Potential difference,

= 1.43 V

Now,

(a)

The energy of photon will be:

E = $\frac{6.626\times 10^{-34}\times 3\times 10^8}{433\times 10^{-9}}$

= $4.59\times 10^{-19} \ J$

or,

= $\frac{4.59\times 10^{-19}}{1.6\times 10^{-19}}$

= $2.87 \ eV$

(b)

As we know,

⇒ $Vq=\frac{hc}{\lambda}-\Phi_0$

By substituting the values, we get

⇒ $1.43\times 1.6\times 10^{19}=\frac{6.626\times 10^{-34}\times 3\times 10^8}{433\times 10^{-9}}-\Phi_0$

⇒ $\Phi_0=2.3\times 10^{-19} \ J$

or,

⇒ $=\frac{2.3\times 10^{-19}}{1.6\times 10^{-19}}$

⇒ $=1.4375 \ eV$

5 0

3 years ago