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just olya [345]
3 years ago
6

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

Physics
1 answer:
Angelina_Jolie [31]3 years ago
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.


I hope some of this information I gave you can help you. I came up with everything myself to help you.
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Volgvan

Answer:

top speed = 17.25

Total height = 281.19 m

Explanation:

given data

mass = 75 kg

thrust = 160 N

coefficient of kinetic friction = 0.1

solution

we get here frictional force acting that is

frictional force = \mu *m*g   .............1

frictional force = 0.1 × 75 × 9.8

frictional force = 73.5 N

and

Net force acting will be F = 160 - 73.5  N

F = 86.5 N

so

Acceleration in the First 15 second  will be

F = ma .........2

86.5 = 75 × a

a = 1.15 m/s²

and

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v = u + at   ..........3

here u is 0

so v will be

V = 1.15 × 15

v = 17.25

and

now we get travels distance S  in 15 s

s = u × t + 0.5 × a × t²  

here u is 0

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s = 0.5 ×  1.15 × 15²  

s = 129.37 m

and

now  acceleration acting is

F =  \mu *m*g  

m a =  \mu *m*g

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here it is negative it mean downward nature of acceleration

and

now we get distance s by this formula

V² - u² = 2 a s    

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s = 151.82 m

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Answer:

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Explanation:

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Before sinking the can  and lead inside it they are floating in the water.

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