Assignment: Ultrasound and Infrasound Research Exploration Ultrasound and infrasound are categories of sound with different frequencies. These categories of sound can be used for a variety of different applications. In this research assignment, you will take a look at the industrial applications of ultrasound and infrasound. Your essay should do the following things: • Define and describe in detail (and in your own words) ultrasound and infrasound • Describe how ultrasound and infrasound are used in specific industrial applications and provide detailed examples. When conducting your research, remember to gather information from multiple sources. Start your research by defining ultrasound and infrasound. Then find ideas about industrial applications. The essay should be about 350-450 words, which is 1-1.5 typed pages, double-spaced, using 12-pt. font. Check the rubric to review how you will be graded for this assignment. plz no funny answers

Answers

Answer 1
Answer:

Answer:

Infrasound vs. Ultrasound: Infrasound is sound that is below the lower limit of human hearing, below 20 Hz, and ultrasound is above the upper limit of human hearing, above 20,000 Hz.  Individuals use infrasound - this recurrence run for checking seismic tremors and volcanoes, graphing rock and oil developments underneath the earth. Infrasound is described by a capacity to get around hindrances with little scattering.  

For instance, a few creatures, for example, whales, elephants and giraffes convey utilizing infrasound over significant distances. Torrential slides, volcanoes, seismic tremors, sea waves, water falls and meteors produce infrasonic waves. Symptomatic ultrasound, additionally called sonography or demonstrative clinical sonography, is an imaging technique that utilizes high-recurrence sound waves to create pictures of structures inside your body. The pictures can give important data to diagnosing and treating an assortment of ailments and conditions.

Explanation:

idk how many words this is but its a start for u to add on to and i hope this helps and its in my own words - pls mark me brainiest

Answer 2
Answer:

Answer:

Ultrasound vs. Infrasound Research Exploration

Beyond the limit of human hearing, ultrasound is above 20,000 Hz. Under the limit of human hearing, infrasound is below 20 Hz. Individuals use infrasound - this recurrence run for checking seismic tremors and volcanoes as well as graphing rock and oil developments beneath the earth. Infrasound is described by a capacity to get around hindrances with little scattering.  

For instance, creatures like whales, elephants and giraffes convey utilizing infrasound over significant distances. Torrential slides, volcanoes, seismic tremors, sea waves, waterfalls and meteors produce infrasonic waves. Symptomatic ultrasound, additionally called sonography or demonstrative clinical sonography, is an imaging technique that utilizes high-recurrence sound waves to create pictures of structures inside your body. The pictures can give important data to diagnosing and treating an assortment of ailments and conditions.

(Not turned in yet, but this is what i have so far. Good luck 8th graders <33)

-Sav xx


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The __ is responsible for determining the frequency of vibration of the air column in the pool with in a wind instrumen . A. Type of reed use . B. Effective length of the tube. C humidity of the air

Answers

Answer;

B. Effective length of the tube.

Explanation;

  • Most wind instruments are of either the open-open tube or the open closed tube type. Open ends of the tubes makes contact with the outside, atmospheric air.
  • At an open end, when the air is disturbed in the wind instruments, the air vibrates with a variety of frequencies, however the frequencies that correspond to the natural frequencies will persist.
  • Effective Length of the tube is responsible for determining the frequency of vibration of the air column in the pool with in a wind instrument.

The effective length of the tube is responsible for determining the frequency of vibration of the air column in the tube within a wind instrument.

A 1.00 l sample of a gas at 25.0◦c and 1.00 atm contains 50.0 % helium and 50.0 % neon by mass. what is the partial pressure of the neon. express your answer in atmospheres.

Answers

V = Volume of sample of gas = 1.00 L = 0.001 m³

T = temperature of the gas = 25.0 ⁰C = 25 + 273 = 298 K

P = pressure = 1.00 atm = 101325 Pa

n = number of moles of gas

using ideal gas equation

PV = n RT

101325 (0.001) = n (8.314) (298)

n = 0.041

n₁ = number of moles of helium

n₂ = number of moles of neon

m₁ = mass of helium = n₁ (4) = 4 n₁

m₂ = mass of neon = n₂ (20.2) = 20.2 n₂

given that :

m₁ = m₂

4 n₁ = 20.2 n₂

n₁ = 5.05 n₂

also

n₁ +  n₂ = n

5.05 n₂ +  n₂ = 0.041

n₂ = 0.0068

mole fraction of neon is given as

mole fraction = n₂ /n = 0.0068/0.041 = 0.166

P₂ = partial pressure of neon = (mole fraction) P

P₂ = (0.166) (1)

P₂ = 0.166 atm



The partial pressure of the neon is  0.16 atm.

Explanation:  

 A 1.00 l sample of a gas at 25.0◦c and 1.00 atm contains 50.0 % helium and 50.0 % neon by mass. what is the partial pressure of the neon. express your answer in atmospheres.

The partial pressure is the notional pressure of constituent gas if it is occupied alone by the entire volume of the original mixture at the same temperature.

50.0 % helium and 50.0 % neon by mass.

P = 1 atm

Mass of the gases = M

Moles of helium = (0.5M)/(4 g/mol)\nMoles of neon = (0.5M)/(20 g/mol)

By using ideal gas equation we get:

PV=nRT\nn=(PV)/(RT)=(1.00 L* 1.00 atm)/(0.0820 L atm/mol K* 298 K)=0.0409 mol

where n =total number of moles of gases

0.0409 mol=(0.5M)/(4 g/mol)+(0.5M)/(20 g/mol)

M = 0.2728 g

Then we can find the mass and mole of gases

The gas molar volume is the volume of one mole of the gas at a given temperature and pressure.

Mass of helium gas = (0.2728g)/(2)=0.1364 g

Moles of helium gas =n_1=(0.1364 g)/(4 g/mol)=0.0341 mole

Mass of neon gas  = (0.2728g)/(2)=0.1364 g

Moles of neon gas = n_2=(0.1364 g)/(20 g/mol)=0.00682mole

Partial pressure of the neon gas:P* \chi_2=P* (n_2)/(n_1+n_2)=1 atm* (0.00682)/(0.00682+0.0341)=0.16 atm

Therefore the partial pressure of the neon gas is 0.16 atm.

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What are minerals made of a single element called?

Answers

Minerals of a single element are called Native elements. These include gold, silver, copper, etc. They can be divided into three groups such as metals, semi-metals, and nonmetals. Hope this helped.

The doppler effect is the change in observed frequency due toa. the type of medium the wave is in
b. the type of wave
c. the original frequency of the source
d. the motion of the source or observer

Answers

d. the motion of the source or observer

I believe the answer is D, im not 100% sure though.

Select the type of volcanoes that frequently erupt violently?A. Cinder cone

B. Stratovolcano

C. Shield volcano

* it’s not B

Answers

the correct answer would be

C

Using a cathode ray tube, Thomson confirmed thatatoms must have an overall positive charge.

atoms are made of particles that have a negative charge.

electrons in atoms must have varying amounts of energy.

electrons are only found in the nucleus.

Answers

Using a cathode ray tube, Thomson confirmed that atoms are made of particles that have a negative charge., therefore the correct answer is option B

What are atomic models?

There are some models that are used to explain the arrangements of subatomic particles inside the atom based on the atomic theory of atom are known as the atomic models.

J.J. Thomson discovered using cathode ray tubes that something emits from the negative plate (referred to as the cathode) and travels towards the positive plate (referred to as the anode), and that when zinc sulfide (also known as a phosphor) coating is applied to the perforated anode, it exhibits green fluorescence at a specific pressure and temperature values.

Thus,by using a cathode ray tube, Thomson confirmed that atoms are made of particles that have a negative charge., therefore the correct answer is option B

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Option (B) is correct .i.e. atoms are made of particles that have negative charge.

Further explanation:

Using cathode ray tube, J.J Thomson found that something emits from the negative plate (called as cathode) and goes towards the positive plate (called as anode), and on applying zinc sulfide (also called phosphor) coating on the perforated anode, it shows green fluorescence at some particular values of pressure and temperature.

William Crookes called them the cathode rays as were originated from cathode. Further, Atoms are overall neutral having equal amounts of positive and negative charge as he explained in his plum pudding model.

Although, electrons in atoms have varying energy but this was not confirmed by J.J. Thomson, also electrons can’t be inside nucleus because this would make atoms very unstable.

Some of the properties of those cathode rays are as follows:

(1). They are made up of material particles (they made the light wheels rotate that come in their path).

(2). The green fluorescence was also observed when he putted two more plates perpendicular to the earlier path and the rays followed a curved path towards the new anode.

(3). They has an ionizing property (they ionizes the gas present in the cathode ray tube, through which they pass).

Thus, Option (B) is correct because the cathode rays are emitted from the metallic plate which’s atoms have particles having negative charge.

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Answer Details:

Grade: High School

Subject: Physics

Chapter: Modern physics

Keywords:

Atoms, cathode, cathode rays, cathode ray tubes, J.J. Thomson, negatively charged particles, nucleus, energy, varying energy, electrons, green fluorescence, phosphor, anode, protons, pressure, temperature, William Crookes.