# Rt Simulation Examination 2

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This is a Simulated Examination for Gulf RT Examinations taken from Last Month's HAAD Feedbacks.
This examination contains 100 of the most UPDATED EXAMS from Abu Dhabi, KSA, and UAE.
Take this examination for 120 minutes. You need to get 86% to pass the HAAD. 60% to pass MOH, DHA, or Prometrics.
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• 1.

### The dB is defined as the _______________ of two intensities

• A.

Sum

• B.

Difference

• C.

Product

• D.

Ratio

D. Ratio
Explanation
The dB is defined as the ratio of two intensities. It measures the relative difference between the two intensities, rather than their absolute values. This allows for a more convenient and standardized way of expressing the difference in power or amplitude between two signals.

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• 2.

### If the power of a sound wave is increased by a factor of 8 how many decibles is this?

• A.

3 dB

• B.

6 dB

• C.

9 dB

• D.

8 dB

C. 9 dB
Explanation
3dB means 2 times bigger 6dB is 3dB + 3dB. Therefore 6dB means 2 x 2 or 4 times bigger. Therefore 8 dB 2x2x2=8 or 3dB + 3dB + 3dB = 9 dB bigger

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• 3.

### An ultrasound system is set at 0 dB and is transmitting at full intensity.  What is the output power when the system is transmitting at 50% of full intensity?

• A.

-3dB

• B.

-50 dB

• C.

10 dB

• D.

3 dB

• E.

-10 dB

A. -3dB
Explanation
50% of full intensity is 1/2. 0 dB is full intensity so -3dB would be half

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• 4.

### An ultrasound system is set at 0 dB and is tranmitting at full intensity. What is the output power when the system is transmitting at 10% of full intensity?

• A.

- 3dB

• B.

-50 dB

• C.

40 dB

• D.

3 dB

• E.

-10 dB

E. -10 dB
Explanation
0 dB = full intensity 10% of full intensity would be negative or 1/10 or -10dB

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• 5.

### What may be a unit of amplitude?

• A.

Cm

• B.

Hz

• C.

Msec

• D.

Watts

• E.

None of the above

A. Cm
Explanation
The unit of amplitude is typically measured in centimeters (cm). Amplitude refers to the maximum displacement or distance from the equilibrium position in a wave or oscillation. It is a measure of the intensity or strength of the wave and is usually represented by the height or depth of the wave. Therefore, cm is a suitable unit to measure the amplitude as it represents the distance or displacement.

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• 6.

• A.

2 kHz

• B.

2.5 kHz

• C.

7.5 kHz

• D.

14 kHz

D. 14 kHz
• 7.

### According to the AIUM statement on bioeffects, there have been no confirmed bioeffects below intensities of __________ watts per square centimeter SPTA.

• A.

10

• B.

0.01

• C.

0.1

• D.

1

• E.

100

C. 0.1
Explanation
According to the AIUM statement on bioeffects, there have been no confirmed bioeffects below intensities of 0.1 watts per square centimeter SPTA. This means that any ultrasound exposure below this intensity level has not been found to cause any harmful effects on biological tissues or organisms.

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• 8.

### Which of the following is not a measure of area?

• A.

Square cm

• B.

Meters squared

• C.

Cubic meters

• D.

Feet x feet

C. Cubic meters
Explanation
Cubic meters is not a measure of area because it is a measure of volume. Area is a two-dimensional measurement that describes the amount of space an object or surface occupies, while volume is a three-dimensional measurement that describes the amount of space an object or substance occupies in all three dimensions. Therefore, cubic meters, which represents the volume of an object or substance, is not a measure of area.

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• 9.

### Which of the following is true of all waves?

• A.

They travel through a medium

• B.

All carry energy from one side to another

• C.

Their amplitudes do not change

• D.

They travel in a straight line

B. All carry energy from one side to another
Explanation
All waves carry energy from one side to another. This is a fundamental characteristic of waves, regardless of the type of wave or the medium through which they travel. As waves propagate, they transfer energy without necessarily transferring matter. This is evident in various types of waves, such as sound waves, light waves, and water waves. The energy carried by waves can be observed through the effects they produce, such as the movement of particles, the sensation of sound or light, or the displacement of water.

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• 10.

### Is Intensity an acoustic variable?

• A.

True

• B.

False

B. False
Explanation
Intensity is not an acoustic variable. Acoustic variables are properties that describe sound waves, such as frequency, wavelength, and amplitude. Intensity, on the other hand, is a measure of the power or energy carried by a sound wave per unit area. It is related to the amplitude of the wave but is not considered an acoustic variable itself. Therefore, the statement "Intensity is an acoustic variable" is false.

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• 11.

### A force is applied to a surface.  If the force is tripled and the surface area over which the force is applied is also tripled, what is the new pressure?

• A.

Three times larger than the original

• B.

One third of the original

• C.

Six times more than the original

• D.

Unchanged

D. Unchanged
Explanation
When the force applied to a surface is tripled and the surface area over which the force is applied is also tripled, the new pressure remains unchanged. This is because pressure is defined as force divided by surface area. When both the force and surface area are tripled, the ratio between them remains the same, resulting in an unchanged pressure.

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• 12.

### Which of the following units are appropriate to describe the period of an acoustic wave?  More than one answer maybe correct.

• A.

Minutes

• B.

Microseconds

• C.

Meters

• D.

Mm/us

A. Minutes
B. Microseconds
Explanation
The period of an acoustic wave refers to the time it takes for one complete cycle of the wave. It is commonly measured in units of time, such as minutes and microseconds. These units are appropriate because they directly relate to the time it takes for the wave to complete a cycle. On the other hand, meters and mm/us are units of distance and velocity respectively, and do not directly describe the period of an acoustic wave.

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• 13.

### Two waves arrive at the same location and interfere.  The resultant sound wave is smaller that either of the two original waves.  What is this called?

• A.

Constructive interference

• B.

Angular interaction

• C.

Destructive interference

• D.

In-phase waves

C. Destructive interference
Explanation
Destructive interference occurs when two waves combine and their amplitudes cancel each other out, resulting in a smaller amplitude wave. This can happen when the peaks of one wave align with the troughs of the other wave, causing them to subtract from each other. In this case, the resultant sound wave is smaller than either of the two original waves, indicating destructive interference.

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• 14.

### Ultrasound is defined as a sound with a frequency of

• A.

Greater thatn 20,000 kHz

• B.

Less thatn 1 kHz

• C.

Greater than 10 MHz

• D.

Greater thatn 0.02 MHz

D. Greater thatn 0.02 MHz
Explanation
Ultrasound refers to sound waves with frequencies higher than the upper limit of human hearing, which is typically around 20,000 Hz. The answer "greater than 0.02 MHz" is correct because 0.02 MHz is equivalent to 20,000 Hz. Therefore, any frequency higher than 0.02 MHz would fall within the ultrasound range.

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• 15.

### Compare two sound waves. A and B.  The frequency of a wave A is one-third that of wave B.  How does the period of a wave A compare the period of wave B?

• A.

A is one-third as long a B

• B.

A is the same as wave B

• C.

A is three times as long as B

• D.

Cannot be determined

C. A is three times as long as B
Explanation
The period of a wave is the time it takes for one complete cycle of the wave to occur. Since the frequency of wave A is one-third that of wave B, it means that wave A takes three times longer to complete one cycle compared to wave B. Therefore, the period of wave A is three times as long as the period of wave B.

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• 16.

### Which of the following wave is ultrasonic and least useful in diagnotic ultrasound?

• A.

2 MHz

• B.

2000 Hz

• C.

24 kHz

• D.

7.5 MHz

C. 24 kHz
Explanation
Ultrasonic waves are sound waves with frequencies higher than the upper limit of human hearing, which is typically around 20 kHz. The given options include frequencies of 2 MHz, 2000 Hz, 24 kHz, and 7.5 MHz. Out of these options, 24 kHz is the only frequency that falls within the audible range of human hearing, making it least useful in diagnostic ultrasound. Diagnostic ultrasound typically utilizes frequencies in the range of 2 MHz to 15 MHz, which allows for better resolution and imaging of internal structures.

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• 17.

### True of False:  With standard diagnostic imaging instrumentation, the sonographer has the ability to vary the amplitude of a sound wave produced by the transducer.

• A.

Ture

• B.

False

A. Ture
Explanation
With standard diagnostic imaging instrumentation, the sonographer does have the ability to vary the amplitude of a sound wave produced by the transducer. This is typically done by adjusting the power output of the ultrasound machine, which controls the strength of the sound waves being emitted. By increasing or decreasing the amplitude, the sonographer can adjust the intensity of the sound waves and optimize the image quality for different types of examinations or patient conditions.

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• 18.

### The power of an ultrasound wave can be reported with which of the following units? More than one answer maybe correct.

• A.

Watts/square centimeter

• B.

DB/Cm

• C.

Watts

• D.

Kg/cm^2

C. Watts
Explanation
The power of an ultrasound wave can be reported in watts because watts is the unit of power. Power is the rate at which energy is transferred or the amount of energy per unit time. Since ultrasound waves carry energy, their power can be measured in watts.

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• 19.

### The maximum value of the density of an acoustic wave is 60 lb/in^2 while the minimum density is 20 lb/in^2.  What is the amplitude of the wave?

• A.

20 lb/in^2

• B.

40 lb/in^2

• C.

60 lb/in^2

• D.

None of the above

A. 20 lb/in^2
Explanation
The amplitude of a wave may be calculated by subracting the minimum value of the acoustic variable from its maximum and then dividing that number in half. In this case, the maximum minus the minimum is 60-20=40. Half of 40 is 20

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• 20.

### The final amplitude of an acoustic wave is reduced to one-half of its original value.  The final power is _______ the original power.

• A.

The same as

• B.

One-half

• C.

Double

• D.

None of the above

D. None of the above
Explanation
Changes in the power of a wave are proportional to changes in the waves amplitude squared. 1/2 squared = 1/4 (1/2 + 1/2 = 1/4) When 1/2 of the wave's original amplitude remains, then only 1/4 of the original power remains.

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• 21.

### The amplitude of an acoustic wave decreases from 27 pascals to 9 pascals.  If the initial power in the wave was 27 watts, what is the wave final power?

• A.

3 watts

• B.

9 watts

• C.

1 watt

• D.

None of the above

A. 3 watts
Explanation
The initial power in the wave is given as 27 watts. The amplitude of the wave decreases from 27 pascals to 9 pascals. The power of a wave is directly proportional to the square of its amplitude. Therefore, if the amplitude decreases by a factor of 3 (from 27 to 9), the power will decrease by a factor of 3 squared, which is 9. Therefore, the final power of the wave is 3 watts.

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• 22.

### What are the units of intensity?

• A.

Watts

• B.

Watts/cm

• C.

Watts/cm^2

• D.

DB

C. Watts/cm^2
Explanation
The units of intensity are typically measured in watts per square centimeter (watts/cm^2). Intensity refers to the amount of power per unit area, and watts/cm^2 is the appropriate unit to express this measurement. Watts alone do not account for the area over which the power is distributed, and watts/cm is not a valid unit for intensity as it does not include the squared term for area. dB (decibel) is a unit used to measure the relative intensity or power level of a sound or signal, but it is not the standard unit for intensity.

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• 23.

### If the power in an ultrasound beam is unchanged, while at the same time, the beam area doubles, then the beams's intensity

• A.

Doubles

• B.

Is halved

• C.

Is quartered

• D.

Remains the same

B. Is halved
Explanation
When the beam area doubles while the power remains the same, the intensity of the ultrasound beam is halved. Intensity is defined as power divided by area, so if the area increases while the power remains constant, the intensity decreases. In this case, since the area doubles, the intensity is halved.

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• 24.

### What happens to an acoustic beams's intensity when the power in the beam increases by 25% while the cross sectional area of the beam remain the same?

• A.

It increased by 25%

• B.

Increases by 75%

• C.

It increases by 50%

• D.

Decreases by 25%

A. It increased by 25%
Explanation
When the power in the acoustic beam increases by 25% while the cross-sectional area remains the same, the intensity of the beam also increases by 25%. This is because intensity is directly proportional to power, and since the power has increased by 25%, the intensity will also increase by the same percentage.

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• 25.

### When the power in an acoustic beam is doubled and the cross sectional area of the beam is halved, then the intensity of the beam is

• A.

Doubled

• B.

Halved

• C.

Quartered

• D.

Four time larger

D. Four time larger
Explanation
When the power in an acoustic beam is doubled, it means that the amount of energy transmitted by the beam has increased. However, when the cross-sectional area of the beam is halved, it means that the same amount of energy is now spread over a smaller area. This results in a higher concentration of energy, leading to an increased intensity of the beam. Therefore, the intensity of the beam is four times larger than before.

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• 26.

### What determines the intensity of an ultrasound beam after it has traveled through the body?

• A.

The sound wave's source

• B.

The medium throught which the sound travels

• C.

Both A and B

• D.

Neither choice A nor B

C. Both A and B
Explanation
The intensity of an ultrasound beam after it has traveled through the body is determined by both the sound wave's source and the medium through which the sound travels. The source of the sound wave determines the initial intensity of the beam, while the medium through which it travels can affect the attenuation or absorption of the sound wave, leading to changes in intensity. Therefore, both factors play a role in determining the intensity of the ultrasound beam after it has passed through the body.

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• 27.

### What happens to the intensity of an ultrasound beam when the beam's cross-sectional area remains unchanged while the amplitude of the wave triples.

• A.

It triples

• B.

Increases ninefold

• C.

Remains the same

• D.

None of the above

B. Increases ninefold
Explanation
When the amplitude of an ultrasound wave triples, the intensity of the ultrasound beam increases ninefold. This is because intensity is directly proportional to the square of the amplitude. Therefore, tripling the amplitude will result in the intensity being multiplied by 3^2, which equals 9.

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• 28.

### The wavelength of a cycle in an ultrasound wave can be reported with which units?

• A.

Units of time (seco, min, etc)

• B.

Units of distance (feet, cm)

• C.

Units of area (m^2, etc

• D.

Mm only

B. Units of distance (feet, cm)
Explanation
Ultrasound waves are a type of mechanical wave that travels through a medium, such as air or water. The wavelength of a wave is the distance between two consecutive points that are in phase, or the distance over which the wave completes one full cycle. In the case of ultrasound waves, the wavelength is measured in units of distance, such as feet or centimeters. This is because the wavelength represents the physical length of the wave, indicating how far the wave travels in one complete cycle. Therefore, the correct answer is units of distance (feet, cm).

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• 29.

### Which of the following terms best describes the relationship between frequency and wavelength for sound traveling in soft tissue?

• A.

Receiprocal

• B.

Direct

• C.

Related

• D.

Inverse

D. Inverse
Explanation
The term "inverse" best describes the relationship between frequency and wavelength for sound traveling in soft tissue. Inverse means that as one variable increases, the other variable decreases, and vice versa. In this case, as the frequency of the sound waves increases, the wavelength decreases, and as the frequency decreases, the wavelength increases. This relationship is consistent with the behavior of sound waves in soft tissue.

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• 30.

### (TRUE OR FALSE) The power in the wave increases as it travels through the mass. (An ultrasound pulse propagates from soft tissue through a mass.  Sound's propagation speed in the mass is 1575 m/s.)

• A.

True

• B.

False

B. False
Explanation
False. The power in the wave does not increase as it travels through the mass. The power of a wave is determined by the amplitude of the wave, which does not change as the wave propagates through a medium. Therefore, the power of the wave remains constant regardless of the medium it travels through.

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• 31.

### The frequency closest to the lower limits of US is:

• A.

19,000kHz

• B.

10,000 MHz

• C.

20,000 msec

• D.

15,000 Hz

D. 15,000 Hz
Explanation
The frequency closest to the lower limits of the US is 15,000 Hz. This is because the given options include frequencies measured in kHz (kilohertz), MHz (megahertz), and msec (milliseconds), while the only option measured in Hz (hertz) is 15,000 Hz. Hertz is the standard unit for measuring frequency, and 15,000 Hz is the closest option to the lower limits of the US frequency range.

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• 32.

• A.

30 KHz

• B.

8 MHz

• C.

8,000 Hz

• D.

3,000 kHz

• E.

15 Hz

A. 30 KHz
• 33.

### _____________ is the distance covered by one cycle

• A.

Pulse length

• B.

Wavelength

• C.

Cycle distance

• D.

Cycle duration

B. Wavelength
Explanation
Wavelength is the distance covered by one cycle. It is a measure of the length between two consecutive points of a wave that are in phase. In other words, it represents the distance between two identical points on a wave, such as two crests or two troughs. Therefore, wavelength is the correct answer as it accurately describes the distance covered by one complete cycle of a wave.

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• 34.

### If the frequency of US is increased from 0.77 MHz to 1.54 MHz, what happens to the wavelength?

• A.

Doubles

• B.

Halved

• C.

Remains the same

• D.

4 times greater

B. Halved
Explanation
When the frequency of a wave increases, the wavelength decreases. This is because wavelength and frequency are inversely proportional to each other. Therefore, if the frequency of US is increased from 0.77 MHz to 1.54 MHz, the wavelength will be halved.

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• 35.

### If the frequency is decreased, the numberical value of the radial resolution is

• A.

Increased

• B.

Decreased

• C.

Unchanged

A. Increased
Explanation
When the frequency is decreased, the numerical value of the radial resolution is increased. Radial resolution refers to the ability to distinguish two closely spaced objects along the radial axis. A higher numerical value indicates a greater ability to differentiate between these objects, meaning that the resolution is improved. Therefore, when the frequency is decreased, the radial resolution is increased.

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• 36.

### If we increase the frequency, the near zone length is

• A.

Increased

• B.

Decreased

• C.

Unchanged

A. Increased
Explanation
When the frequency is increased, the near zone length is increased. The near zone refers to the region close to the source of the waves, where the wavefronts are not yet fully developed. As the frequency increases, the wavelength decreases, causing the wavefronts to become closer together. This results in a longer near zone length, as the waves take more time to fully develop and spread out. Therefore, increasing the frequency leads to an increased near zone length.

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• 37.

### The digital scan converter has a number of pixels assigned to each bit.

• A.

True

• B.

False

B. False
Explanation
The statement is false because a digital scan converter does not have pixels assigned to each bit. In a digital scan converter, pixels are assigned to each sample or data point, not to each bit. The number of pixels assigned to each sample or data point depends on the resolution of the scan converter.

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• 38.

### The more bits per pixel

• A.

The better the resolution

• B.

The higher the pizel density

• C.

• D.

Higer reliability

C. The more shades of gray
Explanation
The correct answer is "the more shades of gray." This means that as the number of bits per pixel increases, the resolution improves, resulting in a higher pixel density. This allows for a greater range of shades of gray to be displayed, which ultimately leads to higher image quality and more accurate representation of colors and details. Additionally, having more shades of gray enhances the reliability of the image by reducing banding or color artifacts.

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• 39.

### What is the actual time that an US machine is creating a pulse?

• A.

Duty factor

• B.

Period

• C.

Pulse period

• D.

Pulse duration

D. Pulse duration
Explanation
The pulse duration refers to the actual time that a US (ultrasound) machine is creating a pulse. It measures the length of time that the pulse is being emitted from the machine. The duty factor, period, and pulse period are not directly related to the actual time of pulse creation, making them incorrect options.

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• 40.

### The angle of incidence of an US beam is perpendicular to an interface.  The two media have the same propagation speeds.  What process cannot occur?

• A.

Refraction

• B.

Reflection

• C.

Transmission

• D.

Attenuation

A. Refraction
Explanation
When the angle of incidence of an ultrasound (US) beam is perpendicular to an interface and the two media have the same propagation speeds, the beam does not change direction as it passes through the interface. This means that there is no bending of the beam, which is the process of refraction. Therefore, the process that cannot occur in this scenario is refraction.

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• 41.

### What is the most typical Doppler shift measured clinically?

• A.

3.5 MHz

• B.

3,500,000 Hz

• C.

2kHz

• D.

1,000 kHz

• E.

20,000 Hz

C. 2kHz
Explanation
The most typical Doppler shift measured clinically is 2kHz. The Doppler effect is used in medical imaging to measure blood flow velocity and direction. In this context, a Doppler shift of 2kHz indicates a moderate flow velocity, which is commonly observed in clinical settings. This frequency shift is within the range that can be easily detected and measured using Doppler ultrasound techniques.

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• 42.

### If the frequency is decreased, the numerical value of the radial resolution is

• A.

Increased

• B.

Decreased

• C.

Unchanges

A. Increased
Explanation
When the frequency is decreased, the numerical value of the radial resolution is increased. Radial resolution refers to the ability to distinguish between two points in an image along the radial direction. A higher numerical value indicates a greater ability to distinguish between these points, which means that the radial resolution is increased. This is because decreasing the frequency results in a longer wavelength, which increases the distance between the points that can be resolved. Therefore, the radial resolution is directly affected by the frequency, and decreasing it leads to an increase in the numerical value of the radial resolution.

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• 43.

### The digital scan converter has a number of pixels assigned to each bit.

• A.

True

• B.

False

B. False
Explanation
The statement is false because the digital scan converter does not assign a specific number of pixels to each bit. A digital scan converter is a device that converts analog ultrasound signals into digital signals for display on a monitor. It does this by sampling the analog signal at regular intervals and assigning a binary value (bit) to each sample. The number of pixels in the display is determined by the resolution of the monitor, not by the number of bits in the digital scan converter.

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• 44.

### The more bits per pixel

• A.

The better the resolution

• B.

The higher the pixel density

• C.

• D.

Higher reliability

C. The more shades of gray
Explanation
The more bits per pixel refers to the number of bits used to represent each pixel in an image. A higher number of bits allows for a greater range of shades of gray to be displayed. This means that the image will have more levels of gray between black and white, resulting in a more detailed and realistic representation of the image. Therefore, the statement "the more shades of gray" is the correct answer as it accurately reflects the relationship between the number of bits per pixel and the quality of the image.

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• 45.

### Increasing the frequency increases the penetration depth.

• A.

True

• B.

False

B. False
Explanation
Increasing the frequency does not increase the penetration depth. In fact, higher frequencies result in decreased penetration depth. This is because higher frequencies have shorter wavelengths, which causes the energy to be absorbed and scattered more quickly as it interacts with the material being penetrated. As a result, lower frequencies are generally used when deeper penetration is required, such as in medical imaging or geological surveys.

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• 46.

### All of the following will improve temporal resolution EXCEPT

• A.

Increased line density

• B.

Single rather than multi focus

• C.

Higher frame rate

• D.

Shallower depth of view

• E.

Slower speed of sound in a medium

A. Increased line density
Explanation
Increased line density refers to the number of scan lines used in an ultrasound image. The more scan lines there are, the higher the spatial resolution, which allows for better visualization of small structures. Therefore, increased line density will actually improve temporal resolution, as it allows for more frequent updates of the image over time. Therefore, the correct answer is "increased line density."

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• 47.

### What componet of a transducer changes electrial to mechanical and mechanical back to electrical engery?

• A.

PZT

• B.

Scan converter

• C.

Demodulator

• D.

A. PZT
Explanation
PZT stands for lead zirconate titanate, which is a type of ceramic material commonly used in transducers. PZT is known for its piezoelectric properties, meaning it can convert electrical energy into mechanical energy and vice versa. In the context of a transducer, the PZT component is responsible for converting electrical signals into mechanical vibrations, which are then transmitted into the body or medium being imaged. These mechanical vibrations then bounce back and are converted back into electrical signals by the PZT, allowing the transducer to receive and detect echoes. Therefore, PZT is the component that changes electrical to mechanical and mechanical back to electrical energy in a transducer.

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• 48.

### What determines the frequency of a sound beam from a pulsed transducer?

• A.

Voltage applied to the PZT

• B.

PZT thickness

• C.

Frequency of the system

• D.

PRF of the pulser

B. PZT thickness
Explanation
The frequency of a sound beam from a pulsed transducer is determined by the thickness of the PZT material. The PZT thickness affects the resonant frequency of the transducer, which in turn determines the frequency of the sound beam produced.

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• 49.

### What measures the output of the a transducer?

• A.

• B.

Display

• C.

Hydrometer

• D.

Hydrophone

D. HydropHone
Explanation
A hydrophone is a transducer that measures the output of a transducer. It is specifically designed to detect and measure underwater sound waves, making it the most suitable option for measuring the output of a transducer in an underwater environment. A receiver is a general term and does not specifically measure the output of a transducer. A display is used to show information but does not directly measure the output. A hydrometer is used to measure the specific gravity or density of a liquid, so it is not the correct option for measuring the output of a transducer.

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• 50.

### Which type of artifact appears most commonly with highly reflective objects?

• A.

Reverberations

• B.

Mirroring

• C.

• D.

Enhancement

• E.

Defocusing

Explanation
Highly reflective objects tend to create strong contrasts between light and dark areas, resulting in the appearance of shadows. Shadowing is the process of casting shadows and is commonly observed with highly reflective objects. This occurs when the light source is blocked or partially blocked by the object, causing a shadow to be formed on the surface or surrounding area. Therefore, shadowing is the most common artifact that appears with highly reflective objects.

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• Mar 19, 2023
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• Jul 21, 2016
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