# Physics 112 Quiz 1 Ch 16

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physics chapter 16 quiz. Study for test please.

• 1.

### A source emits a sound of constant frequency, f. If this sound enters an environment where the speed of sound is greater f stays the same and:

• A.

Lambda becomes longer.

• B.

Lambda becomes shorter.

• C.

Lambda stays the same if f is unchanged.

• D.

No. the speed of sound is a "constant" so f must change along with lambda.

• E.

No. the speed of sound, wavelength and frequency will all be unchanged.

A. Lambda becomes longer.
Explanation
When the sound enters an environment where the speed of sound is greater, the wavelength (lambda) becomes longer. This is because the speed of sound is directly proportional to the wavelength. As the speed of sound increases, the wavelength also increases to maintain a constant frequency (f). This is known as the wavelength-frequency relationship, where wavelength is inversely proportional to frequency. Therefore, if the speed of sound increases, the wavelength must also increase in order to keep the frequency constant.

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• 2.

### A singer practicing the vocal scale singing "do, re,mi,fa,so,la,ti,do" produces sounds

• A.

At progressively longer lambda and lower frequency

• B.

At progressively shorter lambda and lower frequency

• C.

At progressively longer lambda and higher frequency

• D.

At progressively shorter lambda and higher frequency.

• E.

At progressively higher frequencies but constant wavespeed and wavelength.

D. At progressively shorter lambda and higher frequency.
Explanation
As the singer practices the vocal scale singing "do, re, mi, fa, so, la, ti, do", the sounds produced will have progressively shorter wavelengths and higher frequencies. This is because as the pitch increases, the frequency of the sound waves also increases, resulting in shorter wavelengths.

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• 3.

### The wavelength emitted during a persons speech depend on several factors, including the dimensions of their mouth and related cavities. if a person has inhaled helium gas, the speed of the sound waves through this helium gas will be > so when they speak, the same wavelengths and this greater sound speed result in speech at frequencies that are

• A.

Lower than for their normal speech in air

• B.

The same as for their normal speech in air.

• C.

Greater than for their normal speech in air

• D.

Unable to be heard by stationary observers

• E.

More than one of these is correct

C. Greater than for their normal speech in air
Explanation
When a person inhales helium gas, the speed of sound waves through the helium gas is greater compared to air. The wavelength of sound depends on the speed of sound, so with the increased speed in helium, the wavelength of the person's speech will also increase. Since frequency is inversely proportional to wavelength, the increase in wavelength will result in a decrease in frequency. Therefore, the person's speech will have lower frequencies, making it sound deeper or lower than their normal speech in air.

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• 4.

### At a fireworks display an exploding aerial bomb produces a sound intensity of 1.0E-2 w/m at a distance of 100 m from the explosion. the surface area for a sphere is 4pie R^2. what would be the total power output of this explosion, assuming it is uniform and unfocussed.

• A.

3.2E-7 W

• B.

1.0E-2 W

• C.

3.1E2 W

• D.

1.3E3 W

• E.

None of these

D. 1.3E3 W
Explanation
The sound intensity of 1.0E-2 W/m at a distance of 100 m can be used to calculate the power output of the explosion. The surface area of a sphere is 4Ï€R^2, so the power output can be found by multiplying the sound intensity by the surface area at that distance. By plugging in the values, we get 1.0E-2 W/m * 4Ï€(100^2) = 1.3E3 W. Therefore, the total power output of the explosion is 1.3E3 W.

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• 5.

### To have resonance and a standing wave on the sting of a violin or guitar, for the displacement of the string there should be

• A.

Nodes at both ends

• B.

Loops or antinodes at both ends

• C.

A loop or antinode at one end and a node at the other end.

• D.

Nodes but no loops or antinodes since this motion would destroy the standing wave.

• E.

None of these

A. Nodes at both ends
Explanation
In order to have resonance and a standing wave on the string of a violin or guitar, there should be nodes at both ends. Nodes are points of zero displacement in a standing wave, and they occur at the ends of the string. This is because the ends of the string are fixed and cannot vibrate, resulting in zero displacement at those points. Having nodes at both ends allows for the formation of a standing wave pattern, which is necessary for resonance to occur.

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• 6.

### While drinking soda from a bottle, you pause periodically and irritate the people around you by blowing across the top of the bottle to make it toot. the more soda you have removed:

• A.

The lower the pitch of the resulting sound

• B.

The higher the pitch of the resulting sound

• C.

The pitch stays the same the stand waves just get longer

• D.

The pitch stays the same the standing waves just get shorter

• E.

None of these

A. The lower the pitch of the resulting sound
Explanation
As you drink soda from the bottle, the amount of soda decreases, leaving more empty space inside the bottle. Blowing across the top of the bottle creates a sound by causing the air inside to vibrate. When there is more soda in the bottle, the air column above it is shorter, resulting in a higher frequency and therefore a higher pitch sound. As you remove more soda, the air column becomes longer, causing a decrease in frequency and a lower pitch sound. Therefore, the lower the pitch of the resulting sound corresponds to the more soda you have removed.

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• 7.

### Ultrasound refers to sound waves of very

• A.

High intensities

• B.

High frequencies

• C.

Long wavelengths

• D.

All of these

• E.

None of these

B. High frequencies
Explanation
Ultrasound refers to sound waves of high frequencies. High frequency sound waves have shorter wavelengths and are capable of penetrating through different materials. Ultrasound is commonly used in medical imaging to create images of internal organs and tissues. It is also used in various industrial applications such as cleaning, measuring distance, and detecting flaws in materials. Therefore, the correct answer is high frequencies.

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• 8.

### When you hold a seashell to your hear the "sound of the sea" can be heard. this is due to the:

• A.

Marine environment in which the shell was formed

• B.

Shell blocking all other sound from reaching you.

• C.

Action of this shellular telephone that transmits sounds from distant shores to you.

• D.

Sounds whose intensities are too small to be audible under normal conditions are enhanced by reflection within the shell producing noise like the ocean.

• E.

Sounds whose intensities are too small to be audible under normal conditions are enhanced by transmission through the solid walls of the shell, producing noise like the ocean.

D. Sounds whose intensities are too small to be audible under normal conditions are enhanced by reflection within the shell producing noise like the ocean.
Explanation
When you hold a seashell to your ear, you hear the "sound of the sea" because sounds whose intensities are too small to be audible under normal conditions are enhanced by reflection within the shell, producing noise similar to the ocean. The shape and structure of the shell act as a natural amplifier, reflecting and amplifying the sound waves that enter the shell. This phenomenon creates the illusion of hearing the sound of the sea when in reality it is just the amplified ambient noise around us.

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• 9.

### The external auditory canal of the human ear is about .027 m long with diamier of .007 m. If this is approximated as atube, closed at one end, with sound waves speed of 344 m/s this tube would have a resonant frequency of:

• A.

0.027 m

• B.

Less than 10 hz

• C.

3.2E3 hz

• D.

6.4E3 hz

• E.

1.3E4 hz

C. 3.2E3 hz
Explanation
The resonant frequency of a tube closed at one end can be calculated using the formula f = (n * v) / (4 * L), where f is the frequency, n is the harmonic number, v is the speed of sound, and L is the length of the tube. In this case, the length of the tube is given as 0.027 m and the speed of sound is given as 344 m/s. Since the tube is closed at one end, the harmonic number is 1. Plugging these values into the formula, we get f = (1 * 344) / (4 * 0.027) = 3200 Hz, which is equivalent to 3.2E3 Hz. Therefore, the resonant frequency of the tube is 3.2E3 Hz.

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• Current Version
• Jan 28, 2024
Quiz Edited by
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• Feb 16, 2009
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