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Ronch [10]
4 years ago
7

Blank Complete the following paragraph pertaining to the popular audio file formats.

Engineering
2 answers:
erastova [34]4 years ago
7 0

Answer:

So #1 is A and #2 is c.

Explanation:

<u><em>#1.  </em></u>MP3 (MPEG-1 Audio Layer-3) is a standard technology and format for compression a sound sequence into a very small file (about one-twelfth the size of the original file) while preserving the original level of sound quality when it is played.

<u><em>#2.</em></u> AIFF (Audio Interchange File Format) is one of the two most-used audio file formats used in the Apple Macintosh operating system. The other is Sound Designer II (SDII). ... An AIFF file contains the raw audio data, channel information (monophonic orstereophonic), bit depth, sample rate, and application-specific data areas.

Aleks04 [339]4 years ago
3 0

Answer:

A. and B

Explanation:

1. Compresses the size of a file, reducing it to around one-twelfth of the original file. A. MP3

MP3 is an audio format that has an audio quality very similar to a CD Audio track, but with a smaller size.

2. AIFF is widely used in the Apple Macintosh operating system.

AIFF (Audio Interchange File Format) along with SDII (Sound Designer II) are the most common audio files formats used in Apple Macintosh operating systems.

Feel free to ask for more if needed or if you did not understand something.

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SCORPION-xisa [38]

Answer:

F_{thrust} ≅ 111 KN

Explanation:

Given that;

A medium-sized jet has a 3.8-mm-diameter i.e diameter (d) = 3.8

mass = 85,000 kg

drag co-efficient (C) = 0.37

(velocity (v)= 230 m/s

density (ρ) = 1.0 kg/m³

To calculate the thrust; we need to determine the relation of the drag force; which is given as:

F_{drag} = \frac{1}{2} × CρAv²

where;

ρ = density of air wind.

C = drag co-efficient

A = Area of the jet

v = velocity of the jet

From the question, we can deduce that the jet is in motion with a constant speed; as such: the net force acting on the jet in the air = 0

SO, F_{drag}-F_{thrust} = 0

We can as well say:

F_{drag}= F_{thrust}

We can now replace F_{thrust} with F_{drag} in the above equation.

Therefore, F_{thrust} = \frac{1}{2} × CρAv²

The A which stands as the area of the jet is given by the formula:

A=\frac{\pi d^2}{4}

We can now have a new equation after substituting our A into the previous equation as:

F_{thrust} = \frac{1}{2} × Cρ (\frac{\pi d^2}{4})v^2

Substituting our data from above; we have:

F_{thrust} = \frac{1}{2} × (0.37)(1.0kg/m^3)(\frac{\pi(3.8m)^2 }{4})(230m/s)^2

F_{thrust} = \frac{1}{8}   (0.37)(1.0kg/m^3)({\pi(3.8m)^2 })(230m/s)^2

F_{thrust} = 110,990N

F_{thrust}  in N (newton) to KN (kilo-newton) will be:

F_{thrust} = (110,990N)*\frac{1KN}{1,000N}

F_{thrust} = 110.990 KN

F_{thrust} ≅ 111 KN

In conclusion, the jet engine needed to provide 111 KN thrust in order to cruise at 230 m/s at an altitude where the air density is 1.0 kg/m³.

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expeople1 [14]

Answer:

answer is

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