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abruzzese [7]
3 years ago
9

You are recording a friend's 3-minute song with 24 bits per sample at 96 kHz sampling rate for a 5.1 surround sound system (6 ch

annels). How much space will you save by recording this song with 16-bit, 44.1 kHz sampling for a stereo + subwoofer system (3 channels)? Note that you are making these recordings without using any compression. Please give you answer in terms of Gibibits and round to the nearest hundredth.
Engineering
1 answer:
djyliett [7]3 years ago
8 0

Answer:

save space = 1.9626 Gibibits

Explanation:

given data

recording song time = 3 minute = 180 seconds

song per sample =  24 bits per sample

frequency = 96 kHz

surround sound system = 5.1

solution

first we get here required space for 24 bits at  96 kHz that is

required space = 24 × 96 × 10³ × 6 × 180

required space = 2.48832 × 10^{9}  bits

as 1 Gib bit = 2^{30} bits

so required space = 2.48832 × 10^{9}  bits ÷  2^{30} bits

required space = 2.3174 Gibibits   ...............1

and

space required to save record 16 bit at 44.1 kHz

space required = 16 × 44.1 × 10³ × 3 × 180

space required = 0.381024 × 10^{9}  bits

space required = 0.381024 × 10^{9}  bits  ÷  2^{30} bits  

space required = 0.3548 Gibibits    ...........2

so

we get here that save space in 16 bit at 44.1 kHz

save space = 2.3174 Gibibit - 0.3548 Gibibits

save space = 1.9626 Gibibits

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a) \eta_{th} = 10.910\,\%, b) Yes.

Explanation:

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\eta_{th} =\left(1-\frac{253.15\,K}{284.15\,K}  \right) \times 100\,\%

\eta_{th} = 10.910\,\%

b) The claim of the inventor is possible since real efficiency is lower than maximum thermal efficiency.

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4 years ago
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Scorpion4ik [409]
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3 0
3 years ago
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One who waits too long to complain has indicated satisfaction with the agreement despite the initial lack of true consent is the
katovenus [111]

Answer: Doctrine of ratification

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7 0
3 years ago
A transmission line (TL) of length L and conductance per unit length G' is connected to an ideal constant voltage generator V. T
Andru [333]

Answer:

The Current will decrease by a factor of 2

Explanation:

Given the conditions, it should be noted that the current in the circuit is determined by the LOAD. In other words, the amount of current generator will be producing depends upon the load connected to it.

Now, as the question says, the load is reduced to half its original value, we can write:

P1 = \sqrt{3} (V) (I1) Cos\alpha ----- (1)

P2 = \sqrt{3} (V) (I2) Cos\alpha\\

Since, P2 = P1/2,

P1/2 = \sqrt{3} (V) (I2) Cos\alpha ----- (2)\\

Dividing equations (1) and (2), we get,

P1 / (P1/2) = I1/ I2

I2 = I1 / 2\\

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7 0
3 years ago
4) A steel tape is placed around the earth at the equator when the temperature is 0 C. What will the clearance between the tape
snow_lady [41]

Answer:

2102.1 m

Explanation:

Temperature at the equator = 0⁰

Radius of the earth = 6.37x10⁶

Required:

We how to find out what the clearance between tape and ground would be if temperature increases to 30 degrees.

Final temperature = ∆T = 303-273 = 30

S = 11x10^-6

The clearance R = Ro*S*∆T

=6.37x10⁶x 11x10^-6x30

= 2102.1m

Or 2.102 kilometers

Thank you

4 0
3 years ago
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