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grin007 [14]
3 years ago
14

Determine the size of memory needed for CD recording of a piece of music, which lasts for 26 minutes, is done with a 20-bit Anal

og-to-Digital Converter (ADC) in stereo (2 channels), at the rate of 44.1 kSa/s, with the compression factor 6 (allow 10% error margin).
Engineering
1 answer:
kompoz [17]3 years ago
7 0

Answer: the size of memory needed for the CD recording is 28.7 MB

Explanation:

so in the case of stereo, the bitrate is;

⇒ 26 × 60 × 44.1 × 10³ × 2

=   137592 × 10³  

for 10 bit

⇒ 137592 × 10³ × 10

= 1375920 × 10³ bits

now divide by 8 (convert to bytes)

⇒ (1375920 × 10³) / 8

= 171,990,000 BYTE

divide by 1000 (convert to kilobytes)

= 171,990,000 / 1000

= 171,990 KILOBYTES

now Given that, the compression ratio is 6      

so  

171,990 / 6

= 28665 KB

we know that. 1 MB = 1000 KB

x MB = 28665 KB

x MB = 28665 / 1000

⇒ 28.665 MB ≈ 28.7 MB

Therefore the size of memory needed for the CD recording is 28.7 MB

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What is the minimum efficiency of a functioning current-model catalytic converter? a. 60% b. 75% c. 80% d. 90%
slamgirl [31]

Answer:

d. 90%

Explanation:

As we know that internal combustion engine produce lot's of toxic gases to reduce these toxic gases in the environment a device is used and this device is know as current modeling converter.

Generally the efficiency of current model catalytic converter is more than 90%.But the minimum efficiency this converter is 90%.

So option d is correct.

d. 90%

7 0
3 years ago
Which of the following positions would be responsible for attaching I-beams to a crane?
prohojiy [21]

Answer:

Rigger

Explanation:

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4 0
3 years ago
Design a ductile iron pumping main carrying a discharge of 0.35 m3/s over a distance of 4 km. The elevation of the pumping stati
snow_tiger [21]

Answer:

D=0.41m

Explanation:

From the question we are told that:

Discharge rate V_r=0.35 m3/s

Distance d=4km

Elevation of the pumping station h_p= 140 m

Elevation of the Exit point h_e= 150 m

Generally the Steady Flow Energy Equation SFEE is mathematically given by

h_p=h_e+h

With

P_1-P_2

And

V_1=V-2

Therefore

h=140-150

h=10

Generally h is give as

h=\frac{0.5LV^2}{2gD}

h=\frac{8Q^2fL}{\pi^2 gD^5}

Therefore

10=\frac{8Q^2fL}{\pi^2 gD^5}

D=^5\frac{8*(0.35)^2*0.003*4000}{3.142^2*9.81*10}

D=0.41m

8 0
3 years ago
How to find the voltage(B Aab) in series parallel circuit? ​
Sindrei [870]

Answer:

  Vab ≈ 3.426 V

Explanation:

First of all, it is convenient to find the equivalent parallel resistance of R5 and R6. That will be ...

  R56 = (R5)(R6)/(R5 +R6) = (1000)(1500)/(1000 +1500) = 600

Then we can call V1 the voltage at the top of R2. The voltage at Va is a divider from V1:

  Va = V1·(R4/(R3+R4)) = V1(560/1030) ≈ 0.543689V1

The voltage at Vb is also a divider from V1:

  Vb = V1·(R7+R8)/(R2 +R56 +R7 +R8) = V1(780/1710) ≈ 0.456140V1

The parallel branches containing Va and Vb have an effective resistance of ...

  (1030)(1710)/(1030+1710) = 642.81

That forms a divider with R1 to give V1:

  V1 = (100 V)642.81/(1000 +642.81) ≈ 39.1287 V

The difference Va-Vb is ...

  Vab = (39.1287 V)(0.543689 -0.456140) ≈ 3.426 V

_____

We have done this using parallel resistance and voltage divider calculations. You can also do it using node voltage equations. Using the same definition for V1 as above, we have ...

  (Vs -V1)/R1 +(Vb -V1)/(R56+R2) +(Va-V1)/R3 = 0

  (V1 -Vb)/(R56 +R2) -Vb/(R7+R8) = 0

  (V1 -Va)/R3 -Va/R4 = 0

The solution of interest is the value of Vab, shown in the attachment. It computes as 154200/45013 V ≈ 3.42568 V.

4 0
3 years ago
A) Total Resistance<br> b) Total Current<br> c) Current and Voltage through each resistor
Varvara68 [4.7K]
C my friend 20 characters suck
3 0
3 years ago
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