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Virty [35]
3 years ago
13

In what situation you would prefer to use a successive approximation ADC over flash ADC?

Engineering
1 answer:
Levart [38]3 years ago
8 0

Answer and Explanation:

Both both successive approximation normally known SAR type and flash type ADC are used for the conversion of analog signal to digital signal successive approximation type ADC are preferred over flash type when we need fix conversion time and better accuracy and if we have don/t matter with speed of the conversion because flash type has better speed of conversion than SAR type

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Given asphalt content test data:
VMariaS [17]

Answer:

hello your question is incomplete attached below is the complete question

A) overall mean = 5.535,  standard deviation ≈ 0.3239

B ) upper limit = 5.85, lower limit = 5.0

C) Not all the samples meet the contract specifications

D) fluctuation ( unstable Asphalt content )

Explanation:

B) The daily average asphalt content has to obtained in order to determine the upper and lower control limits using an average asphalt content of 5.5% +/- 0.5% everyday

The upper limit : 14 may = ( 5.8 + 5.1 ) / 2 = 5.85

The lower limit : 16 may = ( 5.2 + 4.8 ) / 2 = 5.0

attached below is the required plot

C ) Not all the samples meet the contract specifications and the samples that do not meet up are samples from :

15 may and 16 may . this is because their Asphalt contents are 6.2 and 4.8 respectively and sample number 18 and 20

D ) what can be observed is that the ASPHALT content fluctuates between the dates while the contract specification is fixed

5 0
3 years ago
A spherical hot air balloon is initially filled with air having 120 kPa pressure and 24 °C temperature. Initial diameter of the
tino4ka555 [31]

Answer:

v = 1.076 m /s

Explanation:

Initial volume of balloon = 4/3 x  3.14 x (9.905/2)³

=508.56 m³

Final volume of balloon = 4/3 x 3.14 x (16.502/2)³

= 2351.73 m³

Increase in volume = 1843.17 m³

Cross sectional area of inlet  A  = 3.14 x( 1.458/2)²

A = 1.6687 m²

Volume rate of flow of air = cross sectional area x velocity of inflow

= 1 .6687 V [ V is velocity of inflow ]

Total time taken = Increase in volume / rate of flow of air

17.108 X 60 = 1843.17 / 1.6687 V

V = \frac{1843.17}{1.6687\times17.108\times60}

v = 1.076 m /s

8 0
3 years ago
Why is a metal work enclosure dangerous?
dedylja [7]

Answer:

Why is a metal work enclosure dangerous? Metalworkers are not only exposed to pollutants from metal cut ting and polishing procedures, but they are also exposed to metalworking fluids (MWF).

8 0
2 years ago
When selecting the appropriate gear for intended direction of travel, automatic transmission equipped vehicles should be placed
True [87]

Answer:

Automatic transmissions should be in Drive and Manual transmissions should be in first gear.

6 0
3 years ago
A tank contains initially 2500 liters of 50% solution. Water enters the tank at the rate of 25 iters per minute and the solution
sergejj [24]

Answer:

Percentage of solution=32. 96%

Explanation:

Given that initially tank have 50% solution.It means that amount of solution=0.5 x 2500 =1250 lts

Lets take the amount of solution at time t = A

So

\dfrac{dA}{dt}=rate\ in\ solution -rate\ out\ solution

\dfrac{dA}{dt}=concentration\times rate\ in\ of\ water -concentration\times rate\ out\ of\ water

\dfrac{dA}{dt}=0\times 25-\dfrac{A}{2500}\times 50

\dfrac{dA}{dt}=-\dfrac{A}{50}

Now by integration

\ln A=-\dfrac{1}{50}t+C

Where C is the constant

Given that at t=0 ,A=1250

So C=\ln 1250

\ln A=-\dfrac{1}{50}t+\ln 1250

When t=20 min

\ln A=-\dfrac{1}{50}\times 20+\ln 1250

A=837.90

So percentage of solution after 20 min

=\dfrac{1250-837.9}{1250}\times 100

Percentage of solution=32. 96%

4 0
3 years ago
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