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2 years ago

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

Engineering

1 answer:

Levart [38]2 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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Prompt the user to enter five numbers, being five people's weights. Store the numbers in an array of doubles. Output the array's

atroni [7]

Answer:

See below in the explanation section the Matlab script to solve the problem.

Explanation:

prompt='enter the first weight w1: ';

w1=input(prompt);

wd1=double(w1);

prompt='enter the second weight w2: ';

w2=input(prompt);

wd2=double(w2);

prompt='enter the third weight w3: ';

w3=input(prompt);

wd3=double(w3);

prompt='enter the fourth weight w4: ';

w4=input(prompt);

wd4=double(w4);

prompt='enter the first weight w5: ';

w5=input(prompt);

wd5=double(w5);

x=[wd1 wd2 wd3 wd4 wd5]

format short

6 0

2 years ago

Steam at 1400 kPa and 350°C [state 1] enters a turbine through a pipe that is 8 cm in diameter, at a mass flow rate of 0.1 kg⋅s−

sergeinik [125]

Answer:

Power output, $P_{out} = 178.56 kW$

Given:

Pressure of steam, P = 1400 kPa

Temperature of steam, $T = 350^{\circ}C$

Diameter of pipe, d = 8 cm = 0.08 m

Mass flow rate, $\dot{m} = 0.1 kg.s^{- 1}$

Diameter of exhaust pipe, $d_{h} = 15 cm = 0.15 m$

Pressure at exhaust, P' = 50 kPa

temperature, T' = $100^{\circ}C$

Solution:

Now, calculation of the velocity of fluid at state 1 inlet:

$\dot{m} = \frac{Av_{i}}{V_{1}}$

$0.1 = \frac{\frac{\pi d^{2}}{4}v_{i}}{0.2004}$

$0.1 = \frac{\frac{\pi 0.08^{2}}{4}v_{i}}{0.2004}$

$v_{i} = 3.986 m/s$

Now, eqn for compressible fluid:

$\rho_{1}v_{i}A_{1} = \rho_{2}v_{e}A_{2}$

Now,

$\frac{A_{1}v_{i}}{V_{1}} = \frac{A_{2}v_{e}}{V_{2}}$

$\frac{\frac{\pi d_{i}^{2}}{4}v_{i}}{V_{1}} = \frac{\frac{\pi d_{e}^{2}}{4}v_{e}}{V_{2}}$

$\frac{\frac{\pi \times 0.08^{2}}{4}\times 3.986}{0.2004} = \frac{\frac{\pi 0.15^{2}}{4}v_{e}}{3.418}$

$v_{e} = 19.33 m/s$

Now, the power output can be calculated from the energy balance eqn:

$P_{out} = -\dot{m}W_{s}$

$P_{out} = -\dot{m}(H_{2} - H_{1}) + \frac{v_{e}^{2} - v_{i}^{2}}{2}$

$P_{out} = - 0.1(3.4181 - 0.2004) + \frac{19.33^{2} - 3.986^{2}}{2} = 178.56 kW$

4 0

2 years ago

You have designed a treatment system for contaminant Z. The treatment system consists of a pipe that feeds into a CSTR. The pipe

IRISSAK [1]

Answer:

0.667 per day.

Explanation:

Our values here are

$Q=10m^3/dV_p=15m^3\\V_{cstr}=60m^3\\c_p=2500mg/L\\c_{cstr}=500mg/L$

Degradation constant=k and is unknown.

We calculate the concentration through the formula,

$cc_{cstr} =\frac{c_{in}}{1+K(V/Q)} \\cc_{cstr}=\frac{c_p}{1+K*\frac{V_{csrt}}{Q}}$

Replacing values we have

$1+k(\frac{60}{10})=\frac{2500}{500}\\1+k=5\\K(6)=5-1\\K(6)=4\\K=2/3\\K=0.667/day$

That is the degradation constant of Z-contaminant

3 0

3 years ago

The denominator of a fraction is 4 more than the numenator. If 4 is added to the numenator and 7 is added to the denominator, th

kati45 [8]

Answer:

$\frac{3}{7}$

Explanation:

Lets take the numerator of the fraction to be = x

So the denominator of the fraction is 4 more than the numerator = x+4

The fraction is ; $\frac{x}{4+x}$

Now add 4 to the numerator and add 7 to the denominator as;

$\frac{x+4}{4+x+7} =\frac{x+4}{x+11}$

This new fraction is equal to 1 half =1/2

write the equation as;

$\frac{x+4}{x+11} =\frac{1}{2}$

perform cross-product

2(x+4 )=1( x+11 )

2x+8 = x + 11

2x-x = 11-8

x=3

The original fraction is;

$\frac{x}{4+x} =\frac{3}{3+4} =\frac{3}{7}$

3 0

2 years ago

How can any student outside apply for studying engineering at Cambridge University

telo118 [61]

Admission to the Engineering course at Cambridge is highly competitive, both in terms of the numbers and quality of applicants. In considering applicants, Colleges look for evidence both of academic ability and of motivation towards Engineering. There are no absolute standards required of A Level achievement, but it should be noted that the average entrant to the Department has three A* grades. You need to get top marks in Maths and Physics.All Colleges strongly prefer applicants for Engineering to be taking a third subject that is relevant to Engineering.
Hope that helps and good luck if you are applying. Can you please mark this as brainliest and press the thank you button and if you have any further questions please let me know!!

3 0

2 years ago