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3 years ago

12

Megan is an architect who is creating a building design. Which is a prominent technical aspect that she needs to keep in mind wh

ile choosing materials? A. texture B. durability C. color D. theme

1 answer:

pogonyaev3 years ago

5 0

Answer:

b

Explanation:

did this on my lesson test

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A 500-km, 500-kV, 60-Hz, uncompensated three-phase line has a positivesequence series impedance. z = 5 0.03 1 + j 0.35 V/km and

Anni [7]

Answer:

A) 282.34 - j 12.08 Ω

B) 0.0266 + j 0.621 / unit

C)

A = 0.812 < 1.09° per unit

B = 164.6 < 85.42°Ω

C = 2.061 * 10^-3 < 90.32° s

D = 0.812 < 1.09° per unit

Explanation:

Given data :

Z ( impedance ) = 0.03 i + j 0.35 Ω/km

positive sequence shunt admittance ( Y ) = j4.4*10^-6 S/km

A) calculate Zc

Zc = $\sqrt{\frac{z}{y} }$ = $\sqrt{\frac{0.03 i + j 0.35}{j4.4*10^-6 } }$

= $\sqrt{79837.128< 4.899^o}$ = 282.6 < -2.45°

hence Zc = 282.34 - j 12.08 Ω

B) Calculate gl

gl = $\sqrt{zy} * d$

d = 500

z = 0.03 i + j 0.35

y = j4.4*10^-6 S/km

gl = $\sqrt{0.03 i + j 0.35* j4.4*10^-6} * 500$

= $\sqrt{1.5456*10^{-6} < 175.1^0} * 500$

= 0.622 < 87.55 °

gl = 0.0266 + j 0.621 / unit

C) exact ABCD parameters for this line

A = cos h (gl) . per unit = 0.812 < 1.09° per unit ( as calculated )

B = Zc sin h (gl) Ω = 164.6 < 85.42°Ω ( as calculated )

C = 1/Zc sin h (gl) s = 2.061 * 10^-3 < 90.32° s ( as calculated )

D = cos h (gl) . per unit = 0.812 < 1.09° per unit ( as calculated )

where : cos h (gl) = $\frac{e^{gl} + e^{-gl} }{2}$

sin h (gl) = $\frac{e^{gl}-e^{-gl} }{2}$

7 0

2 years ago

The loneliest people are to kindest

valkas [14]

Answer:

The most damaged people are the wisest is a fact

Explanation:

5 0

3 years ago

Read 2 more answers

A saturated 1.5 ft3 clay sample has a natural water content of 25%, shrinkage limit (SL) of 12% and a specific gravity (GS) of 2

Svetllana [295]

$79 f t^{3}$ is the volume of the sample when the water content is 10%.

<u>Explanation:</u>

Given Data:

$V_{1}=100\ \mathrm{ft}^{3}$

First has a natural water content of 25% = $\frac{25}{100}$ = 0.25

Shrinkage limit, $w_{1}=12 \%=\frac{12}{100}=0.12$

$G_{s}=2.70$

We need to determine the volume of the sample when the water content is 10% (0.10). As we know,

$V \propto[1+e]$

$\frac{V_{2}}{V_{1}}=\frac{1+e_{2}}{1+e_{1}}$ ------> eq 1

$e_{1}=\frac{w_{1} \times G_{s}}{S_{r}}$

The above equation is at $S_{r}=1$ ,

$e_{1}=w_{1} \times G_{s}$

Applying the given values, we get

$e_{1}=0.25 \times 2.70=0.675$

Shrinkage limit is lowest water content

$e_{2}=w_{2} \times G_{s}$

Applying the given values, we get

$e_{2}=0.12 \times 2.70=0.324$

Applying the found values in eq 1, we get

$\frac{V_{2}}{100}=\frac{1+0.324}{1+0.675}=\frac{1.324}{1.675}=0.7904$

$V_{2}=0.7904 \times 100=79\ \mathrm{ft}^{3}$

7 0

3 years ago

If you are interested only in the temperature range of 20° to 40°C and the ADC has a 0 to 3V input range, design a signal condit

mario62 [17]

Explanation:

Temperature range → 0 to 80'c

respective voltage output → 0.2v to 0.5v

required temperature range 20'c to 40'c

Where T = 20'c respective voltage

$\begin{aligned}v_{20} &=0.2+\frac{0.5-0.8}{80} \times 20 \\&=0.2+\frac{0.3}{80} \times 20 \\V_{20} &=0.275 v\end{aligned}$

$\begin{aligned}\text { when } T=40^{\circ} C & \text { . } \\v_{40} &=0.2+\frac{0.5-0.2}{80} \times 40 \\&=0.35 V\end{aligned}$

Therefore, Sensor output changes from 0.275v to 0.35volts for the ADC the required i/p should cover the dynamic range of ADC (ie - 0v to 3v)

so we have to design a circuit which transfers input voltage 0.275volts - 0.35v to 0 - 3v

Therefore, the formula for the circuit will be

$\begin{array}{l}v_{0}=\left(v_{i n}-0.275\right) G \\\sigma=\ldots \frac{3-0}{0.35-0.275}=3 / 0.075=40 \\v_{0}=\left(v_{i n}-0.275\right) 40\end{array}$

The simplest circuit will be a op-amp

NOTE: Refer the figure attached

Vs is sensor output

Vr is the reference volt, Vr = 0.275v

$\begin{aligned}v_{0}=& v_{s}-v_{v}\left(1+\frac{R_{2}}{R_{1}}\right) \\\Rightarrow & \frac{1+\frac{R_{2}}{R_{1}}}{2}=40 \\& \frac{R_{2}}{R_{1}}=39 \quad \Rightarrow\end{aligned}$

choose R2, R1 such that it will maintain required ratio

The output Vo can be connected to voltage buffer if you required better isolation.

3 0

3 years ago

A sensor produces a signal with amplitude 15 mV. A voltage amplifier must amplify the signal such that the amplitude of the outp

Nady [450]

Answer:

42.50 dB

Explanation:

Determine the minimum voltage gain

amplitude of input signal ( Vi ) = 15 mV

amplitude of output signal ( Vo) = 2 V

Vo = 2 v

therefore ; minimum gain = Vo / Vi = 2 / ( 15 * 10^-3 )

= 133.33

Minimum gain in DB = 20 log ( 133.33 )

= 42.498 ≈ 42.50 dB

8 0

3 years ago