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3 years ago

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What would the Select lines need to be to send data for the fifth bit in an 8-bit system (S0 being the MSB and S2 being the LSB)

?
A. S0 = 1, S1 = 0, S2 = 0
B. S0 = 0, S1 = 0, S2 = 0
C. S0 = 0, S1 = 1, S2 = 0
D. S0 = 0, S1 = 1, S2 = 1

1 answer:

Maurinko [17]3 years ago

5 0

Answer:

A. S0 = 1, S1 = 0, S2 = 0

lines need to send data for the fifth bit in an 8 bit system

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The real power delivered by a source to two impedances, ????1=4+????5⁡Ω and ????2=10⁡Ω connected in parallel, is 1000 W. Determi

kirza4 [7]

Answer:

The question is incomplete, below is the complete question

"The real power delivered by a source to two impedance, Z1=4+j5⁡Ω and Z2=10⁡Ω connected in parallel, is 1000 W. Determine (a) the real power absorbed by each of the impedances and (b) the source current."

answer:

a. 615W, 384.4W

b. 17.4A

Explanation:

To determine the real power absorbed by the impedance, we need to find first the equivalent admittance for each impedance.

recall that the symbol for admittance is Y and express as

$Y=\frac{1}{Z}$

Hence for each we have,

$Y_{1} =1/Zx_{1}\\Y_{1} =\frac{1}{4+j5}\\converting to polar \\ Y_{1} =\frac{1}{6.4\leq 51.3}\\ Y_{1} =(0.16 \leq -51.3)S$

for the second impedance we have

$Y_{2}=\frac{1}{10}\\Y_{2}=0.1S$

we also determine the voltage cross the impedance,

P=V^2(Y1 +Y2)

$V=\sqrt{\frac{P}{Y_{1}+Y_{2}}}\\$

$V=\sqrt{\frac{1000}{0.16+0.1}}\\ V=62v$

The real power in the impedance is calculated as

$P_{1}=v^{2}G_{1}\\P_{1}=62*62*0.16\\ P_{1}=615W$

for the second impedance

$P_{2}=v^{2}*G_{2}\\ P_{2}=62*62*0.1\\384.4w$

b. We determine the equivalent admittance

$Y_{total}=Y_{1}+Y_{2}\\Y_{total}=(0.16\leq -51.3 )+0.1\\Y_{total}=(0.16-j1.0)+0.1\\Y_{total}=0.26-J1.0\\$

We convert the equivalent admittance back into the polar form

$Y_{total}=0.28\leq -19.65\\$

the source current flows is

$I_{s}=VY_{total}\\I_{s}=62*0.28\\I_{s}=17.4A$

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3 years ago

A hydraulic jump is formed in a 4m wide outlet just downstream of a control gate, which is located at the upstream end of the ou

cupoosta [38]

Answer:

Width w = 4m

Glow depth = y1 = 20m

Outlet discharge = 40m

V1= velocity of flow = 40/20*4 = 1/2 = 0.5m/s

Froud number = v1/√gy1

= 0.5/√9.81x20 = 0.0356

1. Y2/20 = 1/2[-1+√1+8*(0.0356)²]

Y2 = 0.05

2. Energy loss in the jump = (20-0.05)²/4x20x0.05

= 1985m

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3 years ago

What friction rate should be used to size a duct for a static pressure drop of 0.1 in wc if the duct has a total equivalent leng

skad [1K]

Answer:

0.067wc

Explanation:

The formula is actual static pressure loss = (total equivalent divided by 100) multiplied by rate of friction

We substitute values

actual static pressure = 0.1

Total equivalent length = 150 ft

0.1 = (150ft/100) multiplied by Rate of friction

Friction rate at 100ft = 0.067

So we have that the required friction needed is 0.067wc

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3 years ago

A gas stream contains 18.0 mole% hexane and the remainder nitrogen. The stream flows to a condenser, where its temperature is re

Anna [14]

Answer:

A. 72.34mol/min

B. 76.0%

Explanation:

A.

We start by converting to molar flow rate. Using density and molecular weight of hexane

= 1.59L/min x 0.659g/cm³ x 1000cm³/L x 1/86.17

= 988.5/86.17

= 11.47mol/min

n1 = n2+n3

n1 = n2 + 11.47mol/min

We have a balance on hexane

n1y1C6H14 = n2y2C6H14 + n3y3C6H14

n1(0.18) = n2(0.05) + 11.47(1.00)

To get n2

(n2+11.47mol/min)0.18 = n2(0.05) + 11.47mol/min(1.00)

0.18n2 + 2.0646 = 0.05n2 + 11.47mol/min

0.18n2-0.05n2 = 11.47-2.0646

= 0.13n2 = 9.4054

n2 = 9.4054/0.13

n2 = 72.34 mol/min

This value is the flow rate of gas that is leaving the system.

B.

n1 = n2 + 11.47mol/min

72.34mol/min + 11.47mol/min

= 83.81 mol/min

Amount of hexane entering condenser

0.18(83.81)

= 15.1 mol/min

Then the percentage condensed =

11.47/15.1

= 7.59

~7.6

7.6x100

= 76.0%

Therefore the answers are a.) 72.34mol/min b.) 76.0%

Please refer to the attachment .

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3 years ago

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Answer:

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Explanation:

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2 years ago