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3 years ago

13

Communication "works" to the degree that a wide variety of information is completely and thoroughly shared among the parties, an

d mutual understanding is reached.

1 answer:

Murljashka [212]3 years ago

6 0

Answer:

Hook's law holds good up to. A elastic limit. B. plastic limit. C.yield point. D.Breaking point

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There is evidence that liquid water existed on Mars at some time in the past. What is the evidence found indicating the presence

Norma-Jean [14]

There mag be water on Mars but it’s not very easy to spot, but that is besides the point,

Mars is a smaller planet then Earth, thus it has less gravity, meaning when water evaporates into the atmosphere of the planet the water slowly escapes into space, so there is less and less water on the planet.

Hope this helps!

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3 years ago

28. List the 3 basic parts of a lamp?

bekas [8.4K]

Answer:

cord light bulb

Explanation:

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3 years ago

Read 2 more answers

Question 4 (1 point) Sophia is working in her garden. She made two trips with her wagon along the same path to haul supplies. Th

yaroslaw [1]

Answer:

Option A

Explanation:

Please find the attachment

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3 years ago

A lagoon is designed to accommodate an input flow of 0.10 m^3/s of nonconservative pollutant with concentration 30 mg/L and deca

dexar [7]

Answer:

Volume of the lagoon required for the decay process must be larger than 86580 m³ = 8.658 × 10⁷ L

Explanation:

The lagoon can be modelled as a Mixed flow reactor.

From the value of the decay constant (0.2/day), one can deduce that the decay reaction of the pollutant is a first order reaction.

The performance equation of a Mixed flow reactor is given from the material and component balance thus:

(V/F₀) = (C₀ - C)/((C₀)(-r)) (From the Chemical Reaction Engineering textbook, authored by Prof. Octave Levenspiel)

V = volume of the reactor (The lagoon) = ?

C₀ = Initial concentration of the reactant (the pollutant concentration) = 30 mg/L = 0.03 mg/m³

F₀ = Initial flow rate of reactant in mg/s = 0.10 m³/s × C₀ = 0.1 m³/s × 0.03 mg/m³ = 0.003 mg/s

C = concentration of reactant at any time; effluent concentration < 10mg/L, this means the maximum concentration of pollutant allowed in the effluent is 10 mg/L

For the sake of easy calculation, C = the maximum value = 10 mg/L = 0.01 mg/m³

(-r) = kC (Since we know this decay process is a first order reaction)

This makes the performance equation to be:

(kVC₀/F₀) = (C₀ - C)/C

V = F₀(C₀ - C)/(kC₀C)

k = 0.2/day = 0.2/(24 × 3600s) = 2.31 × 10⁻⁶/s

V = 0.003(0.03 - 0.01)/(2.31 × 10⁻⁶ × 0.03 × 0.01)

V = 86580 m³

Since this calculation is made for the maximum concentration of 10mg/L of pollutant in the effluent, the volume obtained is the minimum volume of reactor (lagoon) to ensure a maximum volume of 10 mg/L of pollutant is contained in the effluent.

The lower the concentration required for the pollutant in the effluent, the larger the volume of reactor (lagoon) required for this decay reaction. (Provided all the other parameters stay the same)

Hope this helps!

5 0

3 years ago

A three-phase wye-connected synchronous generator supplies a network through a transmission line. The network can absorb or deli

Amanda [17]

Answer:

the graph and the answer can be found in the explanation section

Explanation:

Given:

Network rated voltage = 24 kV

Impedance of network = 0.07 + j0.5 Ω/mi, 8 mi

Rn = 0.07 * 8 = 0.56 Ω

Xn = 0.5 * 8 = 4 Ω

If the alternator terminal voltage is equal to network rated voltage will have

Vt = 24 kV/√3 = 13.85 kV/phase

The alternative current is

$I_{a} =\frac{40x10^{6} }{\sqrt{3}*24x10^{3} } =926.2A$

$X_{s} =0.85\frac{13.85}{926.2} =12.7ohm$

The impedance Zn is

$\sqrt{0.56^{2}+4^{2} } =4.03ohm$

The voltage drop is

$I_{a} *Z_{n} =926.2*4.03=3732.58V$

$r_{dc} =\frac{voltage}{2*current} =\frac{13.85}{2*926.2} =7.476ohm$

rac = 1.2rdc = 1.2 * 7.476 = 8.97 Ω

The effective armature resistance is

$Z_{s} =\sqrt{R_{a}^{2}+X_{s}^{2} } =\sqrt{8.97^{2}+12.7^{2} } =15.55ohm$

The induced voltage for leading power factor is

$E_{F} ^{2} =OB^{2} +(BC-CD)^{2}$

if cosθ = 0.5

$E_{F} =\sqrt{(13850*0.5)^{2}+(\frac{3741}{2}-926.2*12.7)^{2} } =11937.51V$

if cosθ= 0.6

EF = 12790.8 V

if cosθ = 0.7

EF = 13731.05 V

if cosθ = 0.8

EF = 14741.6 V

if cosθ = 0.9

EF = 15809.02 V

if cosθ = 1

EF = 13975.6 V

The voltage regulation is

$\frac{E_{F}-V_{t} }{V_{t} } *100$

For each value:

if cosθ = 0.5

voltage regulation = -13.8%

if cosθ = 0.6

voltage regulation = -7.6%

if cosθ = 0.7

voltage regulation = -0.85%

if cosθ = 0.8

voltage regulation = 6.4%

if cosθ = 0.9

voltage regulation = 14%

if cosθ = 1

voltage regulation = 0.9%

the graph is shown in the attached image

for 10% of regulation the power factor is 0.81

8 0

4 years ago