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serg [7]
2 years ago
11

A level loop began and closed on BM_A (elevation = 823.368 ft). The plus and minus sights were kept approximately equal. Reading

s (in feet) listed in the order taken are: 4.388 (BS) on BM_A, 4.538 (FS) and 6.907 (BS) on TP1, 8.8 (FS) and 4.68 (BS) on TP2, 5.978 (FS) and 3.73 (BS) on TP3, 5.245 (FS) and 8.464 (BS) on TP4, and 3.598 (FS) on BM_A.
Distances are: BM_A to TP1 = 375 ft, TP1 to TP2 = 212 ft, TP2 to TP3 = 402 ft, TP3 to TP4 = 257 ft, TP4 to BM_A = 254 ft.

The final (adjusted) height of TP2 is?

What is the observed height difference between TP2 and TP3 ?

The misclosure is?

The number of setups (n) is?

What is the provisional Height of TP3?

The precision (m) for the leveling run is equal to (rounded to nearest integer)

The correction for each set-up is?

Engineering
1 answer:
LiRa [457]2 years ago
8 0

Explanation:

See attached file

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Answer:

The flow velocity reduces to 0.72 m/s

Explanation:

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A_{1}V_{1}=A_{2}V_{2}

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Applying values in the continuity equation and since the width of the channel remains constant 3.0 m we have

3\times 1.8\times 1.2=3\times 3\times V_{2}\\\\\therefore V_{2}=\frac{6.48}{9}=0.72m/s

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Electrical pressure or “force”<br><br> A) current<br> B) resistance <br> C) voltage
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What is the activation energy (Q) for a vacancy formation if 10 moles of a metal have 2.3 X 10^13 vacancies at 425°C?
Yakvenalex [24]

Answer:

Activation\ Energy=2.5\times 10^{-19}\ J

Explanation:

Using the expression shown below as:

N_v=N\times e^{-\frac {Q_v}{k\times T}

Where,

N_v is the number of vacancies

N is the number of defective sites

k is Boltzmann's constant = 1.38\times 10^{-23}\ J/K

{Q_v} is the activation energy

T is the temperature

Given that:

N_v=2.3\times 10^{13}

N = 10 moles

1 mole = 6.023\times 10^{23}

So,

N = 10\times 6.023\times 10^{23}=6.023\times 10^{24}

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The conversion of T( °C) to T(K) is shown below:

T(K) = T( °C) + 273.15  

So,  

T = (425 + 273.15) K = 698.15 K  

T = 698.15 K

Applying the values as:

2.3\times 10^{13}=6.023\times 10^{24}\times e^{-\frac {Q_v}{1.38\times 10^{-23}\times 698.15}

ln[\frac {2.3}{6.023}\times 10^{-11}]=-\frac {Q_v}{1.38\times 10^{-23}\times 698.15}

Q_v=2.5\times 10^{-19}\ J

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Answer:

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Explanation:

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