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serg [7]
3 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]3 years ago
8 0

Explanation:

See attached file

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Two long conducting cylindrical shells are coaxial and have radii of 20 mm and 80 mm. The electric potential of the inner conduc
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If the power factor is corrected to 0.95 lagging, keeping the receiving end MVA constant, what will be the new voltage regulatio
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Answer:

Due to power factor correction; voltage regulation is improved and efficiency is increased.

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Power Factor Vs Voltage Regulation:

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4 years ago
A 35kg block of mass is subjected to forces F1=100N and F2=75N at agive angle thetha= 20° and 35° respectively.find the distance
Talja [164]

Answer:

21 m

Explanation:

Since F₁ = 100 N and acts at an angle of 20° to the horizontal, it has horizontal component F₁' = 100cos20° = 93.97 N and vertical component F₁" = 100sin20° = 34.2 N.

Also, F₂ = 75 N and acts at an angle of -35° to the horizontal, it has horizontal component F₂' = 75cos(-35°) = 75cos35° = 61.44 N and vertical component F₂" = 75sin(-35°) = -75sin35° = -43.02 N

The resultant horizontal force F₃' = F₁' + F₂' = 93.97 N + 61.44 N = 155.41 N

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If f is the frictional force on the block, the net horizontal force on the block is F = F₃' - f.

Since f = μN where μ = coefficient of kinetic friction = 0.4 and N = normal force on the block.

For the block to be in contact with the surface, the vertical forces on the block must balance.

Since the normal force, N must equal the resultant vertical force F₃" and the weight, W = mg of the object for a zero net vertical force,

N = mg + F₃" (since both the weight and the resultant vertical force act downwards)

N = mg + F₃"

Since m = mass of block = 35 kg and g = acceleration due to gravity = 9.8 m/s² and F₃" = 8.82 N

So,

N = mg + F₃"

N = 35 kg × 9.8 m/s² + 8.82 N

N = 343 N + 8.82 N

N = 351.82 N

So, the net horizontal force F = F₃' - f.

F = 155.41 N - 0.4 × 351.82 N

F = 155.41 N - 140.728 N

F = 14.682 N

Since F = ma, where a = acceleration of block,

a = F/m = 14.682 N/35 kg = 0.42 m/s²

To find the distance the block moved, x we use the equation

x = ut + 1/2at² where u = initial speed of block = 0 m/s, t = time = 10 s and a = acceleration of block = 0.42 m/s²

Substituting the values of the variables into the equation, we have

x = ut + 1/2at²

x = 0 m/s × 10 s + 1/2 × 0.42 m/s² × (10 s)²

x = 0 m + 1/2 × 0.42 m/s² × 100 s²

x = 0.21 m/s² × 100 s²

x = 21 m

So, the distance moved by the block is 21 m.

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