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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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44.95 tonnes

Explanation:

According to principle of buoyancy the object will just sink when it's weight is more than the weight of the liquid it displaces

It is given that empty weight of box = 40 tons

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Thus the combined mass of box and stones = (40+M) tonnes..........(i)

Since the box will displace water equal to it's volume V we have volume of box = 25ft*10ft*12ft= 3000ft^{3}

Volume= 84.95m^{3}

Since 1ft^{3} =0.028m^{3}

Now the weight of water displaced = Weight =\rho \times Volumewhererho is density of water = 1000kg/m^{3}

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Equating i and ii we get

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Answer:

Explanation:

Given data:

initial construction co = 0.286 wt %

concentration at surface position cs = 0 wt %

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