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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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A glass tube is inserted into a flowing stream of water with one opening directed upstream and the other end vertical. If the wa
Furkat [3]

Answer:

h=0.46m

Explanation:

From the question we are told that:

Velocity of water V=3m/s

Height=?

Generally, the equation for Water Velocity is mathematically given by

V=\sqrt{2gh}

Therefore Height h is given as

h=\frac{v}{2g}

h=\frac{3^2}{2*9.81}

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5 0
3 years ago
State five applications of thermochromic materials
rusak2 [61]

Explanation:

The end-use industries of thermochromic materials include packaging, printing & coating, medical, textile, industrial, and others. Printing & coating is the fastest-growing end-use industry of thermochromic materials owing to a significant increase in the demand for thermal paper for POS systems. The use of thermochromic materials is gaining momentum for interactive packaging that encourages consumers to take a product off the shelf and use it.

8 0
3 years ago
Knowing that v = –8 m/s when t = 0 and v = 8 m/s when t = 2 s, determine the constant k. (Round the final answer to the nearest
docker41 [41]

Answer:

a)We know that acceleration a=dv/dt

So dv/dt=kt^2

dv=kt^2dt

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v(t)=kt^3/3+C

Puttin t=0

-8=C

Putting t=2

8=8k/3-8

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3 years ago
The position of a particle is given by s = 0.27t
Natali [406]
Sorry bro people do this22.2 pls
8 0
2 years ago
A long rod of 60-mm diameter and thermophysical properties rho= 8000 kg/m3, c= 500 J/kg·K, and k= 50 W/m·K is initially at a uni
Dvinal [7]

Answer:

Tc =    = 424.85 K

Explanation:

Data given:

D = 60 mm = 0.06 m

\rho = 8000 kg/m^3

k = 50 w/m . k

c = 500 j/kg.k

h_{\infty} = 1000 w/m^2

t_{\infity} = 750 k

t_w = 500 K

surface area = As = \pi dL

\frac{As}{L} = \pi D = \pi \timeS 0.06

HEAT FLOW Q  is

Q = h_{\infty} As (T_[\infty} - Tw)

 = 1000 \pi\times 0.06 (750-500)

  = 47123.88 w per unit length of rod

volumetric heat rate

q = \frac{Q}{LAs}

  = \frac{47123.88}{\frac{\pi}{4} D^2 \times 1}

q = 1.66\times 10^{7} w/m^3

Tc = \frac{- qR^2}{4K} + Tw

= \frac{ - 1.67\times 10^7 \times (\frac{0.06}{2})^2}{4\times 56} +  500

   = 424.85 K

7 0
3 years ago
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