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Oduvanchick [21]
3 years ago
6

A horizontal angle was measured by repetition six times with a total station. If the initial display reading was 21o33'18" and t

he final reading was 129o20'04", determine the value of the angle to the nearest second.
Engineering
1 answer:
KIM [24]3 years ago
4 0

Answer:

The value of the angle is  107° 46' 46''

Explanation:

we know that

To find out the value of the angle , subtract the initial display reading from the final reading

Remember that

1°=60'

1'=60''

1°=3,600''

we have

Initial display reading=21° 33' 18''

Convert to seconds

21(3,600)+33(60)+18=77,598''

Final display reading=129° 20' 04''

Convert to seconds

129(3,600)+20(60)+4=465,604''

Find the difference

465,604''-77,598''=388,006''

Convert 388,006'' to degrees

388,006''/3,600=107.7794°

Convert 0.7794° to minutes

0.7794(60)=46.764'

Convert 0.764' to seconds

0.764(60)=45.84''=46''

therefore

The value of the angle is

107° 46' 46''

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Answer:

Following are the conversion to this question:

Explanation:

In point (a):

\to \frac{36459080}{8} = 4557385   + \ \ \ \ \ \ \ \ \ \frac{0}{8}\\\\\to \frac{4557385}{8} = 569673   + \ \ \ \ \ \ \ \ \ \frac{1}{8}\\\\\to \frac{569673}{8} = 71209+ \ \ \ \ \ \ \ \ \ \frac{1}{8}\\\\\to \frac{71209}{8}=8901+\ \ \ \ \ \ \ \ \ \ \ \frac{1}{8}\\\\\to \frac{8901}{8}=1112+ \ \ \ \ \ \ \ \ \  \ \ \ \frac{5}{8}\\\\\to \frac{1112}{8}=139+ \ \ \ \ \ \ \ \ \ \ \frac{0}{8}\\\\\to \frac{139}{8}=17+ \ \ \ \ \ \ \ \ \ \ \frac{3}{8}\\\\\to \frac{17}{8}=2+ \ \ \ \ \ \ \ \ \ \ \ \ \frac{1}{8}\\\\

\to \frac{2}{8}=0+ \ \ \ \ \ \ \ \ \ \frac{2}{8}\\\\ \bold{(36459080)_{10}=(213051110)_8}

In point (b):

\to \frac{20960032010}{16} = 13100020+ \ \ \ \ \ \ \ \ \ \frac{0}{16}\\\\\to \frac{13100020}{16} = 818751+ \ \ \ \ \ \ \ \ \ \frac{4}{16}\\\\\to \frac{818751}{16} = 51171+ \ \ \ \ \ \ \ \ \ \frac{15}{16}\\\\\to \frac{51171}{16}=3198+\ \ \ \ \ \ \ \ \ \ \ \frac{3}{16}\\\\\to \frac{3198}{16}=199+ \ \ \ \ \ \ \ \ \  \ \ \ \frac{14}{1}\\\\\to \frac{199}{16}=12+ \ \ \ \ \ \ \ \ \ \ \frac{7}{16}\\\\\to \frac{12}{16}=0+ \ \ \ \ \ \ \ \ \ \ \frac{12}{16}\\\\ \bold{(20960032010)_{10}=(C7E3F40)_{16}}

In point (c):

\to (2423233303003040)_s=(88757078520)_{10}\\\\\to \frac{88757078520}{25}= 3550283140+ \ \ \ \ \ \ \ \ \ \frac{20}{25}\\\\ \to \frac{3550283140}{25}= 142011325+ \ \ \ \ \ \ \ \ \ \frac{15}{25}\\\\\to \frac{142011325}{25}= 5680453+ \ \ \ \ \ \ \ \ \ \frac{0}{25}\\\\\to \frac{5680453}{25}= 227218+ \ \ \ \ \ \ \ \ \ \frac{3}{25}\\\\\to \frac{227218}{25}= 9088+ \ \ \ \ \ \ \ \ \ \frac{18}{25}\\\\\to \frac{9088}{25}= 363+ \ \ \ \ \ \ \ \ \ \frac{13}{25}\\\\

\to \frac{363}{25}= 14+ \ \ \ \ \ \ \ \ \ \frac{13}{25}\\\\\to \frac{14}{25}= 0+ \ \ \ \ \ \ \ \ \ \frac{14}{25}\\\\\bold{(2423233303003040)_s=(EDDI30FK)_{25}}

Symbols of Base 25 are as follows:

0, 1, 2, 3,4,5,6,7,8,9,A,B,C,D,E,F,G,H,I,J,K,L,M,N, \ and \ O

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Answer:

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Explanation:

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and here mass balance equation will be

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0 = Qcv - Wcv + m × [ ( h1-h2) + \frac{v1^2-v2^2}{2} + g (z1-z2) ]      ....................2

so here Qcv will be

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put here value and we get

Qcv =  1.2353 × [ {-60}+ (310.24-262.96) ]

Qcv =  -15.7130 kW

so here heat transfer rate is -15.71 kW

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