1. Calculate the grade of points A, B and C. El. 126.42m 3.70% 158.75 -4.1% 66.45 126-866 B 2.90% 205.6
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- Reduce the following set of differential leveling notes and perform the arithmetic check. Station BM 130 TP 1 TP 2 TP 3 BM K110 TP 4 BM 132 BS (m) 0.702 0.970 0.559 1.744 1.973 1.927 HI (m) FS (m) 1.111 0.679 2.780 1.668 1.788 0.888 Elevation (m) 168.657 1. Determine the order of accuracy. 2. Adjust the elevation of BM K110. The length of the evel run was 780 m, with setups equally spaced. The elevation of BM 132 is known to be 167.629 m.Part C Compute the linear misclosure. Express your answer in feet to three significant figures. IΫΠ ΑΣΦ ↓↑ vec Submit Part D Compute the relative precision. Express your answer to three significant figures. 1: Request Answer 195| ΑΣΦ. 41 vec Submit Request Answer ? ? ftReduce the following set of differential leveling notes and perform the arithmetic check. Station BM 130 TP 1 TP 2 TP 3 BM K110 TP 4 BM 132 BS (m) 0.702 0.970 0.559 1.744 1.973 1.927 HI (m) FS (m) 1.111 0.679 2.780 1.668 1.788 0.888 Elevation (m) 168.657
- Plz, solve this problem, it is not graded question. ω = P = Reg# − 5180 ( kip or kN or kip/ft or kN/m) Reg # 5395Answer the do (0.0026 L'= L te=) 30 + (-0.0026) s)? (L+= 29.9974 and Explain types of errors in me nents, 1 (L'/L)ME TL = ( 29.9974/30) *500 H TL = 499.9566657 382) Arizontal distance of (500 m) must be lay out with a steel tape, length of the steel tape der standard conditions is (30 m). Field conditions indicate that standard conditions apply except the measured temperature is 50 °C and applied tension 120'N, Determine the correct T= (T-Ts) Lx T= (50-2032) A =T= 0.0lounder P= (P-P₂)L AE P-(120-50)3 The standard temperature 20 °C. 0.02*21*Coefficient of thermal expansion of steel tape = 11.6x10-6 °C 70.005) Standard tension = 50N ∙W C-W² Mass of the tape = 0.05 kg/m 2p² -(0.05 jan Total = 0. olo44 +0.005-0. Total (€) = mans. له میشه راه horizontal distance to be laid out, if Modulus of elasticity (E) = 210*106 N/m2 24 (120) Cross-sectional area of the tape = 0.02 cm² 0.022 022 1.05 +X5 +1·15=3₁3 X5=3-3-22 X5=11 1.1-2.4 -1.3 567 5-1.3= 55.4 fall 1₁4.5-2.28 = -0183Lines of levels between B and C are run over four different routes. B is at elevation 825 m and is higher than C. Route A Distance (km) Difference in Elevation (m) 0.86 0.69 0.75 1.02 6. C 4 8 a. Determine the weight of route B. b. Determine the most probable difference in elevation. c. Determine the most probable elevation of C.
- 1. DIFFERENTIAL notes check. LEVELING. Complete the the differential level customary arithmetic shown below and perform STA BS HI FS ELEVATION BM 10 2.085 137.450 m TP1 2.015 0.982 TP2 1.864 1.428 TP3 0.579 1.527 2.423 1.807 BMII 0.423 TP4 1.446 TP5 T.778 1.725 TP6 2.051 2.339 TP7 2.920 1.005 BM 12 3.186 2.358 TP8 2.805 0.995 TP9 0.774 1.206 BM 13 0.603In a differential leveling two points A and B were observed from an instrument set up midway between these two points. The observed rod readings on A and B were recorded to be 3.25m and 1.10m respectively. The distance between these two points is 123.45m. Assuming uniformity of slope from A to B determine the slope of AB. Select the correct response: O +1.88% +1.74% O -1.74% O -1.88% Compute the approximate area in hectares of a rectangular shaped tract of land whose sides measures 20.50mm by 31.25mm on a map drawn to a scale of 1: 8,000. Select the correct response: O3.23 has O 4.10 has. O 2.18 has. O 1.82 has. On a map of the City of Dasmarinas drawn to a scale of 1:12,000 the map distance from city hall to the town market is 28cm, in another map drawn to a scale of 1: 15,000, the map distance from the city hall to the nearest department store is 24cm, in another map drawn to scale 1;10000, the map distance from the city hall to the nearest police outpost is 15cm, and in another map…2/87 If the resultant of the two forces and couple M passes through point O, determine M. 150 09 Mo =M-400x0.15cos30-320x0.3=0 M=148 Nm CCW N 00
- 1) Complete the missing data and determine the difference in elevation of BM1 and BM2. Show complete solution. STA BS HI FS ELEV (m) BM1 1.097 504.233 TP1 1.661 1.998 ТР2 1.012 2.015 ТРЗ 1.905 2.396 BM2 2.761Module 03- Coursework Activity 03 Complete the Differential Leveling Notes by providing data to the unknown quantities. STA FS BM1 TP1 TP2 BM2 BM3 TP3 BM4 BS 1.256 1.116 1.228 1.189 1.070 1.831 HI 1.886 1.527 2.246 2.017 2.656 2.765 ELEV 127.1331. Determine the elevation at each station. STATION BS HI FS BM1 1.285 TP1 2.314 2.345 TP2 1.562 1.301 TP3 2.721 2.380 TP4 0.895 1.789 TP5 2.645 2.456 TP6 3.115 0.764 TP7 0.865 1.897 BM2 1.239 2.987 TP9 2.413 2.309 TP10 1.234 1.506 TP11 2.345 2.908 TP12 3.456 2.113 TP13 1.654 2.222 TP14 2.321 1.559 TP15 0.789 2.727 TP16 0.456 3.408 TP17 0.223 2.322 TP18 1.546 1.110 BM3 0.557 2.101 TP19 1.353 2.131 TP20 2.458 2.358 TP21 2.221 1.424 TP22 3.105 1.588 TP23 0.557 0.898 TP24 3.455 1.231 TP25 2.626 2.358 TP26 1.325 1.357 TP27 0.898 3.264 BM4 0.566 2.565 TP28 2.356 1.543 TP29 1.259 1.553 TP30 1.110 2.598 TP31 2.369 2.015 TP32 3.122 1.567 TP33 0.897 3.011 TP34 1.335 2.234 TP35 2.131 1.353 BM5 1.357 ELEVATION 445.544 M