STA BS HI FS ELEV BM1 2.883 130.016 127.133 ТР1 2.743 129.710 3.049 126.967 ТР2 2.855 129.876 2.690 127.021 BM2 2.816 129.283 3.409 126.467 BM3 2.697 128.800 3.180 126.103 ТРЗ 3.458 128.440 3.819 124.982 BM4 3.480 127.992 3.928 124.512 ТР4 3.340 127.346 3.986 124.006 ТР5 3.452 127.171 3.627 123.719 ТРб 3.413 126.238 4.346 122.825 BM5 4.117 122.121

Q: Question 1 ELEV. = 130m в L = 1000m D = 200mm f = 0.023 2. L = 1500m L = 300m D = 250mm ELEV.= 40m D…

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Q: 30 30ON M: 0130 M= 0.20 RESB: a, T

A: As there are two different questions posted, according to the guidelines, only the first is being…

Q: 1. Compute and adjust the following set of closed-loop level observations: Pt BS HI FS Elev BM A…

A: To compute and adjust the closed loop level.

Q: Estimate in (m/sec) se= 1572o Pa し-30.48 n = O-000045 M= 0.01 Pa.s 1: 1200

A: step 1-

Q: SOVE FOR EQUILIBRIUM: Problem #3 lo00 300 # 3' 10' 4' Rz

A: This question is related to the Strength of Material or Mechanics. In this question we calculate…

Q: Determine the force in each member of the space truss shown.

A: Write the co-ordinates of the space truss taking B as origin point A(0,0,3) m B(0,0,0) m C(4,0,0) m…

Q: 1. Complete the leveling notes. Show Customary Check. STA BS HI FS ELEV 1.964 |(205.549+y) BM1 1.964…

A: Given: The value of y is 0.99. The differential leveling notes are given below, STA BS HI…

Q: if P=140 N and a=4m and b=2m and tetta= 33.7   find  F(EC) , F(ED), F(AB),F(AD)

A: Given  P=140 N and a=4m and b=2m and tetta= 33.7

Q: -300 200 4S 30k 30 30 E =? Ey=? Y. E,=? Eg=? Bq =? つ V.

A: Strain is given in different orientation 1st we have to resolve these into x and y direction.

Q: (A,B,C,D,E,F,G,H.I.J) 5jäi a Sta. B.S I.F.S F.S H.I Ground Elev. 30+25 A 13.728 12.523 30+50 12.371…

A: Solution; RL+BS=HI 12.15+A=13.728,  A=1.578

Q: central station O. Four surrounding points A, B, C and D were  observed. Angle AOB 63°25’, BOC…

A: AOB 63°25’, BOC 55°45’, COD, 29°15’ and DOA 31°10 MPV of BOC is nearest to 55°45' Option a is…

Q: 1.283 BM-1 1.284 (а) 1.285 0.675 0.674 BM-2 0.677 (b) 0.674 0.677 0.678 Sum Mean INSTRUMENT SET UP…

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Q: TRIAL 2: COUNTERCLOCKWISE DIRECTION: No. of Full Partial Tape Length of Line Perimeter Line Length…

A: In surveying, traverse measurement is one of the works that is always done by the chain measurement…

Q: Follow directions and box your answers.  ANSWER: (a) 1.0     (b) 5/8, - 3/8     (c) 15/8…

A:

Q: Water Resources Engineering: Given the stream section and the following measurements, calculate the…

A: The discharge is given by

Q: 20kn/m 37.784480 A 11.336k 3m 16.25kn/m 16.25km • A B G ↓ ↓ ↓ ↓ 326126 sm →→2 43,049kn 48.5297kn…

A:

Q: #7 Given. A lap traverse ABE DE with a course DE of Bearing $ 11°42′15″E; fillowing angles : A =…

A: Sum of angles of a five sided polygon =540° Sum of angles= 540°0'25'' Angular error = 25" Adjusted…

Q: = 1000 N (C) FDe = 800 N (T) FDC FCE = 900 N (C) FCв 3D 800 N (T) %3D D -600 N 1000 N 36.87° 4 m E…

A:

Q: CCURSE DISTANCE (m) 496 REARING COURSE DISTANCE (T) 1021 1118 661 REARTNG S 12°25' E S 83°44' x N…

A: COURSE DISTANCE (m) BEARING AB 496 N05°30'E BC 851 N46°02'E CD 856 S67°38'E DE 1021 S12°25'E…

Q: 21== Q. No 4). A rectangular channel of X ft width is supplying water at flowrate of Y ft/s. If the…

A: Given that:                        Width of rectangular channel, B=12 ftDischarge in the channel, Q…

Q: Station B.S F.S Elev. B.M1 1.234 42.39 T.P1 0.965 2.732 T.P2 2.332 3.642 Т.РЗ 0.682 3.224 T.P4 2.338…

A: To find closure error and corrected elevation of given points

Q: Question: Cd= 1:0 = 0=1cm @ Derive The Equation Of Motion In Terms Of Height 2 Time For... cd-lo…

A:

Q: dt 1 Find ay if 6y°t + t°y =5 and dx dx Cos t

A: Let Z=6y3t + t3y-5       1 differentiating this funtion with respect to y, we get dzdy=18y2t + t3…

Q: Cm hea of he gven oene ung DPD ond DMD mthod COURSE LENGTH (m) BEARING AB 49585 850.62 N4 02E 855.45…

A: Area by Double Meridian Method (DMD method): DMD method is used to determine the area of the closed…

Q: Computed Lat Computed Dep Correction Adjusted Lat Adjusted Dep Course Length, m AZn Bearing +N -s +E…

A:

Q: Classify the following DE as Linear or Non-linear. 1. +5+ 6y = 0 d'y dx2 dx 2. +x2 dy + x³ dy – xe*…

A: To classify the given differential equations as Linear or Non-linear.   A differential equation (DE)…

Q: A closed travéếrse has the following data: Total Latitude = 1726.8 Total Departure = 1876.3…

A: Given L1=259.2m D1=448.9m Perimeter=2628.5m Length of DE = 518.4m Latitude of DE=259.2m Departure of…

Q: Computed Lat Computed Dep Correction Adjusted Lat Adjusted Dep Course Length, ft Bearing +N -S +E -W…

A: Course Length Bearing(WCB) Computed  Lat Computed Dep Correction   Adjusted  Lat   Adjusted Dep…

Q: 1-1Fthe maghetic quarter directioh (M.Br) for the 1:hes. AB,B EDDE andEF are N55 3 E, S78 00 E, S72…

A: We will draw diagram and solve using mathematical calculations.

Q: 150 kN 24 kN/m A |D BO00 C 000 2 m 1 m 0.5 m0.5 m

A: The degree of static indeterminacy = 4-3 =1. The moment at B is taken as redundant R and the basic…

Q: determine d.? Ey B) Ex 75H de de Ax CO D. 2kps 2 kips 64 75H D) de 10E d6 de A

A: 1) option A  2) option b

Q: Backsight Foresight STA HI Elev Hair Rdg Mean Rdg S Hair Rdg Mean Rdg S 1.153 125.00M 0.936 BM,…

A:

Q: SURVEYING Given the following differential level notes:  STA BS FS BM1 1.73 - TP1 2.5 3.8…

A: Given  Elevation of TP3 = 267.28 m,  To Determine The elevation of BM1.

Q: 70 kips/ft 350kips-ft OB C A 3' 6' 3' 3'

A:

Q: A =? P= ? T= ? 17.0 m 7.0m 0.90 m 3.0m

A:

Q: ROUTE LENGTH OF OBSERVED OBSERVED ADJUSTED ADJUSTED SIDE FORWARD BACKWARD FORWARD BACKWARD BEARING…

A: Route Length Observed Forward Bearing Observed Backward Bearing Adjusted Forward Bearing Adjusted…

Q: #2 4.7 8.629 Bm 15 71.589° 12.124 Tp. - 82116 7.2.32 Tp-3 1.747' 5.643' 2.932' Constraction. site Bm…

A: Plot the data in table

Q: Observed Bearings Length (m) Computed Internal Adjustea values Forward Line Sta Corr Interior Васк…

A: Solution: The table is drawn below:

Q: H.W: 1.55, 1.955, 2.3, 2.655, 3.17, (0.53 ),2.85 , 2.755 ,( 0.3 ), 1.73 , 2.15, R.L of station A=…

A: Given data  Staff readings R.L. Horizontal distance

Q: Route the Pollowing hydrogaph through a river reach for which k-22 hr and x=0-25 at the start of…

A: Since K = 22 hr and x = 0.25 2*K*x = 2*22*0.25 = 11 For best results the routing interval should be…

Q: In the leveling across a wide river on Pampanga, a reciprocal level readings were taken between…

A: Set up near B Average of BS = 2.283+2.284+2.286+2.2834=2.284m   Average of FS…

Q: .. DIFFERENTI AL LEVELING. below and perform Complete the the differential level otes shown…

A:

Q: Point B.S aiel's I.S F.S H.I R.L ie) REMARKS B.M 1.315 16.815 15.5 A 1.49 |3D B 1.362 (TP.) 1571…

A: To find reduced level and height of instrument

Q: 1. DIFFERENTI AL LEVELING. Complete the below and perform differential level notes shown the…

A:

Q: Calculate the interior angles from the following observations: Line PQ QR RS ST TP O O O Bearing S…

A: Draw the rough sketch of the traverse. Then using properties of geometry the internal angles can be…

Q: The subsequent readings were curious about dumpy level with four.zero m leveling team of workers and…

A:

Q: Given the following field records: Line Azimuth (°) from South Distance (m) A-B 205.00 + 0.1 * SN…

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Q: TMarch 23.5 °C TFebruary = 21.75 °C TDew point = 19 °C U₂ = 3.25 m/s

A:

Q: Determine the ff: 1. Force at member BC = [Select] 2. Force at member BD = [Select] 3. Force at…

A:

Q: earing or Azimuth of Line AB: 16710' Sta Observed Interior Angle 0 Adjusted Interior Sta Dist Мean…

A: Given:   Required to find: