The sag correction in surveys is always. 1. positive 2. negative zero 4. None of these 3.
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- i need clear ans and solve very very fast in 20 min and thank you | ᴅʏʙᴀʟᴀ ?✨؛Answer 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 = -0183decimal places. Ifr=h(4h5 - 5), find the percentage is 0.04. %3D error in r at h = 1, if the error in h %3D If R 10 13.12.?
- 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 0.603Reduce 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.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
- Module 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 127.133a) 0.700 The following readings were taken in sequence during a leveling job: Readings (m) 1.250 1.285 1.125 0.810 1.555 -1.400 1.235 0.665 1.905 Remarks BMI R.L. = 101.335 Peg A Peg B 2 Change Point CP2 Change Point CP2 BM2 R.L=103.879 -1.400 Change Point CP1 Change Point CP1 CGS 1905 Inverted staff reading taken to the underside of a bridge Peg C Prepare a standard form of booking for these readings and reduce them using the rise fall method applying all the necessary arithmetic checks and adjust the final Reduced Levels.3. Derive the operational Green-Ampt equation, + K sat F(t)-F(tp) _ \Vƒ\(0−0;) F(tp)+|wf|(-0₁) F(t)+|vf|(0-0₁) K sat t = and from the following two equations, f(t)=-Ksat 1+ Koal (1+1 M ₁ | (0-0₁) F(t) - f(t) = In dF (t) dt = tststw V s where to represents the time of ponding, t represents the total time specified, and -In (a.x+b)) 2 the effective tension at the wetting front. The meanings of the rest of symbols are the same as those defined in class. (Hint: Begin by inverting both sides of the equation, then x. dx X b separate variables and integrate. Also, S a.x+b a a +tp is