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- Given: θ=20o30' Upper Reading=2.800 m Middle Reading=2.000 m (rod reading (RR)= h) Lower Reading=1.200 m Height of Instrument=1.20 m Elevation at point O=140.50 m K=100 C=0 Compute the horizontal distance Determine the vertical distance Calculate the inclined distance find the elevation of point PGiven: A1= 3400, A2=800, A3=706.85834, A4=176.71458 Y1=42.5, Y2=26.666, Y3=52.7324, Y4= 62.5 X1=20, X2=53.333, X3=52.7324, X4=17.5 Total Area: 4730.14376 ȳ=40.604 x̄ (x-bar) = 30.6224 Please, I want you to solve the: 4.) Moment of Inertia About the x-axis Ix 5.) Moment of Inertia About the y – axis Iy 6.) Polar Moment of Inertia about the axes Jo 7.) Moment of Inertia About the centroidal x-axis Ix note: In locating the centroid of the entire section, Varignon's Theorem. Varignon's Theorem locates the centroid of a composite body. I hope you will use the formulas I will be giving.X=9.0 kN. Y=4.0 kN/m
- A 50-m steel tape was used to measure a line and the initial pull = 5.5 kg. Pull at time of measurement = 8.0 kg. Weight of tape = 0.05 kg/m. E = 2.1 x 10^6 kg/cm^2? Cross section = 0.04 sq.cm. Measured distance between two points =488.650 m, in an inclined surface having a gradient of 3%. Find the correct horizontal distance between the two points. A. 486.456 C. 488.625 B. 490.844 D. 488.147 The distance from A to B taken at elevation 1200 m above sea level is 6,750 m. Determine the sea-level distance. A.6748.734 C. 6749.347 B. 6751.266 D.6750.274 ps. with solution, please.Deflection Angle Traverse Given the bearing of each line of a closed traverse ABCDEF, determine the deflection angles of each line. Line Bearing AB S 46°17’ E BC N 23°28’ E CD N 78°55’ W DE S 56°42’ W EF S 13°00’ W FA N 76°05’ E * Sum of Deflection Angle = 360° * Note: EXPRESS YOUR ANSWER UPTO 3 DECIMAL PLACES. (Please answer this problem with accurate and proven solution. Thank u)If you are given two vectors: A = 5i + 8j + 3k and B = 35i + 4j – 9k, what is the angle between the two vectors?explanation not needed
- To determine the horizontal distance between Q and R, a tacheometer was used and was placed at P. The stadia intercepts for Q and R were recorded to be 0.320 m and 0.210 m respectively. Calculate the horizontal distance between Q and R if the angle QPR is equal to 61°30'30". Take stadia multiplication constant as 100 and the stadia addition constant as 0.10 m.For the area shown, determine: a) The centroidal coordinates X and Y. b) The moment of inertia about the X and Y axes (IxeIy) . c) The polar moment of inertia about the origin (Jo) . d) The moment centroidal polar of inertia (Jc).A spherical triangle has the following parts: C=900, c=840, and a=650. Compute angle B. Answer: 83 deg, 87deg If the point (2, 3) is equidistant from (x, -2) and (7, 4), find x.Answer: 3