Read the question carefully and give me right solution according to the question. Station 10+00.00 has an elevation of 110.00 and a grade of -3%. Station 20+00.00 has an elevation of 105.00 and a grade of +2%. What is the stationing at the two grade intersections? options
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Read the question carefully and give me right solution according to the question.
Station 10+00.00 has an elevation of 110.00 and a grade of -3%. Station 20+00.00 has an elevation of 105.00 and a grade of +2%. What is the stationing at the two grade intersections?
options
A) 14+00,
B) 15+00,
C) 16+00,
D) 17+00)
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- From station 0 + 210 with center height of 1.4 m in fill, the ground lines makes a uniform slope of 5% to station 0 + 270 whose center height is 2.8 m in cut . Find the grade of the finish roadway. a. 1% b. 2% c. -2% d. -3%Directions: Answer the following problems. Show your complete solution and draw the necessary figures A downgrade of 4.6% meets an upgrade of 3.20% at Sta. 73+810 where the elevation is 998 m. A parabolic curve AB, 440 m long, connects the grade lines from A at the downgrade B at the upgrade. Find the stationing of B.Read the question carefully and give me right solution according to the question. A −1.10% grade meets a +0.60% grade at station 36 + 00 and elevation 800.00 ft. The +0.60% grade then joins a +2.40% grade at station 39 + 00. What is the station of the midpoint?
- Compute and tabulate full-station elevations for an unequal-tangent vertical curve to fit the requirements shown below. A -3.40% grade meets a +2.75% grade at station 95+00 and elevation 1320.64 ft. Length of first curve 400 ft., second curve 600 ft. Match the correct elevation to the appropriate Station Value. Group of answer choices Sta. 91+00 (BVC) [ Choose ] Sta. 92+00 [ Choose ] Sta. 93+00 [ Choose ] Sta. 94+00 [ ChooseThe data given for a vertical sagcurve on a roadway plan andprofile sheet are as follows: PVIstation @ 32+11.61, PVI elevation =54.18 ft, back tangent gradient g1 =- 4.00 percent, forward tangentgradient g2 = + 7.00 percent, andlength of curve L= 600ft. Determinethe curve elevations at half-station intervals along the curve,and compute the station andelevation of the lowest point. PLEASE PROVIDE COMPLETE AND COMPREHENSIVE SOLUTION AND FINAL ANSWER ASAPYou are going to design a circular curve in a rural road segment where the design vehicleis a passenger car of operation speed is 90 km/h. The width of the lanes of the proposedroad is 3.5m. Based on a survey at the proposed site, it has been found that a multi-storiedbuilding is located towards the inside of the curve at 10 m from the centerline of thehighway. Considering perception reaction time as 2.5 s design the horizontal curve (Rdesign,edesign and fdesign) for that road segment. Record the radius and respective safe sight distancein the tabular format while selecting design radius (word table), plot them in Excel graphand find the Rdesignfor the safe sight distance. If there is no superelevation, what would bethe alternative arrangement?
- A 5° curve intersects s property line CD at point D. The back Tangent intersects the property line at point C which is 105m from PC which is at station 2 + 050. The angle that the property line CD makes with the back tangent is 110°50’.A. COMPUTE the length of the curve from pC to the point of intersection of the line from the center of the curve to point C and the curve.B. Compute distance CDC. What is the station at point D?A grade descending at the rate of -4% intersects another grade ascending at the rate of +8% at station 2+000 and elevation 100m. A vertical curve (symmetrical) is to connect the two such that the curve will clear a boulder located at station1+980 (elevation 101.34m). a. Determine the necessary length of the curve.b. Determine the station of the location of a sewer to be laid out.c. Compute the elevation of station where the sewer is to be place.10. A +2.8% grade at 1 + 200 is to be connected to a -4.2% grade at 1+600 using a symmetrical parabolic curve. The tangents intersect at elevation 178.53m. Compute the distance of the highest point from BVC. a) 200 m c) 160 m b) 240 m d) 178.53 m 11. A +2.8% grade at 1 + 200 is to be connected to a -4.2% grade at 1+600 using a symmetrical parabolic curve. What is the vertical distance of Pl from BVC? a) 4.8 m c) 2.8 m b) 5.6 m d) 8.4 m 12. A sag curve is composed of two parabolic curves. BVC and EVC have elevations 250.30m and 246.30m respectively. The grade of the tangents is -3.5% at BVC (1+300) and +5.5% at EVC (1+800). What is the grade of the common tangent? a) 1.25% c) 0.008% b) 0.08% d) 0.8%
- Two grades +2 % & +2 %, what is required to connect themSingle choice. choose one of the following none of the given circular curve crest vertical curve sag vertical curveUrgent please need answers, must be 4 decimals w complete solution. Thank you The change in grade of a vertical parabolic curve is 0.60%. The grade of the said curve is -4% and +2% respectively. Both tangents intersects at station 12+150.60 at elevation 124.80m above sea level. Determine the elevation of the lowest point of the curve above sea level.The angle between a 5-m full station of a certain curve is 10°. The curve has 7 intermediate stations between P.C. (10+300m) and P.T. After laying out the curve, it was found out that the P.T. falls in a property lot and it should be moved in a way that the P.T. will be free from crossing the lot. As an ENGINEER, what would be your advice for relocation? Compute all the necessary elements of the old curve and new relocated curve.