A 50-m steel tape was used to measure a line and the initial pull = 5.5 kg. Pull attime of measurement = 8.0 kg. Weight of tape = 0.05 kg/m. E = 2.1 x 10^6kg/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 correcthorizontal distance between the two points. A. 486.456 C. 488.625 B. 490.844 D. 488.147 A rectangular lot was being measured using a 30-m metallic tape which was 6 mm too long.The recorded dimension where 70.50 m long by 37.10 m wide. What is the error in area (m^2) introduced due to the erroneous length of tape? A. 1.05 C. 0.86 B. 1.86 D. 2.02 It is required to lay out a distance of 687.78m with a 50m tape that is 0.030 m too long. Compute the distance measured with the tape to make the points the proper distance apart. A. 688.19 m C. 687.19m B. 687.37 m D. 687.94 m4.
A 50-m steel tape was used to measure a line and the initial pull = 5.5 kg. Pull attime of measurement = 8.0 kg. Weight of tape = 0.05 kg/m. E = 2.1 x 10^6kg/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 correcthorizontal distance between the two points.
A. 486.456 C. 488.625
B. 490.844 D. 488.147
A rectangular lot was being measured using a 30-m metallic tape which was 6 mm too long.The recorded dimension where 70.50 m long by 37.10 m wide. What is the error in area (m^2) introduced due to the erroneous length of tape?
A. 1.05 C. 0.86
B. 1.86 D. 2.02
It is required to lay out a distance of 687.78m with a 50m tape that is 0.030 m too long. Compute the distance measured with the tape to make the points the proper distance apart.
A. 688.19 m C. 687.19m
B. 687.37 m D. 687.94 m4.
Compute the sea level distance of a line which measures 324.45 m if it was measured in Mount Apo (elevation 2.954 m).
A. 324.15m C. 324.30 m
B. 324.45m D. 324.60 m
ps. with solution, please.
Step by step
Solved in 4 steps
