Tiêu đề C2 MSTE Reviewer.pdf ✅

C2 MSTE Reviewer ^_^ 1. A surveyor measured a 700 m line using an uncalibrated tape that is known to be 0.015 m too long. He measured the line and came up with an error of -0.2099370189. Determine the tape length of the uncalibrated tape. A. 27 m C. 100 m B. 50 m D. 74 m Solution: Difficulty: 1/5 Correction, C = −emeasured distance C = −(−0.2099370189) = 0.2099370189 Correction due to uncalibrated tape length Too long: C = etape length × MD TL 0.2099370189 = 0.015 ( 700 TL ) TL = 50.015 m ≈ 50m 2. A group of Civil Engineering students was tasked to measure a 500 m walkway. The group has two (2) 50 m uncalibrated steel tapes, with one that is 0.013 m too short and the other that is 0.016 m too long. The group used the tapes alternately until they completed the task, starting with the 0.013 m too short. Determine the actual distance measured by the group. A. 499.985 m C. 498.053 m B. 501.641 m D. 502.164 m Solution: Difficulty: 2/5 Let: 0.013m too short Tape = T1 0.016m too long Tape = T2
C2 MSTE Reviewer ^_^ To measure 500 m, the students need to use a 50-meter tape 10 times. Since they have 2 tapes, they will use each of those 5 times. Therefore, the total error in the measurement will be combination of 5 times that of T1 and 5 times that of T2. CD = MD + ∑C CD = MD − CT1 + CT2 500 = MD − (0.013) ( 50 × 5 50 ) + (0.016) ( 50 × 5 50 ) MD = 499.985 3. Kevin lives in a dorm directly across from the campus. He noticed that it took him 178 paces to reach the campus from his dorm on his first day. On his thirtieth day, he realized that for the past 30 days, the number of paces it took him to reach the campus had been alternately increasing and decreasing by 1 and 2 paces, respectively. Meaning it took him 179 paces on his 2nd day, 177 on his 3rd day, 178 on his 4th day, and so on. If his pace factor is 0.652 m/pace, determine the distance between his dorm and the campus. A. 124.652 m C. 103.414 m B. 111.818 m D. 99.667 m Solution: Difficulty: 3/5 Based on the pattern of his paces, we can conclude that every odd-numbered day (1, 3, 5, etc.) his paces are decreasing by one. Every even-numbered day (2, 4, 6, etc.), his paces are also decreasing by one. Hence, we can form two arithmetic sequence for his paces – for odd- numbered days with an initial value of 178 (pace on first day) and for even-numbered days with an initial value of 179. Therefore, his average pace will be as follows: Average Pace = Sum of paces per day Number of days = Sodd + Seven 30 Sodd + Seven 30 = 15 2 (2(178) + (15 − 1)(−1)) + 15 2 (2(179) + (15 − 1)(−1)) 30 = 171.5 paces Therefore, PF = Distance Ave Pace → Distance = (PF)(Ave Pace) Distance = (0.652)(171.5) = 111.818 m

C2 MSTE Reviewer ^_^ (P2 − 8(9.81))(50000) (0.05 × 102)(200000) = ( 1.075(9.81) 50000 ) 2 (50000) 3 24P2 2 P2 = 197.4 N 6. The common tangent of a compound curve makes an angle of 12° from the tangent passing thru the P.C. and 18° from the tangent passing thru the P.T. If the radius of the second curve is 180 m, find the radius of the first curve if the length of the common tangent is 70 m long A. 354.804 m C. 324.352 m B. 394.76 m D. 263.501 m Solution: Difficulty: 1/5 T1 + T2 = CT R1 tan ( 12 2 ) + 180 tan ( 18 2 ) = 70 R1 = 394.76 m 7. Using Simpson’s 1/3 Rule, approximate the area of a piece of land having irregular boundaries as follows: Station Offset Distance (m) Station Offset Distance (m) 1+120 3.6 1+200 3.6 1+140 4.2 1+220 4.3 1+160 2.7 1+240 3.4 1+180 2.4 1+260 2.5 A. 480.33 m2 C. 444.67 m2 B. 460.67 m2 D. 473.00 m2 Solution: Difficulty: 1/5 A = 20 3 (3.6 + 4(4.2 + 2.4 + 4.3) + 2(2.7 + 3.6) + 3.4) + 3.4 + 2.5 2 = 480.33 m2