Background: Calculated the professor’s mass Elevator at rest: Mean 1089 N (from
ID: 1779596 • Letter: B
Question
Background:
Calculated the professor’s mass
Elevator at rest:
Mean 1089 N (from force-time graph)
Free Body Diagram
Calculate the acceleration of the elevator
Free Body Diagram
Determine the distance between the first and the third floor. (Question)
Statistics for: Latest | Force min: 1128 at 8.640 max: 1159 at 7.440 mean: 1149 median: 1150 std. dev: 6.294 samples: 72 Ay: 31 1150 Statistics for: Latest | Force min: 1081 at 13.86 max: 1098 at 12.88 mean: 1089 median: 1089 std. dev: 3.320 samples: 286 Ay: 17 1100 1050 Statistics for: Latest | Force min: 1083 at 4.120 max: 1098 at 1.640 mean: 1089 median: 1089 std. dev: 2.778 samples: 275 Ay: 15 Statistics for. Latest Force min: 1015 at 17.72 max 1062 at 18.38 mean: 1035 median: 1029 std. dev: 13.46 samples: 87 Ay: 48 1000 20 30 40 50 (32.50, 1155.02) Time (s)Explanation / Answer
Elevator accelrates from 7 to 9 sec at 0.5 m/sec2
Speed at 9 sec = 0.5*(9-7) = 1m/s
Average speed during ths period = 1/2 = 0.5 m/s
Distance covered in this time = 0.5*(9-7) = 1m
Moves at constant speed of 1 m/s, from 9 to 17 sec
distance covered = 1*(17-9) = 8m
During deccelration period of 17 to 19 sec, average speed is again 0.5 m/s
and distance covered = 1 m
Total distance covered = 1+8+1 =10 m = distance between first and third floor.
Related Questions
drjack9650@gmail.com
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.