An elevator descends into a mine shaftat the rate of 10 m/ min.. if the descent starts from 30 m above the ground level, how long will it take to reach 520 m below the ground level?
Answers
Answer:
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Mathematics
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Chapters in NCERT Solutions - Mathematics, Class 7
Exercises in Integers
Question 53
Q7) An elevator descends into a mine shaft at the rate of 6 m/min. If the descent starts from 10 m above the ground level, how long will it take to reach – 350 m.
Solution
Transcript
Solution 7:
Starting position of mine shaft is = 10m above ground
but,
it moves in opposite direction so it travels the distance (-350 m ) below the ground.
Total distance covered by mine shaft = 10m - (-350m)
= 10 + 350
= 360m
Now, taken to cover a distance of 6m by it = 1 minute
Time taken to cover a distance of 1m by it = \frac{1}{6\ }\min
Therefore, tie taken to a cover a distance of 360m = \frac{1}{6}\times\ 360\ =\ 60\ \min\ \
60 minutes = 1 hour
thus, in one hour the mine shraft reaches = 350 below the ground.
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Answer:
Total distance the elevator has to travel = Initial position - Final Position
= 30 - (-520)
= 30 + 520
= 550 m
Now,
Elevator descends at rate of 10m/min
Therefore, elevator travels 10m in = 1 min
Elevator travels 1 m in = 1/10
Elevator travels 550 m in = 1 /10 × 550 min
= 55 minutes
Step-by-step explanation: