An elevator descends into a mine shaft at the rate of 6 min, if the descent starts from 12m below the ground level, how long will it take to reach - 330m ?
Answers
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Clark, for
letting, really,
Answer:
Starting position of mine shaft is = 10m above ground
Starting position of mine shaft is = 10m above groundbut,
Starting position of mine shaft is = 10m above groundbut,it moves in opposite direction so it travels the distance (-350 m ) below the ground.
Starting position of mine shaft is = 10m above groundbut,it moves in opposite direction so it travels the distance (-350 m ) below the ground.Total distance covered by mine shaft = 10m - (-350m)
Starting position of mine shaft is = 10m above groundbut,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
Starting position of mine shaft is = 10m above groundbut,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
Starting position of mine shaft is = 10m above groundbut,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= 360mNow, taken to cover a distance of 6m by it = 1 minute
Starting position of mine shaft is = 10m above groundbut,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= 360mNow, taken to cover a distance of 6m by it = 1 minuteTime taken to cover a distance of 1m by it = \frac{1}{6\ }\min6 1min
Starting position of mine shaft is = 10m above groundbut,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= 360mNow, taken to cover a distance of 6m by it = 1 minuteTime taken to cover a distance of 1m by it = \frac{1}{6\ }\min6 1minTherefore, tie taken to a cover a distance of 360m = \frac{1}{6}\times\ 360\ =\ 60\ \min\ \61× 360 = 60 min
60 minutes = 1 hour
thus, in one hour the mine shraft reaches = 350 below the ground.
