If an elevator descends a mine shaft from a height of 25 m above the ground at the rate of 5 metre per minute what would be its position after 35 minutes
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If an elevator descends a mine shaft from a height of 25 m above the ground at the rate of 5 metre per minute what would be its position after 35 minutes
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Answer:
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
6
1
min
Therefore, tie taken to a cover a distance of 360m = \frac{1}{6}\times\ 360\ =\ 60\ \min\ \
6
1
× 360 = 60 min
60 minutes = 1 hour
thus, in one hour the mine shraft reaches = 350 below the ground.
[tex] \huge \bigstar \underline \bold{ \: Solution}[/tex]
In \: 1 \: minute \: the \: elevator \: goes = 5min1minutetheelevatorgoes=5m
In \: 35 \: minutes \: the \: elevator \: goes = (5 \times 35) = 175min35minutestheelevatorgoes=(5×35)=175m
Final \: position = (175 - 25) = 150m \: under \: the \: ground final position=(175−25)=150m underground
Hope my answer is correct and helps to you