Question

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

need steps

Answer 1

Answer:

an elevator descends into mine shaft at the rate of 5 meter per minute.

so for 1 min elevator descends 5metres.

the position of the elevator after one hour =

we know 1 hour =60 min so 60min x 5metres=300 meters..

but the elevator began to descend from 15 meters above so,

300-15="285metres."

and its position after 45 mins is 45metres x 5 mins=225 meters.

and the elevator has descended from 15 meters above so

225-15 meters="210metres."

hope this helps u.

Step-by-step explanation:

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Answer 2

Answer:

$in \: 1 \: minute \: the \: elevator \: goes = 5m$

$in \: 35 \: minutes \: the \: elevator \: goes = (5 \times 35) = 175m$

$final \: position = (175 - 25) = 150m \: under \: the \: ground$

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