A elevator descend into a mine shaft at the of 6m / min . If the descend start from 10 min above the ground level , how long will it take to reach - 350 ?
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Given an elevator descends into a mine shaft at the rate of 5m/min
Recall that 1 hour = 60 minutes
Position after 1 hour = 60 × 5 = 300 m
Hence the elevator will be at 300 m in the mine
Recall that 1 hour = 60 minutes
Position after 1 hour = 60 × 5 = 300 m
Hence the elevator will be at 300 m in the mine
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☜☆☞hey friend!!! ☜☆☞
here is your answer ☞
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Distance descended is denoted by a negative integer.
Initial height = +10 m
Final depth = −350 m
Total distance to be descended by the elevator = (−350) − (+10) = −360 m
Time taken by the elevator to descend −6 m = 1 min
Thus, time taken by the elevator to descend −360 m = (−360) ÷ (−6)
= 60 minutes = 1 hour
hope it will help you
Devil_king ▄︻̷̿┻̿═━一
here is your answer ☞
→_→→_→→_→→_→→_→
Distance descended is denoted by a negative integer.
Initial height = +10 m
Final depth = −350 m
Total distance to be descended by the elevator = (−350) − (+10) = −360 m
Time taken by the elevator to descend −6 m = 1 min
Thus, time taken by the elevator to descend −360 m = (−360) ÷ (−6)
= 60 minutes = 1 hour
hope it will help you
Devil_king ▄︻̷̿┻̿═━一
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