An elevator descends into a mine shaft at the rate of 7m/min. If the descend start from 10 m above the ground level, how long will it take to reach the shaft 340 m below the ground level?
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Answer:
☞1 hour
Step-by-step explanation:
The elevator starts from 20 m above the ground.
Therefore , total distance to be covered=(400+20)m=420 m.
Speed=7 m/min.
Therefore, time=D/S
=420/7 min
=60 min
=1 hr
☞Ans>The time taken is 1 hour.
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Step-by-step explanation:
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