An elevator descends into a mineshaft at the rate of 6 m/min. If the elevator starts from 10m above the ground level the time it will take to reach 350 rn is
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Initially the levator is at the 10m above the ground level, Hence total distance travelled is 350+10 = 360m,
rate of travel is 6m/min.
hence time taken is
total distance 360 upon 6 = 60mins or 1 hour
rate of travel
Explanation:
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