An elevator descends into a mine shaft at the rate of 5 metres per minute . What will be its position after quarter an hour
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5 *15 =75 is the answer
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Answer:
300 m below ground level.
Step-by.-step explanation:
Sol: Since the elevator is going down, the distance covered by it will be
represented by a negative integer.
Change in position of the elevator in one minute = – 5 m
Position of the elevator after 60 minutes = (–5) × 60 = – 300 m
300 m below ground level.
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