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3. An elevator descends into a mine shaft at the rate of 6 m/min. If it descends from 10 m above the ground level, how long it will take to reach -350 m(below the ground)? |
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Answer» Answer: Step-by-step explanation: SPEED of elevator = 6 m/min Converting speed into m/s = 6/60 = 0.1 m/s Distance to be covered by the elevator = 350 + 10 = 360 m We know that Speed = Distance/Time Let the time be t 0.1 = 360 / t t = Distance/Speed t = 360/ 0.1 t = 3600 sec t = 1 hour So, the time taken by the elevator to reach the desired POSITION = 1 hour Hope it helps you PLEASE Mark me as a brainliest |
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