Question

An elevator defends into a mine shaft at the rate of 6m/min if the descent starts from 10m above the ground level hoe long will it take to reach -350 m deep from the ground​

Answer 1

$\huge \pink \star{ \green{ \boxed{ \boxed{ \boxed{ \purple{\mathfrak{QUESTION}}}}}}} \pink\star$

An elevator defends into a mine shaft at the rate of 6m/min if the descent starts from 10m above the ground level hoe long will it take to reach -350 m deep from the ground

$\huge \pink \star{ \green{ \boxed{ \boxed{ \boxed{ \orange{\mathfrak{ANSWER}}}}}}} \pink\star$

total distance the elevator has tocover 360 metrevelocity = 6m/min 6/60 m/s0.1 m/s

Use the formula t= S/v

t= S/vt 360/0.1

t= S/vt 360/0.1t 3600 seconds = 60

t= S/vt 360/0.1t 3600 seconds = 60minutes = 1 hour

HOPE IT IS HELPFUL TO YOU!!

Answer 2

Step-by-step explanation:

Total distance elevator has to travel= Initial position- final position

= 10+(-350)

= 10+ 350

= 360m

Now,

Elevator descends at rate 6m/min

Elevator travels 6min in = 1min

Elevator travels 1min in= 1\6 min

Elevator travels 360m in= 1/6×360min

= 60min

= 1hr

It will take the elevator 1hr