Question

A lift in mine moves down 24m in 87min. If it moves in uniform rate ,find at what distance below the surface it will be after 6 mins .If the lift starts from a height 10 m above the grounf find how deep the lift will go from the surface after 70 mins.​

Answer 1

Answer:

9.32m

Step-by-step explanation:

If 24m is moved in 87 min, then speed of lift is 24/87 = 0.276 m/min

So, after 6 min, the lift is 0.276x6 = 1.656m below.

If the lift starts 10m above, then it touches the ground after 10/0.276 = 36.232 min

Then, the rest 70-36.232 = 33.768 min is spent below. So, it moves 0.276x 33.768 = 9.32m at this time.

Hence, after 70 mins the lift goes 9.32m deep.

√\_____Anushka❤

Answer 2

Heya mate your answer is----------------------------------------------------

Answer:

a) 1 655 m under ground b) 9.31 m under ground

Step-by-step explanation:

V = 24/87 m/min

V = 0.275 m/min

T = 6 min

S = V × T

S = 0.275 × 6

S = 1.655 m

b) same velocity but time is 70 min and lift starts from a height of 10 m above ground

S = V × T

S = 0.275 × 70

S = 19.31 m

hence we know that the lift starts from a heigt of 10 m from the ground

so. S = 19.31 -10

S = 9.31 m

plz mark brainliest ♥️

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