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2 years ago

If you climb to the top of Mt. Everest, you will be 8850 m (about 5.50 mi) above sea level.

Physics

1 answer:

Umnica [9.8K]2 years ago

5 0

Answer:

9.773m/s2

Explanation:

Given,

h=8848m

The value of sea level is 9.08m/s2

So,

Let g′ be the acceleration due to the gravity on the Mount Everest.

g′=g(1−h2h)

=9.8(1−640000017696)

=9.8(1−0.00276)

9.8×0.99724

=9.773m/s2

Thus, the acceleration due to gravity on the top of Mount Everest is =9.773m/s2

Hope it helped!!!

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Substitute 2.27 m for s (=h₁), 0.301 s for t and 9.8 m/s² for a (=g).

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If the acorn starts from rest and reaches a speed of 6.067 m/s at the top of the scale, it would have fallen a distance h₂ to achieve this speed.

Use the equation of motion,

$v^2=u^2+2as$

Substitute 6.067 m/s for v, 0 m/s for u, 9.8 m/s² for a (=g) and h₂ for s.

$v^2=u^2+2as\\ (6.067m/s)^2=(0m/s)^2+2(9.8m/s^2)h_2\\ h_2=\frac{(6.067m/s)^2}{2(9.8m/s^2)} =1.878 m$

The height h above the ground at which the acorn was is given by,

$h=h_1+h_2=(2.27 m)+(1.878 m)=4.148 m$

The acorn was at a height 4.15m from the ground before dropping down.

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