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3 years ago

If the gravitational field strength at the top of Mount Everest is 9.772 N/kg, approximately how tall (in feet) is

Physics

1 answer:

Delicious77 [7]3 years ago

6 0

Answer:

The height of the Everest mountain is, x = 8514.087 m

Explanation:

Given data,

The gravitational field strength at the top of mount Everest, gₓ = 9.772 m/s²

The formula for gravitational field strength is,

Where, x is the height from the surface of the Earth

Therefore,

(R+x)² = GM/gₓ

x = √(GM/gₓ) - R

Substituting the values,

x = √(6.67408 x 10⁻¹¹ X 5.972 x 10²⁴ / 9.772) - 6.378 x 10⁶

x = 8514.087 m

Therefore, the height of the Everest mountain is, x = 8514.087 m

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2 years ago

A 740-kg boulder is raised from a quarry 119 m deep by a long uniform chain having a mass of 550 kg . This chain is of uniform s

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Answer:

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B) how long does it take to be lifted out at maximum acceleration if it started from rest

Explanation:

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$a_{y} =(t-M_{g} )/M=(2.50m_{c} -M_{g} )/M=(2.50.550kg-1290kg)(9.8m/s^2)/1290kg$

$=0.645m/s^2$

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using $y-y_{0} =v_{oy} t+1/2(a_{y} )t^2$

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$t=\sqrt{2(119m)/0.645m/s^2} =19.20s$

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1 year ago

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3 years ago

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2 years ago