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DENIUS [597]
3 years ago
8

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,

                               <em> gₓ = GM/(R+x)²</em>

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
How does an airplane engine work?
vivado [14]

Answer:

Explanation:

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7 0
3 years ago
Two points charge of 4\mu C and 2\mu C are placed at theopposite corners of a rectangle. What is the potential difference Va- Vb
bulgar [2K]

Answer:

Va-Vb=168KV

Explanation:

From the question we are told that

Two points charge of 4\mu C and 2\mu C

Generally we find the  Va and Vb individually to find there difference

Given a rectangle with two equal sides each,Assume lengths for bot sides

Length L=0.3

Breath B=0.4

Diagonal D=\sqrt{0.3^2+0.4^2} =0.5

at  opposite sides

Mathematically Va can represented as

Va =k(\frac{4*10^_-_6}{0.3} +\frac{-2*10^_-_6}{0.5} )

Va =9*10^9(\frac{4*10^_-_6}{0.3} +\frac{-2*10^_-_6}{0.5} )

Va =9*10^9(0.00001333333-0.000004} )

Va =84000V

Va =84KV

Mathematically Vb is  represented as

Va =k(\frac{-4*10^_-_6}{0.3} +\frac{2*10^_-_6}{0.5} )

Va =9*10^9(\frac{-4*10^_-_6}{0.3} +\frac{+2*10^_-_6}{0.5} )

Va =9*10^9(-0.00001333333+0.000004} )

Va =-84000V

Va =-84KV

Therefore

Va-Vb=84-(-84)\\Va-Vb=84+84\\Va-Vb=168KV

7 0
3 years ago
A fighter jet is launched from an aircraft carrier with the aid of its own engines and a steam-powered catapult. The thrust of i
kirza4 [7]

Answer:

2.5 x 10⁷ J

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F = thrust of the engine = 2.3 x 10⁵ N

d = distance traveled = 87  m

Work done by the engine is given as

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W' = Net work done

W'' = work done by catapult

KE₀ = initial kinetic energy = 0 J

KE = final kinetic energy = 4.5 x 10⁷ J

Net work done is given as

W' = KE - KE₀

W' = 4.5 x 10⁷ J

We know that

W' = W + W''

4.5 x 10⁷ = 2.001 x 10⁷ + W''

W'' = 2.5 x 10⁷ J

8 0
3 years ago
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Answer:

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Explanation:

3 0
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