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3 years ago

11

About 80 percent of earth volume is made up of what?

2 answers:

kolbaska11 [484]3 years ago

4 0

About 80% of the earth's volume is made of mantle.

<span>The </span>mantle<span> is a layer inside a </span>terrestrial planet<span> and some other </span>rocky planetary bodies<span>. For a mantle to form, the planetary body must be large enough to have undergone the process of </span>planetary differentiation<span> by </span>density<span>. The mantle lies between the </span>core<span> below and the </span>crust<span> above. The terrestrial planets (</span>Earth<span>, </span>Venus<span>, </span>Mars<span> and </span>Mercury<span>), the </span>Moon<span>, two of </span>Jupiter<span>'s </span>moons<span> (</span>Io<span> and </span>Europa<span>) and the </span>asteroid Vesta<span> each have a mantle made of </span>silicate<span> rock.</span><span>Interpretation of spacecraft data suggests that at least two other moons of Jupiter (</span>Ganymede<span> and </span>Callisto<span>), as well as </span>Titan<span> and </span>Triton<span> each have a mantle made of </span>ice<span> or other </span>solid volatile<span> substances </span>up of Mantle

Hope this helped.

lora16 [44]3 years ago

3 0

The mantle makes up 80% of the earth's volume.

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Answer:

The magnitude of the vector $\vec{C}$ is 107.76 m

Explanation:

To find the components of the vectors we can use:

$\vec{A} = | \vec{A} | \ ( \ cos(\theta) \ , \ sin (\theta) \ )$

where $| \vec{A} |$ is the magnitude of the vector, and θ is the angle over the positive x axis.

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$\vec{B} = 82.0 \ m \ ( \ cos( 180 \° - 49 \°) \ , \ sin (180 \° - 49 \°) \ )$

$\vec{B} = 82.0 \ m \ ( \ cos( 131 \°) \ , \ sin (131 \°) \ )$

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Now, we can perform vector addition. Taking two vectors, the vector addition is performed:

$(a_x,a_y) + (b_x,b_y) = (a_x+b_x,a_y+b_y)$

So, for our vectors:

$\vec{C} = ( \ -48.497 \ m \ , \ - 28 \ m \ ) + ( \ -53.797 \ m \ , ) = ( \ -48.497 \ m \ -53.797 \ m , \ - 28 \ m \ + \ 61.886\ m \ )$

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To find the magnitude of this vector, we can use the Pythagorean Theorem

$|\vec{C}| = \sqrt{C_x^2 + C_y^2}$

$|\vec{C}| = \sqrt{(- 102.294 \ m)^2 + (\ 33.886 m \)^2}$

$|\vec{C}| =107.76 m$

And this is the magnitude we are looking for.

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