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2 years ago

Which term describes the object that a satellite revolves around?

Physics

2 answers:

m_a_m_a [10]2 years ago

6 0

The object that a satellite revolves around is the <em>central body</em> of the system. <em>(C)</em>

For example:

-- The central body of the solar system is the Sun.

-- The central body for TV satellites, GPS satellites, weather satellites, and the International Space Station is the Earth.

-- The central body for Phobos and Deimos is Mars.

This should be a pretty easy question to answer by elimination, when you notice that "Orbit", "Period", and "Rotation" are not "Bodies".

jeka57 [31]2 years ago

6 0

Answer:

Central body

Explanation:

A P E X

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a closed system does not allow matter or energy to pass through

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How many elections are shared between one nitrogen atom and one carbon atom?

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The route followed by a hiker consists of three displacement vectors, X, Y and Z. Vector X is along a measured trail and is 1430

poizon [28]

Answer:

magnitude : 1635.43 m
Angle: 130°28'20'' north of east

Explanation:

First, we will find the Cartesian Representation of the $\vec{X}$ and $\vec{Y}$ vectors. We can do this, using the formula

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

where $| \vec{A} |$ its the magnitude of the vector and θ the angle. For $\vec{X}$ we have:

$\vec{X}= 1430 m \ ( \ cos( 42 \°) \ , \ sin (42 \°) \ )$

$\vec{X}= ( \ 1062.70 m \ , \ 956.86 m \ )$

where the unit vector $\hat{i}$ points east, and $\hat{j}$ points north. Now, the $\vec{Y}$ will be:

$\vec{Y}= - 2200 m \hat{j} = ( \ 0 \ , \ - 2200 m \ )$

Now, taking the sum:

$\vec{X} + \vec{Y} + \vec{Z} = 0$

This is

$\vec{Z} = - \vec{X} - \vec{Y}$

$(Z_x , Z_y) = - ( \ 1062.70 m \ , \ 956.86 m \ ) - ( \ 0 \ , \ - 2200 m \ )$

$(Z_x , Z_y) = ( \ - 1062.70 m \ , \ 2200 m \ - \ 956.86 m \ )$

$(Z_x , Z_y) = ( \ - 1062.70 m \ , \ 1243.14 m\ )$

Now, for the magnitude, we just have to take its length:

$|\vec{Z}| = \sqrt{Z_x^2 + Z_y^2}$

$|\vec{Z}| = \sqrt{(- 1062.70 m)^2 + (1243.14 m)^2}$

$|\vec{Z}| = 1635.43 m$

For its angle, as the vector lays in the second quadrant, we can use:

$\theta = 180\° - arctan(\frac{1243.14 m}{ - 1062.70 m})$

$\theta = 180\° - arctan( -1.1720)$

$\theta = 180\° - 45\°31'40''$

$\theta = 130\°28'20''$

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