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finlep [7]
2 years ago
10

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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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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