It takes Mars about ____ Earth years to orbit the Sun.

2 answers:

7 0

<span>At this distance, and with an orbital speed of 24.077 km/s, Mars takes 686.971 Earth days, the equivalent of 1.88 Earth years, to complete a orbit around the Sun. This eccentricity is one of the most pronounced in the Solar System, with only Mercury having a greater one (0.205).

686.971 rounds to 687
HOPE I HELPED!</span>

Feliz [49]3 years ago

4 0

the anser is 1.8 on e2020

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Misha Larkins [42]

Explanation:

$\underline{\boxed{\large{\bf{Option \; A!! }}}}$

Here,

$\rm { R_1}$ = 2Ω
$\rm { R_2}$ = 2Ω
$\rm { R_3}$ = 2Ω
$\rm { R_4}$ = 2Ω

We have to find the equivalent resistance of the circuit.

Here, $\rm { R_1}$ and $\rm { R_2}$ are connected in series, so their combined resistance will be given by,

$\longrightarrow \rm { R_{(1,2)} = R_1 + R_2} \\$

$\longrightarrow \rm { R_{(1,2)} = (2 + 2) \; Omega} \\$

$\longrightarrow \rm { R_{(1,2)} = 4 \; Omega} \\$

Now, the combined resistance of $\rm { R_1}$ and $\rm { R_2}$ is connected in parallel combination with $\rm { R_3}$ , so their combined resistance will be given by,

$\longrightarrow \rm {\dfrac{1}{ R_{(1,2,3)}} = \dfrac{1}{R_{(1,2)}} + \dfrac{1}{R_3} } \\$

$\longrightarrow \rm {\dfrac{1}{ R_{(1,2,3)}} = \Bigg ( \dfrac{1}{4} + \dfrac{1}{2} \Bigg ) \;\Omega} \\$

$\longrightarrow \rm {\dfrac{1}{ R_{(1,2,3)}} = \Bigg ( \dfrac{1 + 2}{4} \Bigg ) \;\Omega} \\$

$\longrightarrow \rm {\dfrac{1}{ R_{(1,2,3)}} = \Bigg ( \dfrac{3}{4} \Bigg ) \;\Omega} \\$

Reciprocating both sides,

$\longrightarrow \rm {R_{(1,2,3)}= \dfrac{4}{3} \;\Omega} \\$

Now, the combined resistance of $\rm { R_1}$ , $\rm { R_2}$ and $\rm { R_3}$ is connected in series combination with $\rm { R_4}$ . So, equivalent resistance will be given by,

$\longrightarrow \rm {R_{(1,2,3,4)}= R_{(1,2,3)} + R_4} \\$