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3 years ago

calculate the eccentricity of jupiter's orbit if the aphelion is 816.62 million km and the perihelion is 741.52 million km

Physics

1 answer:

schepotkina [342]3 years ago

8 0

<h2>Answer:0.048</h2>

Explanation:

Let the length of perihelion be $p$

Let the length of aphelion be $a$

Let the length of semi major axis of the elliptical path be $s$

As we know that semi major axis is the average of perihelion and aphelion.

So, $s=\frac{p+a}{2} =\frac{1558.14}{2}=779.07$ $millionKm$

Let the eccentricity of the elliptical path be $e$

$e=1-\frac{p}{s}=1-\frac{741.52}{779.07}=0.048$

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You wish to produce an emf of 41.0 mV using an inductor whose inductance is 13.0 H. You start with a current of 1.50 mA through

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Answer:

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Therefore, the current through the inductor at the end of 2.60 s is 9.7 mA.

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Answer:

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Explanation:

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A charge Q is uniformly spread over one surface of a very large nonconducting square elastic sheet having sides of length d. At

yaroslaw [1]

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