Describe how a unique position of the sun , earth, and moon creates a lunar eclipse

2 answers:

8 0

A eclipse of the sun can only occur at a new moon when the moon passes between earth and sun. If the moons shadow happens to fall upon earths surface at a time, you'll see some portion of the suns disk covered or aka eclipsed by the moon.. I hope that helps

Naddika [18.5K]2 years ago

7 0

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How were lion fish introduced to the areas around Florida

mr_godi [17]

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<h3>I think this will answer your question. This is information is not mine and this rightfully belongs to <u>columbia.edu.</u></h3><h3><u /></h3>

This brightly colored fish is native to the Indo-Pacific from Australia north to southern Japan and south to Micronesia. The lionfish is usually found in coral reefs of tropical waters, hovering in caves or near crevices. Native regions as well as Savannah, Georgia; Palm Beach and Boca Raton, Florida; Long Island, New York; Bermuda and possibly Charleston. In southern Florida and off the coast of the Carolinas in early to mid 1990s.

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sammy [17]

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The planet Mars has a mass of 6.1 × 1023 kg and radius of 3.4 × 106 m. What is the acceleration of an object in free fall near t

Oksanka [162]

The acceleration due to gravity of Mars is $3.5\ m / s^{2}$

<u>Explanation:</u>

As per universal law of gravity, the gravitational force is directly proportional to the product of masses and inversely proportional to the square of the distance between them. But in the present case, the gravity need to be determined between Mars and the object on Mars. Since the mass of Mars is greater than the mass of any object. Thus,

$\text {Gravitational force of planet}=\frac{G M m}{R^{2}}$

Here, G is the gravitational constant, R is the radius of Mars and M, m is the mass of Mars and the object respectively..

Also, according to Newton’s second law of motion, the acceleration of any object will be equal to the ratio of force exerted on it to the mass of the object.

So in order to determine the acceleration due to gravity of Mars, divide the gravitational force of Mars by mass of object on the surface of Mars.

$Acceleration\ due\ to\ gravity\ of\ mars =\frac{\text {Gravitational force of Mars}}{m \text { of object }}$

$Acceleration\ due\ to\ gravity\ of\ mars =\frac{G M m}{R^{2} \times m}=\frac{G M}{R^{2}}$

$\text { Acceleration due to gravity of mars }=\frac{6.67259 \times 10^{-11} \times 6.1 \times 10^{23}}{3.4 \times 3.4 \times 10^{12}}=\frac{40.703 \times 10^{12}}{11.56 \times 10^{12}}$