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3 years ago

From Kepler's third law, an asteroid with an orbital period of 8 years lies at an average distance from the Sun equal to:

Physics

1 answer:

bazaltina [42]3 years ago

7 0

Answer:

The correct option is (B).

Explanation:

The Kepler's third law of motion gives the relationship between the orbital time period and the distance from the semi major axis such that,

$T^2\propto a^3\\\\T^2=ka^3$

It is mentioned that, an asteroid with an orbital period of 8 years. So,

$(8)^2=ka^3\\\\64=ka^3\\\\a=(64)^{\dfrac{1}{3}}\\\\a=4\ AU$

So, an asteroid with an orbital period of 8 years lies at an average distance from the Sun equal to 4 astronomical units.

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Phoenix [80]

Answer:

Option c is correct

Explanation:

There are two types of collisions-elastic collision and inelastic collision.

In elastic collision, both kinetic energy and total momentum are conserved. On the other hand, in inelastic collision, total momentum is conserved but kinetic energy is not conserved. Thus, option b and d are incorrect.

Total energy is always conserved in both types. Thus, option a is incorrect.

In a perfectly inelastic collision, objects stick together. This happens because maximum kinetic energy is dissipated and used in bonding of the two objects. Thus, correct option is c.

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3 years ago

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Alika [10]

Answer:

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Explanation:

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krok68 [10]

Answer:

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Explanation:

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2 years ago

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3 years ago

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