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zmey [24]
3 years ago
15

A planet travels in an elliptical path around a star, as shown in the figure. As the planet gets closer to the star, the gravita

tional force that the star exerts on the planet increases. Which statement of reasoning best supports and correctly identifies what happens to the magnitude of the force that the planet exerts on the star as the planet gets closer to the star?
The force remains constant because the mass of the planet remains constant.
A

The force increases because it is part of a Newton’s third law pair of forces with the force that the star exerts on the planet.
B

The force decreases because the planet increases its speed as it gets closer to the star.
C

The force fluctuates such that it increases and decreases because the planet does not travel in a perfectly circular path.

Chemistry
1 answer:
Soloha48 [4]3 years ago
3 0

Answer:

The force increases because it is part of a Newton’s third law pair of forces with the force that the star exerts on the planet.

Explanation:

Force between two objects can be expressed by an equation:

F = G • m1 • m2 / r^2,

where m1 and m2 are objects' masses, r is the distance between them, and G is a gravitational constant.

That means that greater the masses or lesser the distance, the force will be greater, and vice versa.

This force exists between any two objects, but is generally extremely weak, so it's best observed with big and large objects with great mass, such as planets and stars.

This force, whatever its magnitude may be, always works on both objects, following the third Newton's law.

So, whatever the force the stat exerts on the planet is, the planet will exert the same amount of force on the star.

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What is the mass of a sample of alcohol (specific heat = 2.4 J/gC), if it requires 4780 J of heat to raise the temperature by 5.
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Hence, it is found out that by raising the temperature of the given product, the mass of alcohol would be 367 g.

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We are required to find the mass of alcohol m = ?

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The specific heat given is represented by = c = 2.4 J/gC

The temperature given is ΔT = 5.43° C

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The formula is c = \frac{q}{mt}

We can drive value of m bu shifting m on the left hand side,

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