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Nezavi [6.7K]
3 years ago
15

A distant planet with a mass of (7.2000x10^26) has a moon with a mass of (5.0000x10^23). The distance between the planet and the

moon is (6.10x10^11). What is the gravitational force between the two objects?
A distant planet with a mass of (7.2000x10^26) has a moon with a mass of (5.0000x10^23). The distance between the planet and the moon is (6.10x10^11). What is the gravitational force between the two objects?
Physics
1 answer:
BARSIC [14]3 years ago
8 0

Answer:

Explanation:

This is a simple gravitational force problem using the equation:

F_g=\frac{Gm_1m_2}{r^2} where F is the gravitational force, G is the universal gravitational constant, the m's are the masses of the2 objects, and r is the distance between the centers of the masses. I am going to state G to 3 sig fig's so that is the number of sig fig's we will have in our answer. If we are solving for the gravitational force, we can fill in everything else where it goes. Keep in mind that I am NOT rounding until the very end, even when I show some simplification before the final answer.

Filling in:

F_g=\frac{(6.67*19^{-11})(7.2000*10^{26})(5.0000*10^{23})}{(6.10*10^{11})^2} I'm going to do the math on the top and then on the bottom and divide at the end.

F_g=\frac{2.4012*10^{40}}{3.721*10^{23}} and now when I divide I will express my answer to the correct number of sig dig's:

Fg= 6.45 × 10¹⁶ N

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Our values are given as

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\alpha = 58.89rad/s^2

With the expression of the acceleration found it is now necessary to replace it on the torque equation and the respective moment of inertia for the disk, so

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\tau = (\frac{1}{2}(17*10^{-3})(7*10^{-2})^2)(58.89)

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Therefore the torque exerted on it is 2.45*10^{-3}N\cdot m

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defon

Answer:

A

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