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3 years ago

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The Moon's center is 3.9x10 m from Earth's center. The Moon is 1.5x10^8 km from the Sun's center. If the mass of the Moon is 7.3

x10^22 kg, find the ratio of the gravitational forces exerted by Earth and the Sun on the Moon

1 answer:

nika2105 [10]3 years ago

3 0

Explanation:

It is given that The Moon's center is 3.9x10⁸ m from Earth's center. The moon 1.5x10⁸ km from the Sun's center. We need to find the ratio of the gravitational forces exerted by Earth and the Sun on the Moon.

The gravitational force is given by :

$F=\dfrac{Gm_em_m}{r^2}$

It means $F\propto \dfrac{1}{r^2}$

So,

$\dfrac{F_1}{F_2}=\dfrac{r_2}{r_1}$

r₁ = 3.9x10⁸ km

r₂= 1.5x10⁸ km

So,

$\dfrac{F_1}{F_2}=\dfrac{1.5\times 10^8}{3.9\times 10^8}\\\\\dfrac{F_1}{F_2}=\dfrac{5}{13}$

Hence, the ratio of the gravitational forces exerted by Earth and the Sun on the Moon is 5:13.

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2 years ago

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3 years ago

In a 49 s interval, 595 hailstones strike a glass window of an area of 0.954 m at an angle of 25° to the window surface. Each ha

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Average force on the window: 0.32 N

Explanation:

The average force exerted on the window is given by Newton's second law

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3 0

3 years ago

Help pls i need this right now

pantera1 [17]

Answer:

The x-component of $F_{3}$ is 56.148 newtons.

Explanation:

From 1st and 2nd Newton's Law we know that a system is at rest when net acceleration is zero. Then, the vectorial sum of the three forces must be equal to zero. That is:

$\vec F_{1} + \vec F_{2} + \vec F_{3} = \vec O$ (1)

Where:

$\vec F_{1}$ , $\vec F_{2}$ , $\vec F_{3}$ - External forces exerted on the ring, measured in newtons.

$\vec O$ - Vector zero, measured in newtons.

If we know that $\vec F_{1} = (70.711,70.711)\,[N]$ , $\vec F_{2} = (-126.859, 46.173)\,[N]$ , $F_{3} = (F_{3,x},F_{3,y})$ and $\vec O = (0,0)\,[N]$ , then we construct the following system of linear equations:

$\Sigma F_{x} = 70.711\,N - 126.859\,N +F_{3,x} = 0\,N$ (2)

$\Sigma F_{y} = 70.711\,N + 46.173\,N+F_{3,y} = 0\,N$ (3)

The solution of this system is:

$F_{3,x} = 56.148\,N$ , $F_{3,y} = -116.884\,N$

The x-component of $F_{3}$ is 56.148 newtons.

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