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shutvik [7]
4 years ago
13

Use the law of universal gravitacion to predict what would happen to Earth if the moon were replaced with an object that had the

mass of Jupiter.
Physics
1 answer:
Zina [86]4 years ago
8 0
Verrrrry interesting !
If the moon were replaced by something with a vastly greater mass
but at the same distance, then ...

-- The period of its revolution around the Earth would be much shorter.
That is, it would orbit the Earth in much less than 27.3 days.  We might
see it go through a complete set of phases in 2 weeks, or even 1 week.

-- The ocean tides would be much greater.  Low tides would be
much lower, and high tides would be much higher.

-- Sadly, the land tides, and the forces on the Earth's internal structure,
would also be much greater.  That means great increases in earthquake
and volcanic activity.

-- The Earth and moon both revolve around their common center of
mass. Under the current arrangement ... with the Earth having 80 times
the mass of the Moon ... that point is inside the Earth, and it looks a lot
like the Moon is orbiting a stationary Earth.
When the new body arrives to replace the lightweight Moon, that point
will be a lot closer to the new companion ... maybe even inside it. 
Then, it will look a lot like the monster is the stationary one, and the
Earth is orbiting it.
I actually don't believe that we would SEE that change, or feel it.

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A single point charge is placed at the center of an imaginary cube that has 10 cm long edges. The electric flux out of one of th
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Answer:

Explanation:

<u>Given:</u>

Length of each side of the cube, L=10\ cm

The Elecric flux through one of the side of the cube is, \phi =-1 kNm^2/C.

The net flux through a closed surface is defined as the total charge that lie inside the closed surface divided by \epsilon_0

Since Flux is a scalar quantity. It can added to get total flux through the surface.

\phi_{total}=\dfrac{Q_{in}}{\epsilon_0}\\6\times {-1}\times 10^3=\dfrac{Q_{in}}{\epsilon_0}\\\\Q_{in}=-6}\times 10^3\epsilon_0\\Q_{in}=-6}\times 10^3\times8.85\times 10^{-12}\\Q_{in}=-53.1\times10^{-9}\ C

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you stop the stopwatch at 4.0 s, but you notice a short time later that the same ant is at 0.81 m on the meter stick. Assuming t
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The time elapsed since you stopped the stopwatch is 0.41 s.

<em>Your question is not complete, it seems to be missing the following information;</em>

"The velocity of the ant is 2 m/s"

The given parameters;

  • velocity of the ant, v = 2 m/s
  • change in position of the ant, Δx = 0.81 m
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Velocity is defined as the change in displacement per change in time of motion of an object.

v = \frac{\Delta x}{\Delta t} = \frac{\Delta x}{t_2 - t_1} \\\\t_2 -t_1 = \frac{\Delta x}{v} \\\\t_2 - 4 = \frac{0.81}{2} \\\\t_2 - 4 = 0.405\\\\t_2 = 0.405 + 4\\\\t_2 = 4.405 \approx 4.41 \ s

The time elapsed since you stopped the stopwatch is calculated as;

t_{elapsed} = 4.41 \ s - 4\ s = 0.41 \ s

Thus, the time elapsed since you stopped the stopwatch is 0.41 s.

Learn more here: brainly.com/question/18153640

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