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SashulF [63]
3 years ago
10

AP Physics Question 3

Physics
1 answer:
antiseptic1488 [7]3 years ago
7 0

The answer is B

The gravitational force exerted from the planet on Moon is eight times larger than the gravitational force exerted from the planet on Moon .

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One of the foci for the moon's orbit would be the
suter [353]
The question is oversimplified, and pretty sloppy.

Relative to the Earth . . .
The Moon is in an elliptical orbit around us, with a period of
27.32... days, and with the Earth at one focus of the ellipse.

Relative to the Sun . . .
The Moon is in an elliptical orbit around the Sun, with a period
of 365.24... days, and with the Sun at one focus of the ellipse,
and the Moon itself makes little dimples or squiggles in its orbit
on account of the gravitational influence of the nearby Earth.

I'm sorry if that seems complicated.  You know that motion is
always relative to something, and the solar system is not simple.
5 0
3 years ago
Read 2 more answers
They occupy the space surrounding the nucleus of the atom, they have a -1 charge
kipiarov [429]

Answer:

Electrons.

Explanation:

Look it up.

4 0
3 years ago
Explain how wind erosion changes land forms
valkas [14]

Answer:

the wind carries abrasive materials

Explanation:

such as sand and salt over time theses small particles slowly strip way at the land form sculpting it by eroding the softer layers first

8 0
2 years ago
What is the acceleration of a 50 kg object pushed with a net force of 500 newtons?
vodomira [7]
Using the formula F=ma
500N=50kg (a)
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3 0
4 years ago
The potential energy of a particle experiencing a certain kind of force is given by U(x) = 2x + 8 x measured in joules (J), for
masha68 [24]

Answer:

Explanation:

Given

Potential Energy of a particle is given by U(x)=2x+\frac{8}{x}

For minimum or maximum Potential Energy differentiate U(x) w.r.t x

\frac{\mathrm{d} U}{\mathrm{d} x}=2-\frac{8}{x^2}

putting \frac{\mathrm{d} }{\mathrm{d} x}=0

2-\frac{8}{x^2}=0

x^2=\frac{8}{2}

x^2=4

x=\sqrt{4}=\pm 2

To know minimum value check \frac{\mathrm{d^2} U}{\mathrm{d} x^2}

at x=-2

\frac{\mathrm{d^2} U}{\mathrm{d} x^2}=\frac{16}{x^3}=-\frac{16}{8}=-2

so at x=-2 potential energy is minimum U=-8\ J

8 0
3 years ago
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