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3 years ago

Discuss the force that exists between the Earth and the moon by referring to the mass of each.

1 answer:

7 0

The word gravity is used to describe the gravitational pull (force) an object experiences on or near the surface of a planet or moon. The gravitational force is a force that attracts objects with mass towards each other. Any object with mass exerts a gravitational force on any other object with mass.

Hope it answers your question!

Brainliest would be nice but of course you don’t gotta :)

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Calculate the location xcm of the center of mass of the Earth-Moon system. Use a coordinate system in which the center of the Ea

Jet001 [13]

Answer:

The center of mass of the Earth-Moon system is 4.673 kilometers away from center of Earth.

Explanation:

Let suppose that planet and satellite can be treated as particles. The masses of Earth and Moon ( $m_{E}$ , $m_{M}$ ) are $5.972\times 10^{24}\,kg$ and $7.349\times 10^{22}\,kg$ , respectively. The distance between centers is 384,403 kilometers. The location of the center of mass can be found by using weighted averages:

$\bar x = \frac{x_{E}\cdot m_{E}+x_{M}\cdot m_{M}}{m_{E}+m_{M}}$

If $x_{E} = 0\,km$ and $x_{M} = 384,403\,km$ , then:

$\bar x = \frac{(0\,km)\cdot (5.972\times 10^{24}\,kg)+(384,403\,km)\cdot (7.349\times 10^{22}\,kg)}{5.972\times 10^{24}\,kg+7.349\times 10^{22}\,kg}$

$\bar x = 4.673\,km$

The center of mass of the Earth-Moon system is 4.673 kilometers away from center of Earth.

8 0

3 years ago

A Carnot engine operates between reservoirs at 600 and 300 K. If the engine absorbs 100 J per cycle at the hot reservoir, what i

PIT_PIT [208]

Answer:

W =50 J

Explanation:

given data:

T_h=600 k

T_L=300 k

Q_h=100 J

required:

<u>solution:</u>

║W║=║Q_h║(1-T_L/T_h)

=50 J

7 0

3 years ago

Read 2 more answers

NASA has asked your team of rocket scientists about the feasibility of a new satellite launcher that will save rocket fuel. NASA

kkurt [141]

Answer:

The answer is " $q=0.0945\,C$ ".

Explanation:

Its minimum velocity energy is provided whenever the satellite(charge 4 q) becomes 15 m far below the square center generated by the electrode (charge q).

$U_i=\frac{1}{4\pi\epsilon_0} \times \frac{4\times4q^2}{\sqrt{(15)^2+(5/\sqrt2)^2}}$

It's ultimate energy capacity whenever the satellite is now in the middle of the electric squares:

$U_f=\frac{1}{4\pi\epsilon_0}\ \times \frac{4\times4q^2}{( \frac{5}{\sqrt{2}})}$

Potential energy shifts:

$= U_f -U_i \\\\ =\frac{16q^2}{4\pi\epsilon_0}\left ( \frac{\sqrt2}{5}-\frac{1}{\sqrt{(15)^2+( \frac{5}{\sqrt{2})^2)}}\right ) \\\\ =\frac{16q^2}{4\pi\epsilon_0}\left ( \frac{\sqrt2}{5}-\frac{1}{ 15 +( \frac{5}{2})}}\right )\\\\ =\frac{16q^2}{4\pi\epsilon_0}\left ( \frac{\sqrt2}{5}-\frac{1}{ (\frac{30+5}{2})}}\right )\\\\$

$=\frac{16q^2}{4\pi\epsilon_0}\left ( \frac{\sqrt2}{5}-\frac{1}{ (\frac{35}{2})}}\right )\\\\=\frac{16q^2}{4\pi\epsilon_0}\left ( \frac{\sqrt2}{5}-\frac{1}{17.5}}\right )\\\\ =\frac{16q^2}{4\pi\epsilon_0}\left ( \frac{ 24.74- 5 }{87.5}}\right )\\\\ =\frac{16q^2}{4\pi\epsilon_0}\left ( \frac{ 19.74- 5 }{87.5}}\right )\\\\ =\frac{4q^2}{\pi\epsilon_0}\left ( 0.2256 }\right )\\\\= \frac{0.28 \times q^2}{ \epsilon_0}\\\\=q^2\times31.35 \times10^9\,J$

Now that's the energy necessary to lift a satellite of 100 kg to 300 km across the surface of the earth.

$=\frac{GMm}{R}-\frac{GMm}{R+h} \\\\=(6.67\times10^{-11}\times6.0\times10^{24}\times100)\left(\frac{1}{6400\times1000}-\frac{1}{6700\times1000} \right ) \\\\ =(6.67\times10^{-11}\times6.0\times10^{26})\left(\frac{1}{64\times10^{5}}-\frac{1}{67\times10^{5}} \right ) \\\\=(6.67\times6.0\times10^{15})\left(\frac{67 \times 10^{5} - 64 \times 10^{5} }{ 4,228 \times10^{5}} \right ) \\\\$

$=( 40.02\times10^{15})\left(\frac{3 \times 10^{5}}{ 4,228 \times10^{5}} \right ) \\\\ =40.02 \times10^{15} \times 0.0007 \\\\$

$\\\\ =0.02799\times10^{10}\,J \\\\= q^2\times31.35\times10^{9} \\\\ =0.02799\times10^{10} \\\\q=0.0945\,C$

This satellite is transmitted by it system at a height of 300 km and not in orbit, any other mechanism is required to bring the satellite into space.

6 0

3 years ago

You are raising up a big bucket of water from a 25.9 m deep well. The combined mass of the water and the bucket is 13.9 kg. The

Mars2501 [29]

The total work done is 5980 Joules and the power expended is 57 Watts.

The work done is the work done in the gravitational field as the bucket is raised up Thus work required to remove the bucket Wb;

Wb = 13.9 kg * 25.9 m * 9.8 m/s^2 = 3530 Joules

Height of the center of mass of chain = 25.9 / 2 = 12.95 m

Work done by the chain Wc;

Wc = 12.95 * 19.3 * 9.8 = 2450 Joules

Total work = 3530 + 2450 = 5980 Joules

Power expended = W / t = 5980 J / 105 sec = 57 J/s = 57 Watts

Learn more about work done:brainly.com/question/13662169

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3 0

2 years ago

What force could be applied to a box to make the net force in the horizontal direction of zero

Kazeer [188]

If there is no friction, the force that moves the box forward horizontally must be matched by the same force.

If there is friction, then the force moving it forward = frictional force + the additional force you need to add.

6 0

4 years ago