Which equation best describes the law of conservation of momentum?

2 answers:

8 0

The law of conservation of momentum states that for two objects colliding in an isolated system, the total momentum before and after the collision is equal. Momentum should be conserved. Hope this answers the question. Have a nice day.

allochka39001 [22]3 years ago

6 0

Answer:

pi = pf

Explanation:

its the equation

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What effect or effects would be most significant if the Moon's orbital plane were exactly the same as the ecliptic plane?

Lilit [14]

Answer:

c. Solar eclipses would be much more frequent.

Explanation:

The ecliptic plane is the apparent orbit that the sun describes around the earth (although it is the earth that orbits the sun), is the path the sun follows in earth's sky.

A solar eclipse occurs when the moon gets between the earth and the sun, so a shadow is cast on the earth because the light from the sun is blocked.

The reason why solar eclipses are not very frequent is because the moon's orbital plane is not in the same plane as the orbit of the earth around the sun, but rather that it is somewhat inclined with respect to it.

So if both orbits were aligned, the moon would interpose between the sun and the earth more frequently, producing more solar eclipses.

So, if the moon's orbital plane were exacly the same as the ecliptic plane solar eclipses would be more frequent.

the answer is: c.

8 0

3 years ago

A 100-kg spacecraft is in a circular orbit about Earth at a height h = 2RE .

maria [59]

To solve this problem it is necessary to apply the concepts related to the conservation of the Gravitational Force and the centripetal force by equilibrium,

$F_g = F_c$

$\frac{GmM}{r^2} = \frac{mv^2}{r}$

Where,

m = Mass of spacecraft

M = Mass of Earth

r = Radius (Orbit)

G = Gravitational Universal Music

v = Velocity

Re-arrange to find the velocity

$\frac{GM}{r^2} = \frac{v^2}{r}$

$\frac{GM}{r} = v^2$

$v = \sqrt{\frac{GM}{r}}$

PART A ) The radius of the spacecraft's orbit is 2 times the radius of the earth, that is, considering the center of the earth, the spacecraft is 3 times at that distance. Replacing then,

$v = \sqrt{\frac{(6.67*10^{-11})(5.97*10^{24})}{3*(6.371*10^6)}}$