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3 years ago

How do meteorites differ from meteors and meteoroids?

Physics

2 answers:

Evgesh-ka [11]3 years ago

8 0

Answer:

When meteoroids enter Earth's atmosphere (or that of another planet, like Mars) at high speed and burn up, the fireballs or “shooting stars” are called meteors. When a meteoroid survives a trip through the atmosphere and hits the ground, it's called a meteorite.

Explanation:

Ira Lisetskai [31]3 years ago

6 0

Answer: Meteoroid: Small particle from a comet or asteroid orbiting the Sun. Meteor: The light phenomena which results from a meteoroid entering the Earth's atmosphere and vaporizing; this is basically a shooting star. Meteorite: A meteoroid that survives through the Earth's atmosphere and lands on the Earth's surface.

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Dos cargas Q1=2pc y Q2=4pc estan separadas por una distancia de 6cm ¿con que fuerza se atraen?

noname [10]

Here we can use coulomb's law to find the force between two charges

As per coulombs law

]tex]F = \frac{kq_1q_2}{r^2}[/tex]

here we have

$k = 9 * 10^9$

$q_1 = 2pC$

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$r = 6cm = 0.06 m$

now by using the above equation we have

$F = \frac{9*10^9 * 2*10^{-12} * 4*10^{-12}}{0.06^2}$

$F = 2 * 10^{-11} N$

so here the force between two charges is of above magnitude and this will be repulsive force between them as both charges are of same sign.

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Answer:

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A small, positively charged ball is moved close to a large, positively charged ball. Which describes how the small ball likely r

Angelina_Jolie [31]

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Suppose that an asteroid traveling straight toward the center of the earth were to collide with our planet at the equator and bu

vlada-n [284]

Answer:

$\frac{1}{10}$ M

Explanation:

To apply the concept of angular momentum conservation, there should be no external torque before and after

As the asteroid is travelling directly towards the center of the Earth, after impact ,it does not impose any torque on earth's rotation, So angular momentum of earth is conserved

⇒ $I_{1} \times W_{1} =I_{2} \times W_{2}$

$I_{1}$ is the moment of interia of earth before impact
$W_{1}$ is the angular velocity of earth about an axis passing through the center of earth before impact
$I_{2}$ is moment of interia of earth and asteroid system
$W_{2}$ is the angular velocity of earth and asteroid system about the same axis

let $W_{1}=W$

since $\text{Time period of rotation}∝$ $\frac{1}{\text{Angular velocity}}$

⇒ if time period is to increase by 25%, which is $\frac{5}{4}$ times, the angular velocity decreases 25% which is $\frac{4}{5}$ times

therefore $W_{1}$ $=$ $\frac{4}{5} \times W_{1}$

$I_{1}=\frac{2}{5} \times M\times R^{2}$ (moment of inertia of solid sphere)

where M is mass of earth

R is radius of earth

$I_{2}=\frac{2}{5} \times M\times R^{2}+M_{1}\times R^{2}$

(As given asteroid is very small compared to earth, we assume it be a particle compared to earth, therefore by parallel axis theorem we find its moment of inertia with respect to axis)

where $M_{1}$ is mass of asteroid