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2 years ago

What is an asteroid that is pulled off course by a planet’s gravity and orbits around that planet called?

1 answer:

4 0

The definition of the celestial bodies allows us to find that the correct answer for a body that is captured and is in planetary orbit is:

Moon

Asteroids are small rocky bodies that rotate around the Sun, when this body enters the atmosphere of a planet and reaches the surface it is called meteoroids.

A meteorite is a fragment of meteoroid, which has been divided in space or the atmosphere during the entrance to the planet, in general they are smaller

A meteor is the atmospheric phenomenon that occurs when the pattern meteorite or meteoroid enters, that is, it does not correspond to a celestial body.

An asteroid satellite or Moon is a celestial object that revolves captures and around another asteroid, this concept can be extended to an asteroid revolving captures and around a planet

A satellite is a celestial body that orbits a planet, its origin is varied and could be formed during the formation of the planet itself, or by capturing a nearby body during the initial formation of the solar system.

Let's examine the different answers

Moon.

True. A body captured by a planet is generally called the Moon.

Meteoriode.

False. A meteoroid is a body that enters the atmosphere of the plant and reaches its surface.

Meteorite

False. It is a fragment of meteoroid that manages to reach the surface of the planet.

Meteor

False. Atmospheric phenomenon visible when passing a meteoroid or meteorite.

In conclusion, using the definition of celestial bodies we can find that the correct answer for a body that is captured and is in planetary orbit is:

Moon

Learn more here: brainly.com/question/3889451

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Light of wavelength 400 nm is incident on a single slit of width 15 microns. If a screen is placed 2.5 m from the slit. How far

olganol [36]

Answer:

0.0667 m

Explanation:

λ = wavelength of light = 400 nm = 400 x 10⁻⁹ m

D = screen distance = 2.5 m

d = slit width = 15 x 10⁻⁶ m

n = order = 1

θ = angle = ?

Using the equation

d Sinθ = n λ

(15 x 10⁻⁶) Sinθ = (1) (400 x 10⁻⁹)

Sinθ = 26.67 x 10⁻³

y = position of first minimum

Using the equation for small angles

tanθ = Sinθ = y/D

26.67 x 10⁻³ = y/2.5

y = 0.0667 m

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A uniform meterstick of mass 0.20 kg is pivoted at the 40 cm mark. where should one hang a mass of 0.50 kg to balance the stick?

Tcecarenko [31]

The weight of the meterstick is:
$W=mg=0.20 kg \cdot 9.81 m/s^2 = 1.97 N$
and this weight is applied at the center of mass of the meterstick, so at x=0.50 m, therefore at a distance
$d_1 = 0.50 m - 0.40 m=0.10 m$
from the pivot.
The torque generated by the weight of the meterstick around the pivot is:
$M_w = W d_1 = (1.97 N)(0.10 m)=0.20 Nm$

To keep the system in equilibrium, the mass of 0.50 kg must generate an equal torque with opposite direction of rotation, so it must be located at a distance d2 somewhere between x=0 and x=0.40 m. The magnitude of the torque should be the same, 0.20 Nm, and so we have:
$(mg) d_2 = 0.20 Nm$
from which we find the value of d2:
$d_2 = \frac{0.20 Nm}{mg}= \frac{0.20 Nm}{(0.5 kg)(9.81 m/s^2)}=0.04 m$

So, the mass should be put at x=-0.04 m from the pivot, therefore at the x=36 cm mark.

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