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3 years ago

HELPPPPP WILL GIVE BRAINLIEST!!!

Physics

2 answers:

Lady bird [3.3K]3 years ago

7 0

Answer:

3.round object that orbits the Sun but lacks the ability to clear the neighborhood around its orbit.

Explanation:

in 2006 the IAU, said that a dwarf planet is round object that has not cleared the area round a object and that is why Pluto, Ceres, and Eris are dwarf planet.

Vsevolod [243]3 years ago

3 0

Answer:

A dwarf planet is a "round object that orbits the Sun but lacks the ability to clear the neighborhood around its orbit." (option 3)

Explanation:

A dwarf planet is defined as a celestial body that is characterized by:

Being in orbit around the Sun (rotate around this)
Its mass is enough for its own gravity to exceed the stiff body strength. In other words: it has a spherical or almost spherical shape determined by hydrostatic equilibrium.
It is not a planet's satellite.
It has failed to clean the neighborhoods of its orbit. In general, a star reaches a certain level of evolution and incisively influences the surrounding stars, either by attracting them, moving them away or making them revolve around it. This is not the case with dwarf planets, which causes other relatively independent stars to exist around their orbit. In other words, the planet revolves around the Sun accompanied by other celestial bodies that perform the same orbit (it is not gravitationally dominant as it shares its orbital space with other bodies of similar size).

In summary, a dwarf planet is a "round object that orbits the Sun but lacks the ability to clear the neighborhood around its orbit." (option 3)

You might be interested in

Which instrument is used to measure depth of ocean ?

skelet666 [1.2K]

Echo sounding is a type of SONAR used to determine the depth of water by transmitting sound pulses into water. The time interval between emission and return of a pulse is recorded, which is used to determine the depth of water along with the speed of sound in water at the time.

5 0

2 years ago

The moon has a diameter of 3.48 x 106 m and is a distance of 3.85 x 108 m from the earth. The sun has a diameter of 1.39 x 109 m

Mrrafil [7]

Answer:

0.00903 rad

0.00926 rad

6.268\times 10^{-6}

Explanation:

s = Diameter of the object

r = Distance between the Earth and the object

Angle subtended is given by

$\theta=\frac{s}{r}$

For the Moon

$\theta_m=\dfrac{3.48\times 10^6}{3.85\times 10^8}\\\Rightarrow \theta_m=0.00903\ rad$

The angle subtended by the Moon is 0.00903 rad

For the Sun

$\theta_s=\dfrac{1.39\times 10^9}{1.5\times 10^{11}}\\\Rightarrow \theta_s=0.00926\ rad$

The angle subtended by the Sun is 0.00926 rad

Area ratio is given by

$\frac{A_m}{A_s}=\dfrac{\pi r_m^2}{\pi r_s^2}\\\Rightarrow \frac{A_m}{A_s}=\dfrac{d_m^2}{d_s^2}\\\Rightarrow \frac{A_m}{A_s}=\dfrac{(3.48\times 10^{6})^2}{(1.39\times 10^9)^2}\\\Rightarrow \frac{A_m}{A_s}=6.268\times 10^{-6}$

The area ratio is $6.268\times 10^{-6}$

3 0

3 years ago

A mass of 0.5 kg hangs motionless from a vertical spring whose length is 1.10 m and whose unstretched length is 0.50 m. Next the

ser-zykov [4K]

Answer:

The maximum length during the motion is $L_{max} = 1.45m$

Explanation:

From the question we are told that

The mass is $m =0.5 kg$

The vertical spring length is $L = 1.10m$

The unstretched length is $L_{un} = 1.30m$

The initial speed is $v_i = 1.3m/s$

The new length of the spring $L_{new} = 1.30 m$

The spring constant k is mathematically represented as

$k = -\frac{F}{y}$

Where F is the force applied $= m * g = 0.5 * 9.8=4.9N$

y is the difference in weight which is $=1.10-0.50=0.6m$

The negative sign is because the displacement of the spring (i.e its extension occurs against the force F)

Now substituting values accordingly

$k = \frac{4.9}{0.6}$

$= 8.17 N/m$

The elastic potential energy is given as $E_{PE} = \frac{1}{2} k D^2$

where D is this the is the displacement

Since Energy is conserved the total elastic potential energy would be

$E_T = initial \ elastic\ potential \ energy + kinetic \ energy$

$E_T = \frac{1}{2} k D_{max}^2 = \frac{1}{2} k D^2 + \frac{1}{2} mv^2$

Substituting value accordingly

$\frac{1}{2} *8.17 *D_{max}^2 =\frac{1}{2} * 8.17*(1.30 - 0.50)^2 + \frac{1}{2} * 0.5 *1.30^2$

$4.085 * D_{max}^2 = 3.69$

$D^2_{max} = 0.9033$

$D_{max} = 0.950m$

So to obtain total length we would add the unstretched length

So we have

$L_{max} = 0.950 + 0.5 = 1.45m$

5 0

2 years ago

Read 2 more answers

What is the mass moment of inertia of a 20kg sphere with a radius of 0.2m about a point on the sphere's perimeter

Kobotan [32]

Answer:

I = M R^2 is the moment of inertia about a point that is a distance R from the center of mass (uniform distributed mass).

The moment of inertia about the center of a sphere is 2 / 5 M R^2.

By the parallel axis theorem the moment of inertia about a point on the rim of the sphere is I = 2/5 M R^2 + M R^2 = 7/5 M R^2

I = 7/5 * 20 kg * .2^2 m = 1.12 kg m^2

7 0

2 years ago

WHAT DOES DENSITY HAVE TO DO WITH PLATE TECTONICS? Explain

Korvikt [17]

Answer: The reason for the differences in density is the composition of rock in the plates. When two plates come in contact with each other through plate tectonics, scientists can use the density of the plates to predict what will happen. Whichever plate is more dense will sink, and the less dense plate will float over it.

Explanation:

Hope this helps ( not copied and pasted, this answer was done by me so I don't know if it's good or not)

5 0

2 years ago