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4 years ago

HELPPPPP WILL GIVE BRAINLIEST!!!

Physics

2 answers:

Lady bird [3.3K]4 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]4 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)

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A baseball with mass of 0.145 kg is thrown straight down at the ground. At a particular speed, it has a drag force of 0.4 N acti

Tju [1.3M]

Answer:

It is -7.0 m/s2

Explanation:

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2 years ago

A transformer connected to a 120 V (rms) ac line is to supply 13,000 V (rms) for a neon sign. To reduce shock hazard, a fuse is

TiliK225 [7]

Answer with Explanation:

We are given that

$V_1=120 V$

$V_2=13000 V$

$I_{2}=8.5 mA=8.5\times 10^{-3} A$

$1 mA=10^{-3} A$

a.We know that

$\frac{N_2}{N_1}=\frac{V_2}{V_1}$

$\frac{N_2}{N_1}=\frac{13000}{120}=108.3$

b. $P_{avg}=I_2V_2$

$P_{avg}=13000\times 8.5\times 10^{-3}=110.5 Watt$

c. $I_1=\frac{P_{avg}}{V_1}$

$I_1=\frac{110.5}{120}=0.92 A$

5 0

4 years ago

Read 2 more answers

How much potassium nitrate, KNO3, would completely dissolve in 100g of water at 40℃?

const2013 [10]

Ok I know this from other stuff potassium nitrate would completely dissolve in a 100 g on was at 30 c would be 60 but this is 40 so I’m not really sure and I don’t what to ok give you a bad grade but if I had to guess I would go with 65 grams

8 0

3 years ago

A point charge Q is held at a distance r from the center of a dipole that consists of two charges ±qseparated by a distance s. T

atroni [7]

Answer:

The magnitude of the force on the dipole due to the charge Q = $\rm \dfrac{1}{\epsilon_o}\times \dfrac{1}{4\pi }\dfrac{2qQs}{r^3}.$

The magnitude of the torque on the dipole = $\rm \dfrac{1}{\epsilon_o}\times \dfrac{1}{4\pi}\dfrac{2qQs^2}{r^3}.$

Explanation:

Given that a point charge Q is held at a distance r from the center of a dipole that consists of two charges ±q, separated by a distance s and the charge Q is located in the plane that bisects the dipole.

The magnitude of the electric field that the dipole exerts at the position where the charge Q is held is given by

$\rm E = \dfrac{k2qs}{(r^2+s^2)^{3/2}}.$

where,

k is the Coulomb's constant, having value = $\dfrac{1}{4\pi \epsilon_o}$

$\epsilon_o$ is the electrical permittivity of free space.

Also, r>>s, therefore, $\rm r^2+s^2\approx r^2.$

$\rm E = \dfrac{k2qs}{(r^2)^{3/2}}=\dfrac{k2qs}{r^3}.$

The magnitude of the electric force F on a charge q placed at a point and the magnitude of the electric field E at that point are related as

$\rm F=qE$

Therefore, the electric force on the charge Q due to the dipole is given by

$\rm F=Q\dfrac{k2qs}{r^3}=\dfrac{1}{4\pi \epsilon_o}\dfrac{2qQs}{r^3}.$

According to Newton's third law of motion, the magnitude of the force exerted by the dipole on the charge Q is same as the magnitude of the force exerted by the charge on the dipole.

Thus, the magnitude of the force on the dipole due to the charge Q = $\dfrac{1}{\epsilon_o}\times \dfrac{1}{4\pi }\dfrac{2qQs}{r^3}.$

The magnitude of the torque on the dipole is given by

$\rm \tau = Fs\ \sin\theta$

Since the charge Q is placed in the plane that bisects the dipole, therefore, $\theta = 90^\circ$ .

$\rm \tau = \dfrac{1}{4\pi \epsilon_o}\dfrac{2qQs}{r^3}\cdot s\cdot 1=\dfrac{1}{4\pi \epsilon_o}\dfrac{2qQs^2}{r^3}.$

4 0

3 years ago

A quantity that has only one magnitude

Len [333]

Answer:

scalar quantity

Explanation:

Vector quantities have two characteristics, a magnitude and a direction. Scalar quantities have only a magnitude.

3 0

4 years ago