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anygoal [31]
4 years ago
6

Help please iwudhdbbdbdxkx

Physics
1 answer:
Ghella [55]4 years ago
4 0

Answer:

created

Explanation:

According to law of conservation of mass " Matter can neither be created nor destroyed"

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Which statements describe elliptical galaxies? *Select the two correct answers.*
Alecsey [184]

Elliptical galaxies are galaxies that are shaped like an ellipse. An ellipse is oval in shape, which means elliptical galaxies are oval in shape.

The <u>two correct statements</u> that describe elliptical galaxies are:

B. They are the most common galaxy type.

C. They typically contain very little gas and dust.

  • Elliptical galaxies are galaxies where you will find little or no gas and dust. The galaxies that are numerous in number are the elliptical galaxies.

  • In the elliptical galaxy, you can't find any <u>spiral component </u>in it.

Therefore, from the above explanation, we can see that <u>option B and C is the correct option.</u>

To learn more, visit the link below:

brainly.com/question/13645005

6 0
3 years ago
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What is a virgin i dk
anastassius [24]

Answer:

BYE- a person who has never had sexual intercourse.

Explanation:

3 0
4 years ago
Can someone please help me with some physics​
Dimas [21]

Answer:

sure I will helpy you iru

6 0
3 years ago
What si unit replaces pound
Artist 52 [7]

The SI unit of mass is the kilogram, and
the SI unit of force is the Newton.

8 0
4 years ago
Electrons (mass m, charge –e) are accelerated from rest through a potential difference V and are then deflected by a magnetic fi
adell [148]

Answer:

r=\dfrac{1}{B}\sqrt{\dfrac{2Vm}{e}}

Explanation:

Let m and e are the mass and charge of an electron. It is accelerated from rest through a potential difference V and are then deflected by a magnetic field that is perpendicular to their velocity. Let v is the velocity of the electron. It can be calculated as :

\dfrac{1}{2}mv^2=eV

v=\sqrt{\dfrac{2eV}{m}}

When the electron enters the magnetic field, the centripetal force is balanced by the magnetic force as :

\dfrac{mv^2}{r}=evB

r=\dfrac{mv}{eB}

or

r=\dfrac{1}{B}\sqrt{\dfrac{2Vm}{e}}

So, the radius of the resulting electron trajectory is \dfrac{1}{B}\sqrt{\dfrac{2Vm}{e}}. Hence, this is the required solution.

8 0
3 years ago
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