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docker41 [41]
2 years ago
5

What is in the center of our galaxy?

Physics
2 answers:
irakobra [83]2 years ago
8 0

Answer:

The Galactic center also known as "A supermassive black hole"

Explanation:

"Its the rotational center of the Milky way" -google

sattari [20]2 years ago
7 0
The answer would be the sun
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Recall: earth applies _____________ on Earth or equivalently
Olin [163]

Answer:

gravitational attraction

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3 years ago
Electrons are added into the outermost what in groups 1 & 2
aivan3 [116]

S orbital.

Group 1 elements have a general configuration ns^{1}, where n represents the highest occupied Principal Energy Level. For example, Lithium has the valence configuration 2s^{1} whereas Cesium has 6s^{1}. Both of them belong to Group 1 of Periodic Table.

Group 2 elements have a general configuration of ns^{2}. For example, Magnesium has 3s^{2} as its outer shell configuration while Strontium has the same as 5s^{2}.

We see that in both the cases, the outermost S orbital is being filled.

3 0
3 years ago
What is the speed of sound in air when the temperature is 20°C?
Anarel [89]
The answer to your question is 343 m/s
5 0
3 years ago
Read 2 more answers
Which method will correctly determine whether the forces on an object are balanced or unbalanced?
MaRussiya [10]

For this case, the first thing you should do is define a reference system.

Once the system is defined, we must follow the following steps:

1) Do the sum of forces in a horizontal direction

2) Do the sum of forces in vertical direction

The forces will be balanced if for each direction the net force is equal to zero.

The forces will be unbalanced if for each direction the net force is nonzero.

Answer:

Add the forces in the horizontal and vertical directions separately.

6 0
3 years ago
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Starting from zero, the electric current takes 2 seconds to reach half its maximum possible value in an RL circuit with a resist
Leno4ka [110]

Answer:

time=4s

Explanation:

we know that in a RL circuit with a resistance R, an inductance L and a battery of emf E, the current (i) will vary in following fashion

i(t)=\frac{E}{R}(1-e^\frac{-t}{\frac{L}{R}}), where imax=\frac{E}{R}

Given that, at i(2)=\frac{imax}{2} =\frac{E}{2R}

⇒\frac{E}{2R}=\frac{E}{R}(1-e^\frac{-2}{\frac{L}{R}})

⇒\frac{1}{2}=1-e^\frac{-2}{\frac{L}{R}}

⇒\frac{1}{2}=e^\frac{-2}{\frac{L}{R}}

Applying logarithm on both sides,

⇒log(\frac{1}{2})=\frac{-2}{\frac{L}{R}}

⇒log(2)=\frac{2}{\frac{L}{R}}

⇒\frac{L}{R}=\frac{2}{log2}

Now substitute i(t)=\frac{3}{4}imax=\frac{3E}{4R}

⇒\frac{3E}{4R}=\frac{E}{R}(1-e^\frac{-t}{\frac{L}{R}})

⇒\frac{3}{4}=1-e^\frac{-t}{\frac{L}{R}}

⇒\frac{1}{4}=e^\frac{-t}{\frac{L}{R}}

Applying logarithm on both sides,

⇒log(\frac{1}{4})=\frac{-t}{\frac{L}{R}}

⇒log(4)=\frac{t}{\frac{L}{R}}

⇒t=log4\frac{L}{R}

now subs. \frac{L}{R}=\frac{2}{log2}

⇒t=log4\frac{2}{log2}

also log4=log2^{2}=2log2

⇒t=2log2\frac{2}{log2}

⇒t=4

5 0
3 years ago
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