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Sever21 [200]
3 years ago
10

Please help

Physics
1 answer:
adelina 88 [10]3 years ago
5 0

Answer:

I'm pretty sure that the answer is A

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To which class of compounds does ammonia belong?
12345 [234]
Depends what you want . 1 it belongs to hydrides , it is nitrogen hydride . 2 if you want to use the property given ( and i presume you mean pH ) then it is a base . 3 because it is a soluble base , it is called an alkali .
6 0
3 years ago
If two stars have the same absolute magnitude, what can be a reason for the difference in their brightness?
AysviL [449]

B. their distances from the sun.

Explanation:

Absolute Magnitude:

Astronomers defines the absolute magnitude of a stars brightness in terms of how bright a star appears from a standard distance of 10 parsecs. Parsec is a unit of distance in astronomy. 10 parsecs is equal to 32.6 light years.

Apparent Magnitude:

Apparent magnitude of a star refers to how bright the star appears at its distance from the Earth.

If two stars have the same absolute magnitude but their apparent magnitude differs, the reason is that the distance of both the stars from the Earth varies. Hence their brightness differs when measured from Earth. The farther a star is from the Earth, the fainter its brightness.

Keywords: star, brightness, parsec, light years, apparent magnitude, absolute magnitude

Learn more about stars and absolute magnitude from:

brainly.com/question/13002384

brainly.com/question/1384449

#learnwithBrainly

5 0
3 years ago
Match the particles with their characteristics,
In-s [12.5K]
Positive charge=proton
Negative charge=electron
No charge/neutral=neutron
6 0
4 years ago
Read 2 more answers
Two particles, each of charge Q, are fixed at opposite corners of a square that lies in the plane of the page. A positive test c
amid [387]

Answer:

The magnitude of the net force is √2F.

Explanation:

Since the two particles have the same charge Q, they exert the same force on the test charge; both attractive or repulsive. So, the angle between the two forces is 90° in any case. Now, as we know the magnitude of these forces and that they form a 90° angle, we can use the Pythagorean Theorem to calculate the magnitude of the resultant net force:

F_N=\sqrt{F^{2}+F^{2}}\\\\F_N=\sqrt{2F^{2}}\\\\F_N=\sqrt{2}F

Then, it means that the net force acting on the test charge has a magnitude of √2F.

7 0
3 years ago
In midair an M = 145 kg bomb explodes into two pieces of m1 = 115 kg and another, respectively. Before the explosion, the bomb w
Daniel [21]

Answer:

v_2=-133.17m/s, the minus meaning west.

Explanation:

We know that linear momentum must be conserved, so it will be the same before (p_i) and after (p_f) the explosion. We will take the east direction as positive.

Before the explosion we have p_i=m_iv_i=Mv_i.

After the explosion we have pieces 1 and 2, so p_f=m_1v_1+m_2v_2.

These equations must be vectorial but since we look at the instants before and after the explosions and the bomb fragments in only 2 pieces the problem can be simplified in one dimension with direction east-west.

Since we know momentum must be conserved we have:

Mv_i=m_1v_1+m_2v_2

Which means (since we want v_2 and M=m_1+m_2):

v_2=\frac{Mv_i-m_1v_1}{m_2}=\frac{Mv_i-m_1v_1}{M-m_1}

So for our values we have:

v_2=\frac{(145kg)(24m/s)-(115kg)(65m/s)}{(145kg-115kg)}=-133.17m/s

5 0
3 years ago
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