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12345 [234]
3 years ago
10

What are the three main categories of elements?

Physics
2 answers:
o-na [289]3 years ago
8 0

Answer:

Classification of the Elements. These three groups are: metals, nonmetals, and inert gases.

Explanation:

iogann1982 [59]3 years ago
7 0

Answer:If you're referring to Organic Molecules, then they are CARBON, HYDROGEN and OXYGEN.

Explanation:

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At what angle two forces P + Q and (P - Q) act so that their resultant is :
stiv31 [10]

Use resultant formula

\boxed{\sf R=\sqrt{A^2+B^2+2ABcos\theta}}

So

#1

A be p+q and B be p-q

\\ \rm\Rrightarrow R=\sqrt{3p^2+q^2}

\\ \rm\Rrightarrow \sqrt{(p+q)^2+(p-q)^2+2(p+q)(p-q)cos\alpha}=\sqrt{3p^2+q^2}

\\ \rm\Rrightarrow 2p^2+2q^2+2(p^2-q^2)cos\alpha=3p^2+q^2

\\ \rm\Rrightarrow 2cos\alpha=1

\\ \rm\Rrightarrow cos\alpha=\dfrac{1}{2}

\\ \rm\Rrightarrow \alpha=\dfrac{\pi}{3}

#2

\\ \rm\Rrightarrow 2(p^2+q^2)+2(p^2-q^2)cos\beta=2(p^2+q^2)

\\ \rm\Rrightarrow 2cos\beta=0

\\ \rm\Rrightarrow cos\beta=0

\\ \rm\Rrightarrow \beta=\dfrac{\pi}{2}

#3

\\ \rm\Rrightarrow 2(p^2+q^2)+2(p^2-q^2)cos\gamma=p^2+q^2

\\ \rm\Rrightarrow 2(p^2-q^2)cos\gamma=-(p^2+q^2)

\\ \rm\Rrightarrow cos\gamma =\dfrac{q^2-p^2}{2(p^2-q^2)}

\\ \rm\Rrightarrow \gamma=cos^{-1}\left(\dfrac{q^2-p^2}{2(p^2+q^2)}\right)

8 0
2 years ago
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Which is true of a mechanical wave?
mario62 [17]
A mechanical wave must have a medium.

Hope this helps,
♥<em>A.W.E.<u>S.W.A.N.</u></em>♥
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3 years ago
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at room temperature, iron is a solid and mercury is a liquid. based on this information we can infer thata. iron has a higher bo
olchik [2.2K]
A. iron has a higher boiling point thanmercury.
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which of the following would be best to do if you were inside a car and a power line fell on the car​
NARA [144]

Answer:

Do not move stay in your car and wait for someone from the power company to come and help

Explanation:

Plz vote my answer as the brainiest, i rlly need it! hope this helps!

6 0
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Why is it expensive to bring electricity into urban areas?
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Because of the pole and the generator you would have to biuld
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