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MatroZZZ [7]
3 years ago
7

* THE ANSWER IS D * Silicon (chemical symbol Si) is located in Group 14, Period 3. Which is silicon most likely

Physics
2 answers:
laila [671]3 years ago
8 0

Answer:

D. Metalloid with properties of both metals and nonmetals

o-na [289]3 years ago
4 0

Answer:

Hank u so much out don't know how much this rally

Explanation:

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Can someone please explain the quantum theory in simple terms
BaLLatris [955]
Quantum theory is the theoretical basis of modern physics that explains the nature and behavior of matter and energy on the atomic and subatomic level. The nature and behavior of matter and energy at that level is sometimes referred to as quantum physics and quantum mechanics.
6 0
2 years ago
A positively charged non-metal ball A is placed near metal ball B. Prove that if the charge on B is positive but of small magnit
nordsb [41]

Answer:

D

Explanation:

Ball A is a non positively charged non metal while ball B is metal ball.

Given: The ball B positive charge of small magnitude

To prove: Balls will attract each other

IN this condition because of induction negative charges on the sphere B move to the side closer to sphere A,(induction charging) while the positive charges move to the side further away from sphere A. The polarization of charge on B will cause a greater attractive force than the repulsion of the like charges.

Hence the correct answer will be D .

8 0
3 years ago
A gold bar has a density of 19.3 g/mL. If the gold bar has a mass of 6.3 grams, what is the volume?
Natali [406]

Answer:

The correct answer is "0.32 mL".

Explanation:

The given values are:

Density of gold bar,

d = 19.3 g/mL

Mass of gold bar,

m = 6.3 grams

Now,

The volume will be:

⇒  Density = \frac{Mass}{Volume}

or,

⇒  Volume=\frac{Mass}{Density}

On substituting the values, we get

⇒               =\frac{6.3 \ g}{19.3 \ g/mL}

⇒               =0.32 \ mL

7 0
2 years ago
The earth has a vertical electric field at the surface, pointing down, that averages 119 N/C . This field is maintained by vario
velikii [3]

Answer:

q=5.37*10^{5}C

Explanation:

If we assume that the Earth is a spherical conductor, according to  Gauss's Law, the electric field is given by:

E=\frac{kq}{r^2}

Here k is the Coulomb constant, the excess charge on the Earth's surface and r its radius. Solving for q:

q=\frac{Er^2}{k}\\q=\frac{119\frac{N}{C}(6.371*10^6m)^2}{8.99\frac{N\cdot m^2}{C^2}}\\q=5.37*10^{5}C

5 0
3 years ago
0.75 km expressed in centimeters
disa [49]
75000 lol enjoy..............using up 20 characters 
7 0
2 years ago
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