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levacccp [35]
2 years ago
5

What is the coordination number of each atom in Silicon?

Physics
2 answers:
Marrrta [24]2 years ago
6 0

Answer:

the coordination number of silicon is two.

sp2606 [1]2 years ago
3 0
In most silicon-oxygen compounds, the silicon atom is 4-fold coordinated. Exceptionally, silicon is found in 6-fold coordination, as in the mineral thaumasite, which also may be formed in the degradation, and sometimes subsequent weakening, of certain concretes.
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A car traveling on a flat (unbanked), circular track accelerates uniformly from rest with a tangential acceleration of 1.85 m/s2
Tasya [4]

Answer:

Coefficient of friction = 0.836

Explanation:

If v be the speed after one quarter of the circular path

v² = 2as = 2 x 1.85 x 2πr/4 ; v²/r = 1.85 x 3.14 = 5.8

tangential acceleration = 5.8 m/s²

radial acceleration = v² /r = 5.8

total acceleration = √2 x 5.8

m x√2 x 5.8 = m x g xμ

μ = √2 x 5.8 / 9.8 = 0.836

7 0
3 years ago
At what phase would you expect to find extremely high and low tides?
creativ13 [48]

Answer:

full moon and new moon

4 0
2 years ago
Why is the following situation impossible? The object of mass m = 4.00 kg in Figure P6.10 is attached to a vertical rod by two s
Crank
The situation is impossible mainly because we can't see Figure P6.10 .
It would undoubtedly be the same story on an another planet, until we
see the figure and understand what's going on.
4 0
3 years ago
A sleigh weighing 2000 newtons is pulled my a horse a distance of 1.0 kilometer (or 1000 meters) in 45 minutes. what is the powe
hoa [83]
Work = Force* Distance
2000*1000=2000000

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8 0
2 years ago
The de broglie wavelength of an electron with a velocity of 6.00 × 106 m/s is ________ m. The mass of the electron is 9.11 × 10-
WINSTONCH [101]

Answer: 1.212(10)^{-10} m

Explanation:

The de Broglie wavelength \lambda is given by the following formula:

\lambda=\frac{h}{p} (1)

Where:

h=6.626(10)^{-34}\frac{m^{2}kg}{s} is the Planck constant

p is the momentum of the atom, which is given by:

p=m_{e}v (2)

Where:

m_{e}=9.11(10)^{-28}g=9.11(10)^{-31}kg is the mass of the electron

v=6(10)^{6}m/s is the velocity of the electron

This means equation (2) can be written as:

p=(9.11(10)^{-31}kg)(6(10)^{6}m/s) (3)

Substituting (3) in (1):

\lambda=\frac{6.626(10)^{-34}\frac{m^{2}kg}{s}}{(9.11(10)^{-31}kg)(6(10)^{6}m/s)} (4)

Now, we only have to find \lambda:

\lambda=1.2122(10)^{-10} m>>> This is the de Broglie wavelength of the electron

8 0
3 years ago
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