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3 years ago

Which statement best describes metallic bonding?

Physics

2 answers:

aksik [14]3 years ago

3 0

THE ANSWER IS D. A NONMETAL ATOM TRANSFERS ELCTRONS TO A METAL ATOM

ITS A TYPE OF ENERGY BONDING THAT ARISES FROM ELECTROSTATIC. ATTRACTIVE FORCE BETWEEN CONDUCTION ELCTRONS (A NONMETAL ATOM) AND A POSITIVELY CHARGED METAL ATOM ( WHO ATTRACTS THE NONMETALS ELECTRONS BECAUSE ITS POSITIVELY CHARGED)

mario62 [17]3 years ago

3 0

Answer:

D.A nonmetal atom transfers electrons to a metal atom.

Explanation:

Metals are chemical elements that have as their main physical characteristic the ability to lose electrons and, consequently, form metal cations. For this reason, they can perform two types of chemical bonds: the ionic bond and the metallic bond.

The metal bond is established between atoms of a single metal element. This type of bonding occurs only between atoms of a single metal and exclusively because a metal cannot establish chemical bond with another different metallic element.

In the metal bond, the crystalline lattices that form the metals are actually an ionic cluster (composed only of cations and electrons). The electrons present in the valence layer of the metal atoms are delocalized, that is, they leave the valence layer, causing the atom to become a cation (electron deficient).

After being delocalized, the electrons from the metal atoms surround the cations, forming a true "sea of electrons". Each of the electrons in this sea has the ability to move through the crystal lattice of the metal freely.

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A particle's position is given by z(t) = −(6.50 m/s2)t2k for t ≥ 0. (Express your answer in vector form.) a. Find the particle's

blondinia [14]

Answer:

a) $z'(t) =v(t) = -13t$

Now we can replace the velocity for t=1.75 s

$v(1.75s) = -13*1.75 =-22.75 \frac{m}{s}$

For t = 3.0 s we have:

$v(3.0s) = -13*3.0 =-39 \frac{m}{s}$

b) $v_{avg}= \frac{z_f - z_i}{t_f -t_i}$

And we can find the positions for the two times required like this:

$z_f = z(3.0s) = -(6.5 \frac{m}{s^2}) (3.0s)^2=-58.5m$

$z_i = z(1.75s) = -(6.5 \frac{m}{s^2}) (1.75s)^2=-19.906m$

And now we can replace and we got:

$V_{avg}= \frac{-58.5 -(-19.906) m}{3-1.75 s}= -30.875 \frac{m}{s}$

Explanation:

The particle position is given by:

$z(t) = -(6.5 \frac{m}{s^2}) t^2, t\geq 0$

Part a

In order to find the velocity we need to take the first derivate for the position function like this:

$z'(t) =v(t) = -13t$

Now we can replace the velocity for t=1.75 s

$v(1.75s) = -13*1.75 =-22.75 \frac{m}{s}$

For t = 3.0 s we have:

$v(3.0s) = -13*3.0 =-39 \frac{m}{s}$

Part b

For this case we can find the average velocity with the following formula:

$v_{avg}= \frac{z_f - z_i}{t_f -t_i}$

And we can find the positions for the two times required like this:

$z_f = z(3.0s) = -(6.5 \frac{m}{s^2}) (3.0s)^2=-58.5m$

$z_i = z(1.75s) = -(6.5 \frac{m}{s^2}) (1.75s)^2=-19.906m$

And now we can replace and we got:

$V_{avg}= \frac{-58.5 -(-19.906) m}{3-1.75 s}= -30.875 \frac{m}{s}$

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3 years ago

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Ipatiy [6.2K]

Answer: The statement that supports Newton's first law of motion is the one that says that planets can move at a varying speed due to forces exerted in space

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Nataly_w [17]

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yuradex [85]

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