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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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3 years ago

What is one action a person can take to help ensure ability to participate in physical activity thoughout life

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2 years ago

A stretched spring has 5184 J of elastic potential energy and a spring constant of 16,200 N/m. What is the displacement of the s

yawa3891 [41]

Hello!

A stretched spring has 5184 J of elastic potential energy and a spring constant of 16,200 N/m. What is the displacement of the spring ?

Data:

$E_{pe}\:(elastic\:potential\:energy) = 5184\:J$

$K\:(constant) = 16200\:N/m$

$x\:(displacement) =\:?$

For a spring (or an elastic), the elastic potential energy is calculated by the following expression:

$E_{pe} = \dfrac{k*x^2}{2}$

Where k represents the elastic constant of the spring (or elastic) and x the deformation or displacement suffered by the spring.

Solving:

$E_{pe} = \dfrac{k*x^2}{2}$

$5184 = \dfrac{16200*x^2}{2}$

$5184*2 = 16200*x^2$

$10368 = 16200\:x^2$

$16200\:x^2 = 10368$

$x^{2} = \dfrac{10368}{16200}$

$x^{2} = 0.64$

$x = \sqrt{0.64}$

$\boxed{\boxed{x = 0.8\:m}}\end{array}}\qquad\checkmark$

Answer:

The displacement of the spring = 0.8 m

_______________________________

I Hope this helps, greetings ... Dexteright02! =)

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4 years ago

Given a die, would it be more likely to get a single 6 in six rolls, at least two 6s in twelve rolls, or at least one-hundred 6s

vichka [17]

Answer:

Explanation:

In first case we are interested in one time 6 in six rolls

Thus probability = number of chances required/Total chances

= 1/6

Similarly in the second case probability = 2/12 = 1/6

In the same way in last case probability = 100/600 = 1/6

The probability is the same . Thus all the cases has equal chances

4 0

3 years ago

A pitcher throws a 0.140 kg baseball, and it approaches the bat at a speed of 35.0 m/s. The bat does Wnc = 75.0 J of work on the

Eva8 [605]

Answer:

The speed of the ball is 42.5 m/s

Explanation:

The initial kinetic energy of the ball is:

$K_1=\frac{1}{2} m v_0^2=\frac{1}{2}*0.140 kg*(35.0 m/s)^2$ = 85.75 J

The speed of the ball after leaving the bat is:

$K_2=K_1+W_{nc}\\ \frac{1}{2}mV^2= 85.75 J + 75 J\\ (\frac{1}{2}mV^2)2=( 160.75 J)2\\ mV^2= 321.5 J\\ V^2= \frac{321.5 J}{0.140kg} \\ V=\sqrt{\frac{321.5 J}{0.140kg}}$