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2 years ago

Which describes a characteristic of metallic bonds?

Physics

1 answer:

melamori03 [73]2 years ago

8 0

Answer:

arge number of electrons free to move between the charged ions in the lattice.

Explanation:

The metallic bond occurs when an atom with few electrons is united in its last level, therefore the best way to decrease the total energy of the system is to lose all its electrons to remain with the configuration of a noble gas. The electrons that it loses cannot be acquired by other atoms since they all have few electrons, thus leaving a large number of electrons free to move between the charged ions in the lattice.

Some important characteristics emerge from this description of the metallic bond:

* It has many free electrons therefore its electrical conductivity is high

* As the charged ions are fixed, the material can be malleable, bent without breaking since the free electrons create the bond that keeps the system stable.

* As the electrons are free when heating a part of the material, these electrons acquire energy and rapidly propagate it to the other side, giving a high thermal conductivity

* As the temperature increases, the electrons acquire more kinetic energy, which is why there are more collisions between them and consequently the resistivity of the material increases.

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How are electromagnetic waves used in the military?

sweet [91]

It uses electromagnetic radiation waves to enable military communications, navigation, radar, nonintrusive inspection of aircraft, and other equipment. Hope this helps.

7 0

2 years ago

Jupiter's semimajor axis is 7.78×1011 m. The mass of the Sun is 1.99×1030 kg. (a) What is the period of Jupiter's orbit in secon

dedylja [7]

Explanation:

It is given that,

Semi major axis of the Jupiter, $a=7.78\times 10^{11}\ m$

Mass of the sun, $M=1.99\times 10^{30}\ kg$

(a) Let T is the period of Jupiter's orbit. It is given by :

$T^2\propto a^3$

$T^2=\dfrac{4\pi^2}{GM}a^3$

$T^2=\dfrac{4\pi^2}{6.67\times 10^{-11}\times 1.99\times 10^{30}}\times (7.78\times 10^{11})^3$

$T=3.74\times 10^8\ s$

(b) We know that,

$1\ year=3.154\times 10^7\ s$

$1\ s=3.171\times 10^{-8}\ year$

$3.74\times 10^8\ s={3.171\times 10^{-8}}\times {3.74\times 10^8}$

T = 11.859 earth years

Hence, this is the required solution.

7 0

3 years ago

Read 2 more answers

A shopper pushes a grocery cart 20.0 m at constant speed on level ground, against a 35.0 n frictional force. he pushes in a dire

alexandr1967 [171]

Work is calculated as the product of Force, Distance, and angular motion. In this case, the work done by gravity is perpendicular to the motion of the cart, so θ = 90°

and W=Fdcosθ

W=35.0 N x 20.0 m x cos90

W=0 J

This means that work done perpendicular to the direction of the motion is always zero.

5 0

2 years ago

A typical helium-neon laser found in supermarket checkout scanners emits 633-nm-wavelength light in a 1.0-mm-diameter beam with

Fofino [41]

Answer:

Eo = 9.796 x 10^2 N/C

Bo = 3.266 x 10^-6 T

Explanation:

Given

Wavelength λ = 633 nm

Diameter of the beam D = 1.0 mm

Power P = 1.0 mW

Solution

Radius of the beam r = D/2 = 0.5 mm = 0.0005 m

Area of cross section

$A = \pi r^{2} \\A = 3.15 \times 0.0005^{2}\\A = 7.58 \times 10^{-7} m^{2}\\$

Intensity

$I = \frac{P}{A} \\I = \frac{0.001}{7.85\times 10^{-7}} \\I = 1273.885 {W}/{m^{2} }$

Amplitude of Electric Field

$E_{o} = \sqrt{\frac{2I}{ \epsilon_{o}c } } \\E_{o} = \sqrt{\frac{2 \times 1273.88}{ 8.85 \times 10^{-12} \times 3 \times 10^{8} } }\\E_{o} = 9.796 \times 10^{2}N/C$

Amplitude of Magnetic Field

$B_{o} = \sqrt{\frac{2 \mu_{o}I}{c } } \\B_{o} = \sqrt{\frac{2 \times 4 \times \pi \times 10^{-7} \times 1273.88}{ 3 \times 10^{8} } }\\B_{o} = 3.266 \times 10^{-6} T$

5 0

3 years ago

BRAINLIEST PLEASE HELP ASAP DO NOT ANSWER UNLESS YOU KNOW AND DON'T GIVE ME A LINK OR I WILL REPORT YOU

Korvikt [17]

Answer:

In a third class lever, the effort is located between the load and the fulcrum. ... If the fulcrum is closer to the effort, then the load will move a greater distance. A pair of tweezers, swinging a baseball bat or using your arm to lift something are examples of third class levers.

Explanation:

3 0

2 years ago