Each orbit is limited to a maximum of four electrons true or false

Which statement best describes the atoms of elements that form compounds by covalent bonding?

k0ka [10]

Answer:

they share electrons between them.

Explanation:

taking the test rn lol i think its right

4 0

3 years ago

The image shows the electric field lines around two charged particles.

DENIUS [597]

It's either 3 or 4 I know this becuase I have read a book about electricity

8 0

3 years ago

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A 62.0-kg athlete leaps straight up into the air from a trampoline with an initial speed of 9.6 m/s. The goal of this problem is

pochemuha

Answer:

2856.96 J

0

$\frac{1}{2}mv_i^2+mgh_i=\frac{1}{2}mv_f^2+mgh_f$

6.78822 m/s

Explanation:

$v_i$ = Initial velocity = 9.6 m/s

g = Acceleration due to gravity = 9.81 m/s²

h = Height

The athlete only interacts with the gravitational potential energy. Air resistance is neglected.

At height y = 0

Kinetic energy

$K=\frac{1}{2}mv^2\\\Rightarrow K=\frac{1}{2}\times 62\times 9.6^2\\\Rightarrow K=2856.96\ J$

At height y = 0 the potential energy is 0 as

$P=mgy\\\Rightarrow P=mg0=0$

At maximum height her velocity becomes 0 so the kinetic energy becomes zero.

As the the potential and kinetic energy are conserved

The general equation

$K_i+P_i=K_f+P_f\\\Rightarrow \frac{1}{2}mv_i^2+mgh_i=\frac{1}{2}mv_f^2+mgh_f$

Half of maximum height

$\\\Rightarrow mgh_i+\frac{1}{2}mv_f^2=mg\frac{h_i}{2}+\frac{1}{2}mv^2\\\Rightarrow gh_i=g\frac{h_i}{2}+\frac{1}{2}v^2\\\Rightarrow g\frac{h_i}{2}=\frac{1}{2}v^2\\\Rightarrow v=\sqrt{gh}$

$h_i=\frac{v_i^2}{2g}$

$v=\sqrt{gh}\\\Rightarrow v=\sqrt{g\times \frac{v_i^2}{2g}}\\\Rightarrow v=\sqrt{\frac{v_i^2}{2}}\\\Rightarrow v=\sqrt{\frac{9.6^2}{2}}\\\Rightarrow v=6.78822\ m/s$

The velocity of the athlete at half the maximum height is 6.78822 m/s

8 0

2 years ago

Lift a notebook a few inches off a table or other surface. As it moves up, it has kinetic energy. Which force opposed the motion

Greeley [361]

If im correct the answer is gravity
and wind resistance if you lift it fast

~~~hope this helps~~~
~~davatar~~

3 0

2 years ago

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Titanium metal requires a photon with a minimum energy of 6.94×10−19J to emit electrons. If titanium is irradiated with light of

butalik [34]

Answer:

a) $1.59(10)^{-19} J$

b) $2.34(10)^{12} electrons$

Explanation:

The photoelectric effect consists of the emission of electrons (electric current) that occurs when light falls on a metal surface under certain conditions.

If the light is a stream of photons and each of them has energy, this energy is able to pull an electron out of the crystalline lattice of the metal and communicate, in addition, a kinetic energy.

<u>This is what Einstein proposed: </u>

Light behaves like a stream of particles called photons with an energy $E$ :

$E=\frac{hc}{\lambda}$ (1)

So, the energy $E$ of the incident photon must be equal to the sum of the Work function $\Phi$ of the metal and the kinetic energy $K$ of the photoelectron:

$E=\Phi+K$ (2)

Where $\Phi=6.94(10)^{-19} J$ is the minimum amount of energy required to induce the photoemission of electrons from the surface of Titanium metal.

Knowing this, let's begin with the answers:

<h3 /><h3>a) Maximum possible kinetic energy of the emitted electrons ( $K$ )</h3>

From (1) we can know the energy of one photon of 233 nm light:

$E=\frac{hc}{\lambda}$

Where:

$h=6.63(10)^{-34}J.s$ is the Planck constant

$\lambda=233 (10)^{-9} m$ is the wavelength

$c=3 (10)^{8} m/s$ is the speed of light

$E=\frac{(6.63(10)^{-34}J.s)(3 (10)^{8} m/s)}{3 (10)^{8} m/s}$ (3)

$E=8.53(10)^{-19} J$ (4) This is the energy of one 233 nm photon

Substituting (4) in (2):

$8.53(10)^{-19} J=6.94(10)^{-19} J+K$ (5)

Finding $K$ :

$K=1.59(10)^{-19} J$ (5) This is the maximum possible kinetic energy of the emitted electrons

<h3>b) Maximum number of electrons that can be freed by a burst of light whose total energy is $2 \mu J=2(10)^{-6} J$ </h3>

Since one photon of 233 nm is able to free at most one electron from the Titanium metal, we can calculate the following relation:

$\frac{E_{burst}}{E}$

Where $E_{burst}=2(10)^{-6} J$ is the energy of the burst of light

Hence:

$\frac{E_{burst}}{E}=\frac{2(10)^{-6} J}{8.53(10)^{-19} J}=2.34(10)^{12} electrons$ This is the maximum number of electrons that can be freed by the burst of light.

4 0

2 years ago