What is the simplest requirement for an electromagnet

While flying over the grand canyon the pilot

Veronika [31]

We cannot answer the question if we do not have the answers - sorry!!

6 0

3 years ago

A space shuttle travels around the Earth at a constant speed of 28000 kilometers per hour. If it takes 90 minutes to complete on

riadik2000 [5.3K]

Answer:

Explanation:

14

8 0

2 years ago

2) How could a motorcycle (a vehicle with less mass) make a van (a

jeyben [28]

it depends upon what state they are in like in motion or res

8 0

2 years ago

Here is a graph of speed vs time. If the object is moving to the east, which BEST describes the speed and velocity of the graph?

san4es73 [151]

Answer: A JUST DID IT ON USA TESTPREP

Explanation:

5 0

2 years ago

A light source of wavelength λ illuminates a metal with a work function (a.k.a., binding energy) of BE=2.00 eV and ejects electr

slega [8]

<h2>Answer: 1.011 eV</h2>

Explanation:

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

If we consider the light as 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 <u>kinetic energy. </u>

This is what Einstein proposed:

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

$E=h.f$ (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 $\Phi$ is the <u>minimum amount of energy required to induce the photoemission of electrons from the surface of a metal, and </u><u>its value depends on the metal. </u>

In this case $\Phi=2eV$ and $K_{1}=4eV$

So, for the first light source of wavelength $\lambda_{1}$ , and applying equation (2) we have:

$E_{1}=2eV+4eV$ (3)

$E_{1}=6eV$ (4)

Now, substituting (1) in (4):

$h.f=6eV$ (5)

Where:

$h=4.136(10)^{-15}eV.s$ is the Planck constant

$f$ is the frequency

Now, the <u>frequency has an inverse relation with the wavelength </u>

$\lambda_{1}$ :

$f=\frac{c}{\lambda_{1}}$ (6)

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

Substituting (6) in (5):

$\frac{hc}{\lambda_{1}}=6eV$ (7)

Then finding $\lambda_{1}$ :

$\lambda_{1}=\frac{hc}{6eV }$ (8)

$\lambda_{1}=\frac{(4.136(10)^{-15} eV.s)(3(10)^{8}m/s)}{6eV}$

We obtain the wavelength of the first light suorce $\lambda_{1}$ :

$\lambda_{1}=2.06(10)^{-7}m$ (9)

Now, we are told the second light source $\lambda_{2}$ has the double the wavelength of the first:

$\lambda_{2}=2\lambda_{1}=(2)(2.06(10)^{-7}m)$ (10)

Then: $\lambda_{2}=4.12(10)^{-7}m$ (11)

Knowing this value we can find $E_{2}$ :

$E_{2}=\frac{hc}{\lambda_{2}}$ (12)

$E_{2}=\frac{(4.136(10)^{-15} eV.s)(3(10)^{8}m/s)}{4.12(10)^{-7}m}$ (12)

$E_{2}=3.011eV$ (13)

Knowing the value of $E_{2}$ and $\lambda_{2}$ , and knowing we are working with the same work function, we can finally find the maximum kinetic energy $K_{2}$ for this wavelength:

$E_{2}=\Phi+K_{2}$ (14)

$K_{2}=E_{2}-\Phi$ (15)

$K_{2}=3.011eV-2eV$

$K_{2}=1.011 eV$ This is the maximum kinetic energy for the second light source

7 0

3 years ago