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3 years ago

How is a wavelength measured?

Physics

2 answers:

klasskru [66]3 years ago

6 0

Hey there,

Wavelength is measured from crest to crest so the correct answer will be A.

Hope I helped :)

kenny6666 [7]3 years ago

3 0

I believe the correct answer is A) From crest to crest.
~Silver

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An undiscovered planet, many lightyears from Earth, has one moon in a periodic orbit. This moon takes 23.0 days on average to co

Gelneren [198K]

Answer:

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Explanation:

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7 0

2 years ago

If you help me with the correct answer I will give you the brainlist

gulaghasi [49]

Answer:

1. A force is a push or pull upon an object resulting from the object's interaction with another object.

Explanation:

6 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

Which name is given to the type of friction that an objects falling through the air experience

andrey2020 [161]

It is called Air resistance

3 0

3 years ago

A block of inertia m is placed on an inclined plane that makes an angle θ with the horizontal. The block is given a shove direct

Wittaler [7]

Answer:

A) d = v² / (2g (μ cos θ + syn θ) B) μ = tan θ

Explanation:

Part A

We can work this part with the work and energy theorem, where the work of the friction forces is equal to the energy change of the system.

The work is

W = fr .d

With the force of friction it opposes the movement

W = - fr d

The energy at the lowest point is

Em₀ = K = ½ m v²

The energy at the highest point

$Em_{f}$ = U = m g y

The height (y) can be found by trigonometry

sin θ = y / d

y = d sin θ

W = $Em_{f}$ –Em₀

-fr d = mg d sin θ - ½ m v²

The equation for the force of friction is

fr = μ N

From Newton's second law

N - W cos Te = 0

We replace

-μ (mg cos θ) d - mg d sin θ = - ½ m v²

d g (μ cos θ + sin θ) = ½ v²

d = v² / (2g (μ cos θ + syn θ)

Part B

The block is stopped, what is the Angle tet, let's use Newton's second law

fr - W sin θ = 0 ⇒ fr = W sin θ

N - W cos θ= 0 ⇒ N = w cos θ

fr = μ N

μ (mg cos θ) = mg syn θ

μ = syn θ / cos θ

μ = tan θ

7 0

2 years ago