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ZanzabumX [31]
3 years ago
13

What is the speed of a wave that has a frequency of 173 Hz and a wavelength of 2.59 meters? Express your answer to the nearest w

hole number. m/s

Physics
2 answers:
devlian [24]3 years ago
7 0

Answer:

448 m/s is the correct answer.

Explanation:

qwelly [4]3 years ago
6 0

The speed of the wave is about 448 m/s

<h3>Further explanation</h3>

The term of package of electromagnetic wave radiation energy was first introduced by Max Planck. He termed it with photons with the magnitude is :

\large {\boxed {E = h \times f}}

<em>E = Energi of A Photon ( Joule )</em>

<em>h = Planck's Constant ( 6.63 × 10⁻³⁴ Js )</em>

<em>f = Frequency of Eletromagnetic Wave ( Hz )</em>

The photoelectric effect is an effect in which electrons are released from the metal surface when illuminated by electromagnetic waves with large enough of radiation energy.

\large {\boxed {E = \frac{1}{2}mv^2 + \Phi}}

\large {\boxed {E = qV + \Phi}}

<em>E = Energi of A Photon ( Joule )</em>

<em>m = Mass of an Electron ( kg )</em>

<em>v = Electron Release Speed ( m/s )</em>

<em>Ф = Work Function of Metal ( Joule )</em>

<em>q = Charge of an Electron ( Coulomb )</em>

<em>V = Stopping Potential ( Volt )</em>

Let us now tackle the problem !

<u>Given:</u>

λ = 2.59 m

f = 173 Hz

<u>Unknown:</u>

v = ?

<u>Solution:</u>

v = \lambda \times f

v = 2.59 \times 173

v = 448.07 \texttt{ m/s}

v \approx 448 \texttt{ m/s}

\texttt{ }

<h3>Learn more</h3>
  • Photoelectric Effect : brainly.com/question/1408276
  • Statements about the Photoelectric Effect : brainly.com/question/9260704
  • Rutherford model and Photoelecric Effect : brainly.com/question/1458544

<h3>Answer details</h3>

Grade: College

Subject: Physics

Chapter: Quantum Physics

Keywords: Quantum , Physics , Photoelectric , Effect , Threshold , Wavelength , Stopping , Potential , Copper , Surface , Ultraviolet , Light

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a).

ρ= 1.2 \frac{kg}{m^{3} }, A_{t}= 0.7 m^{2}, D_{t}= 0.28

F_{t1} = \frac{1}{2} * D_{t} * A_{t}* p_{t}* v_{t}^{2}

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F_{t1} = \frac{1}{2} * 0.28 * 0.7m^{2} * 1.2\frac{kg}{m^{3} }* 29.44^{2}= 101.92 N

b).

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F_{t1} = \frac{1}{2} * D_{h} * A_{h}* p_{h}* v_{h}^{2}

F_{t1} = \frac{1}{2} * 0.57 * 2.44 m^{2} * 1.2\frac{kg}{m^{3} }* 21.38^{2}= 381.44 N

F_{t1} = \frac{1}{2} * 0.57 * 2.44 m^{2} * 1.2\frac{kg}{m^{3} }* 29.44^{2}= 723.25 N

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