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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 grindstone of radius 4.0 m is initially spinning with an angular speed of 8.0 rad/s. the angular speed is then increased to 1
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The average angular speed of the grindstone is 10 rad/s

\texttt{ }

<h3>Further explanation</h3>

<em>Let's recall </em><em>Angular Speed</em><em> formula as follows:</em>

\boxed{ \omega = \omega_o + \alpha t }

\boxed{ \theta = \omega_o t + \frac{1}{2} \alpha t^2 }

\boxed{ \omega^2 = \omega_o^2 + 2 \alpha \theta }

\boxed{ \theta = \frac{( \omega + \omega_o )}{2} t }

<em>where :</em>

<em>ω = final angular speed ( rad/s )</em>

<em>ω₀ = initial angular speed ( rad/s )</em>

<em>α = angular acceleration ( rad/s² )</em>

<em>t = elapsed time ( s )</em>

<em>θ = angular displacement ( rad )</em>

\texttt{ }

<u>Given:</u>

radius of the grindstone = R = 4.0 m

initial angular speed = ω₀ = 8.0 rad/s

final angular speed = ω = 12 rad/s

elapsed time = t = 4.0 seconds

<u>Asked:</u>

average angular speed = ?

<u>Solution:</u>

<em>Firstly , we will calculate </em><em>angular displacement </em><em>as follows:</em>

\theta = \frac{( \omega + \omega_o )}{2} t

\theta = \frac{ ( 12 + 8.0 ) }{2} \times 4.0

\theta = 10 \times 4.0

\boxed {\theta = 40 \texttt{ rad}}

\texttt{ }

<em>Next , we could calculate the </em><em>average angular speed</em><em> as follows:</em>

\texttt{average angular speed} = \theta \div t

\texttt{average angular speed} = 40 \div 4.0

\boxed{\texttt{average angular speed} = 10 \texttt{ rad/s}}

\texttt{ }

<h3>Learn more</h3>
  • Impacts of Gravity : brainly.com/question/5330244
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  • The Acceleration Due To Gravity : brainly.com/question/4189441
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  • The Ratio of the Moments of Inertia : brainly.com/question/2176655

\texttt{ }

<h3>Answer details</h3>

Grade: High School

Subject: Physics

Chapter: Rotational Dynamics

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A projectile is launched into the air with an initial speed of 40 m/s and a launch angle of 20° above the horizontal. The projec
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V=40m/s

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Projectile travels during 5 seconds and the ramge becomes:

x=V.t=30.64 . 5=153.2m

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A huge tank of glycerine with a density of 1.260 g/cm3 is vertically stationed on a platform which is 15 m above the ground. The
EleoNora [17]

Answer:

The tank is losing 4.976*10^{-4}  m^3/s

v_g = 19.81 \ m/s

Explanation:

According to the Bernoulli’s equation:

P_1 + 1 \frac{1}{2} \rho v_1^2 + \rho gh_1 = P_2 +  \frac{1}{2}  \rho v_2^2 + \rho gh_2

We are being informed that both the tank and the hole is being exposed to air :

∴ P₁ = P₂

Also as the tank is voluminous ; we take the initial volume  v_1 ≅ 0 ;

then v_2 can be determined as:\sqrt{[2g (h_1- h_2)]

h₁ = 5 + 15 = 20 m;

h₂ = 15 m

v_2 = \sqrt{[2*9.81*(20 - 15)]

v_2 = \sqrt{[2*9.81*(5)]

v_2= 9.9 \ m/s  as it leaves the hole at the base.

radius r = d/2  = 4/2 = 2.0 mm

(a) From the law of continuity; its equation can be expressed as:

J = A_1v_2

J = πr²v_2    

J =\pi *(2*10^{-3})^{2}*9.9

J =1.244*10^{-4}  m^3/s

b)

How fast is the water from the hole moving just as it reaches the ground?

In order to determine that; we use the relation of the velocity from the equation of motion which says:

v² = u² + 2gh ₂

v² = 9.9² + 2×9.81×15

v² = 392.31

The velocity of how fast the water from the hole is moving just as it reaches the ground is : v_g = \sqrt{392.31}

v_g = 19.81 \ m/s

4 0
3 years ago
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