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o-na [289]
3 years ago
11

Light of a single frequency falls on a photo-electric material but no electrons are emitted . Electrons may be emitted if the __

___________

Physics
1 answer:
lesya692 [45]3 years ago
8 0
Correct answer is "frequency of the light is increased".

The light that hits the photo-electric material consists of several photons, each one carrying an energy of
E=hf
where h is the Planck constant while f is the light frequency. Part of this energy is used to extract the photo-electrons from the material, and part it is converted into kinetic energy K of the electron:
hf= \phi + K
where \phi is the work function of the material, which is the amount of energy needed to extract the photo-electron.

If no photo-electrons are emitted, it means that the energy given by the photon is not enough to extract the photo-electron. But we see that the energy of the photon is directly proportional to the frequency of the light f, so in order to increase the energy, we should increase the frequency of the light.
You might be interested in
A spring is hung from the ceiling. When a block is attached to its end, it stretches 2.6 cm before reaching its new equilibrium
Kaylis [27]

Answer:

f = 3.09 Hz

Explanation:

This is a simple harmonic motion exercise where the angular velocity is

         w² = \frac{k}{m}

to find the constant (k) of the spring, we use Hooke's law with the initial data

         F = - kx

where the force is the weight of the body that is hanging

        F = W = m g

we substitute

        m g = - k x

        k = - \frac{m g}{x}

we calculate

        k = - \frac{9.8 m}{- 2.6 \ 10^{-2}}

        k = 3.769 10² m

we substitute in the first equation

       w² = \frac{ 3.769 \ 10^2 \ m }{m}

       w = 19.415 rad / s

angular velocity and frequency are related

       w = 2πf

        f = \frac{w}{2\pi }

        f = 19.415 / 2pi

        f = 3.09 Hz

7 0
3 years ago
If the work done to stretch an ideal spring by 4.0 cm is 6.0 j, what is the spring constant (force constant) of this spring?
Alisiya [41]

Answer:

The spring constant is 3750 N/m  

Explanation:

Use the following two relationships:

(Work) = (Force) x (Displacement)

(Force) = (Spring constant) x (Displacement)

=>

(Spring constant) = (Force) / (Displacement) = (Work) / (Displacement)^2

(Spring constant) = 6.0 kg.(m^2/s^2) / 0.0016 m^2 = 3750 N/m

The spring constant is 3750 N/m  

4 0
3 years ago
Q. A mass of 300g is lifted to a<br> height of 10m<br> 205 by a person. Calculate his work done
sergiy2304 [10]
A=mgh
m=300g=0.3kg
g=9,81 m/s^2
h=10m
A=29.43J
3 0
2 years ago
An electron and a proton are held on an x axis, with the electron at x = + 1.000 m
mixas84 [53]

Answer:

  r2 = 1 m

therefore the electron that comes with velocity does not reach the origin, it stops when it reaches the position of the electron at x = 1m

Explanation:

For this exercise we must use conservation of energy

the electric potential energy is

          U = k \frac{q_1q_2}{r_{12}}

for the proton at x = -1 m

          U₁ =- k \frac{e^2 }{r+1}

for the electron at x = 1 m

          U₂ = k \frac{e^2 }{r-1}

starting point.

        Em₀ = K + U₁ + U₂

        Em₀ = \frac{1}{2} m v^2 - k \frac{e^2}{r+1} + k \frac{e^2}{r-1}

final point

         Em_f = k e^2 ( -\frac{1}{r_2 +1} + \frac{1}{r_2 -1})

   

energy is conserved

        Em₀ = Em_f

        \frac{1}{2} m v^2 - k \frac{e^2}{r+1} + k \frac{e^2}{r-1} = k e^2 (- \frac{1}{r_2 +1} + \frac{1}{r_2 -1})              

       

        \frac{1}{2} m v^2 - k \frac{e^2}{r+1} + k \frac{e^2}{r-1} = k e²(  \frac{2}{(r_2+1)(r_2-1)} )

we substitute the values

½ 9.1 10⁻³¹ 450 + 9 10⁹ (1.6 10⁻¹⁹)² [ - \frac{1}{20+1} + \frac{1}{20-1} ) = 9 109 (1.6 10-19) ²( \frac{2}{r_2^2 -1} )

          2.0475 10⁻²⁸ + 2.304 10⁻³⁷ (5.0125 10⁻³) = 4.608 10⁻³⁷ ( \frac{1}{r_2^2 -1} )

          2.0475 10⁻²⁸ + 1.1549 10⁻³⁹ = 4.608 10⁻³⁷     \frac{1}{r_2^2 -1}

          \frac{2.0475 \ 10^{-28} }{1.1549 \ 10^{-37} } = \frac{1}{r_2^2 -1}

          r₂² -1 = (4.443 10⁸)⁻¹

           

          r2 = \sqrt{1 + 2.25 10^{-9}}

          r2 = 1 m

therefore the electron that comes with velocity does not reach the origin, it stops when it reaches the position of the electron at x = 1m

4 0
3 years ago
a ball of mass 16 kg on the end of a string is spun at a constant speed of 2 m/s in a horizontal circle with a radius of 1m. Wha
miss Akunina [59]

The work done by the centripetal force during om complete revolution is 401.92 J.

<h3>What is centripetal force?</h3>

Centripetal force is a force that acts on a body undergoing a circular motion and is directed towards the center of the circle in which the body is moving.

To Calculate the work done by the centripetal force during one complete revolution, we use the formula below.

Formula:

  • W = (mv²/r)2πr
  • W = 2πmv²................... Equation 1

Where:

  • W = Work done by the centripetal force
  • m = mass of the ball
  • v = velocity of the ball
  • π = pie

From the question,

Given:

  • m = 16 kg
  • v = 2 m/s
  • π = 3.14

Substitute these values into equation 1

  • W = 2×3.14×16×2²
  • W = 401.92 J

Hence, The work done by the centripetal force during om complete revolution is 401.92 J.
Learn more about centripetal force here: brainly.com/question/20905151

5 0
2 years ago
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