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gavmur [86]
2 years ago
7

If the frequency of the incoming light is increased, will the energy of the ejected electrons increase, decrease, or stay the sa

me
Physics
1 answer:
Arlecino [84]2 years ago
3 0

Answer:

increase

Explanation:

According to Einstein's photoelectric equation; the energy of a photon striking a metal surface is related to the kinetic energy of the ejected photoelectron by the formula;

KE= hf - hfo

Where h is the planks constant, f and fo refer to the frequency of incident photon and the threshold frequency respectively.

Hence, we can clearly see from the foregoing that the kinetic energy of the ejected photoelectron is proportional to the frequency of the incident photon.

Hence, if the frequency of the incident photon is increased, the kinetic energy of the ejected photoelectron increases also.

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What is the magnitude of the velocity of a 25 kg mass that is moving with a momentum of 100 kg*m/s?
Gekata [30.6K]

Answer:

v= 4 m/s

Explanation:

Momenutm is, by definition, the product of mass and velocity.

p = mv

Let's replace what we know and solve for whatever's left

100 kg\cdot m/s = 25kg \cdot v \rightarrow v= 4 m/s

7 0
3 years ago
8. List three temperature scales. Then write the boiling and freezing
Tomtit [17]

Explanation:

For most temperature scales, the boiling point of water and the freezing point is used to calibrate it.

Three known temperature scales;

  • Kelvin scale
  • Celcius scale
  • Fahrenheit scale

                               Kelvin scale              Celcius scale          Fahrenheit scale

Freezing point             273K                            0°C                        32°F

Melting point                373K                          100°C                     212°F

5 0
2 years ago
Inside a vacuum tube, an electron is in the presence of a uniform electric field with a magnitude of 320 N/C. (a) What is the ma
nignag [31]

(a) The magnitude of the acceleration of the electron is 5.62 x 10¹³ m/s².

(b) The speed of the electron after the given time is  4.78 x 10⁵ m/s.

<h3>Acceleration of the electron</h3>

The acceleration of the electron is calculated as follows;

F = qE

ma = qE

a = qE/m

a = (1.6 x 10⁻¹⁹ x 320)/(9.11 x 10⁻³¹)

a = 5.62 x 10¹³ m/s²

<h3>Speed of the electron</h3>

v = at

v = 5.62 x 10¹³ m/s² x  8.50 x 10⁻⁹ s

v = 4.78 x 10⁵ m/s

Learn more about speed here: brainly.com/question/4931057

#SPJ1

7 0
2 years ago
Where do you see triangulation used on this structure? Explain how triangulation​
Olegator [25]
I like your profile picture:)
3 0
3 years ago
An 67-kg jogger is heading due east at a speed of 2.3 m/s. A 70-kg jogger is heading 61 ° north of east at a speed of 1.3 m/s. F
ololo11 [35]

Answer

given,

mass of jogger  = 67 kg

speed in east direction = 2.3 m/s

mass of jogger 2 = 70 Kg

speed  = 1.3 m/s  in  61 ° north of east.

jogger one

P_1 = m_1 v_1 \hat{i}

P_1 = 67 \times 2.3\hat{i}

P_1 = 154.1 \hat{i}

P_2 = m_2 v_2 \hat{i} +m_2 v_2 \hat{j}

P_2 = 70\times v cos \theta \hat{i} +70\times v sin \theta \hat{j}

P_2 = 70\times 1.3 cos 61^0 \hat{i} +70\times 1.3 sin 61^0\hat{j}

P_2 = 44.12\hat{i} +79.59\hat{j}

now

P = P₁ + P₂

P = 198.22 \hat{i} +79.59 \hat{j}

magnitude

P = \sqrt{198.22^2 + 79.59^2}

P =213.60 kg.m/s

\theta = tan^{-1}\dfrac{79.59}{198.22}

\theta = 21.87

the angle is \theta = 21.87 north of east

7 0
3 years ago
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