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Setler [38]
3 years ago
7

What do we call the minimum energy that is required by an electron to leave the metal target in the photoelectric effect?Select

one of the options below as your answer: A. energy function B. minimum function C. work function D. quanta E. electrical energy
Physics
2 answers:
Liula [17]3 years ago
7 0

Answer:

C. work function

Explanation:

In the photoelectric effect, the energy of the incident photon is used in part to extract the electron from the metal (and this energy is called work function) and the rest is converted into kinetic energy of the electron. In formula:

hf = \phi +K

where

hf is the energy of the incident photon, which is the product between h (the Planck constant) and f (the photon's frequency)

\phi is the work function

K is the kinetic energy of the photoelectron as it leaves the material

Sergio [31]3 years ago
6 0
The work function is what we call the minimum energy that is required by an electron to leave the metal target in the photoelectric effect. 
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Answer:

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8 0
3 years ago
An erect object is placed on the central axis of a thin lens, further from the lens than the magnitude of its focal length. The
Zolol [24]

Answer:

the image is virtual and erect and the lens divergent; therefore the correct answer is C

Explanation:

In a thin lens the magnification given by

      m = h '/ h = - q / p

where h ’is the height of the image, h is the height of the object, q is the distance to the image and p is the distance to the object.

It indicates that the object is straight and is placed at a distance p> f

analyze the situation tells us that the magnification is positive so the distance to the image must be negative, that is, that the image is on the same side as the object.

Consequently the lens must be divergent

The magnification value is

          0.4 = h ’/ h

          h ’= 0.4 h

therefore the erect images

therefore the image is virtual and erect and the lens divergent; therefore the correct answer is C

4 0
3 years ago
A spaceship maneuvering near planet zeta is located at r⃗ =(600i^−400j^+200k^)×103km, relative to the planet, and traveling at v
slava [35]

<u>Answer:</u>

 The spaceship's position when the engine shuts off = (708.15 i - 444.1 j + 200 k)*10^3km

<u>Explanation:</u>

  Initial location of spaceship = (600 i - 400 j + 200 k)*10^3km= (600 i - 400 j + 200 k)*10^6m

  Initial velocity = 9500 i m/s

  Acceleration = (40 i - 20 k)10^3m/s^2

  Time = 35 minute = 35 * 60 = 2100 seconds

 We have equation of motion , s= ut+\frac{1}{2} at^2, s is the displacement, u is the initial velocity, a is the acceleration and t is the time.

  Substituting

       s= (9500 i)*2100+\frac{1}{2}*(40 i - 20 k)*2100^2\\ \\ s=9500*2100 i+20*2100^2i-10*2100^2j\\ \\ s=19.95*10^6i+88.2*10^6i-44.1*10^6j\\ \\ \\s=(108.15i-44.1j)*10^6m

     So final position = ((600 i - 400 j + 200 k)+(108.15i-44.1j))*10^6=(708.15 i - 444.1 j + 200 k)*10^6m

                              =(708.15 i - 444.1 j + 200 k)*10^3km

    The spaceship's position when the engine shuts off = (708.15 i - 444.1 j + 200 k)*10^3km

3 0
3 years ago
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Answer:

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Explanation:

Hope it helped

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3 years ago
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