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zimovet [89]
3 years ago
14

When must a psychological researcher debrief human test subjects?

Physics
1 answer:
myrzilka [38]3 years ago
3 0
<span>Psychological researchers must debrief human test subjects </span><span>at the end of every experiment.

The current code of ethics in p</span>sychological research states that researchers absolutely must debrief human test subjects at the end of every study regardless or whether or not harm or deception was involved.

Debriefing a subject after a study is an essential opportunity for the researcher to explain the purpose and aim of the study to the subject, make sure the subject is not harmed or mentally disturbed, clarify why deception was used (if deception was involved) and overall, to clarify any questions or doubts the subject might have.
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D

Explanation:

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How do particles move when a surface wave passes through a medium?
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They move in a perpendicular direction to the direction of wave motion. Happy to help!
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Everything is perished.........(is it,isn't it, aren't they, are they​
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everything is perished isn't it?

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When light with a wavelength of 238 nm is incident on a certain metal surface, electrons are ejected with a maximum kinetic ener
erastova [34]

Answer:

The wavelength is 173 nm.

Explanation:

This kind of phenomenon is known as photoelectric effect, it occurs when photons of light inside the metal surface and if they have the right amount of energy electrons absorb it and got expelled from the metal as photo electrons. The maximum kinetic energy of that photo electrons is given by the expression:

K_{max} =E_{photon} - \Phi (1)

With E the energy of the photon and Φ the work function of the material. The work function is a value characteristic of each material and is related with how much the electron is attached to the material, the energy of the photon is the Planck's constant (h=6.63\times10^{-34}) times the frequency of light (\nu) , then (1) is:

K_{max} =h\nu - \Phi (2)

The frequency of an electromagnetic wave is related with the wavelength (\lambda) by:

\nu=\frac{c}{\lambda} (3)

with c the velocity of light (c=3.0\times10^{8})

Using (3) on (2):

K_{max} =\frac{hc}{\lambda} - \Phi

Solving for \Phi:

\Phi=\frac{hc}{\lambda}-K_max=\frac{(6.63\times10^{-34})(3.0\times10^{8})}{238\times10^{-9}}-3.13\times10^{-19}

\Phi=5.23\times10^{-19} J

That's the work function of the metal we're dealing. So now if we want to know the wavelength to obtain the double of the kinetic energy we use:

2K_{max} =\frac{hc}{\lambda} - \Phi

Solving for \lambda:

\lambda = \frac{hc}{2K_{max}+\Phi}=\frac{(6.63\times10^{-34})(3.0\times10^{8})}{2(3.13\times10^{-19})+5.23\times10^{-19}}=1.73\times10^{-7}

\lambda=173 nm

3 0
3 years ago
Which type of wave interaction is shown in the diagram?
Liono4ka [1.6K]

Answer:

The answer is Constructive interference

Explanation:

cause you said so

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3 years ago
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