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3 years ago

When must a psychological researcher debrief human test subjects?

1 answer:

3 0

Psychological researchers must debrief human test subjects at the end of every experiment.

The current code of ethics in psychological 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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Electrical outlets in a house maintain a steady voltage, even when the amount of resistance on them changes. How is this possibl

makvit [3.9K]

The situation presented above is possible because the outlets could be operating in a parallel circuit. Electrical outlets in a house maintain a steady voltage, even when the amount of resistance on them changes because it operates with a parallel circuit wherein voltage is constant even if resistance changes.

8 0

3 years ago

An earthquake emits both S-waves and P-waves which travel at different speeds through the Earth. A P-wave travels at 9 000 m/s a

Cerrena [4.2K]

Assuming constant speeds, the P-wave covers a distance d in time t such that

9000 m/s = d/(60 t)

while the S-wave covers the same distance after 1 more minute so that

5000 m/s = d/(60(t + 1))

Now,

d = 540,000 t

d = 300,000(t + 1) = 300,000 t + 300,000

Solve for t in the first equation and substitute it into the second equation, then solve for d :

t = d/540,000

d = 300,000/540,000 d + 300,000

4/9 d = 300,000

d = 675,000

So the earthquake center is 675,000 m away from the seismic station.

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3 years ago

1. Kinetic energy is the energy of _____.

lorasvet [3.4K]

Explanation:

kinetic energy is energy that it possesses due to its motion.

7 0

4 years ago

The latent heat of fusion of water at 0 °C is 6.025 kJ mol'' and the molar heat

Scilla [17]

Answer:

$\Delta H_{tot} = 2258.025\,kJ$

Explanation:

The amount of heat released from water is equal to the sum of latent and sensible heats. Let suppose that water is initially at a temperature of $25^{\circ}C$ . Then:

$\Delta H_{tot} = \Delta H_{s, w} + \Delta H_{f,w} + \Delta H_{s,i}$

$\Delta H_{tot} = n\cdot (c_{w}\cdot \Delta T_{w} + L_{f} + c_{i}\cdot \Delta T_{i})$

Finally, the amount of heat released from water is now computed by replacing variables:

$\Delta H_{tot} = (1\,mol)\cdot \left[\left(75.3\,\frac{kJ}{mol\cdot K} \right)\cdot (25^{\circ}C-0^{\circ}C)+ 6.025\,\frac{kJ}{mol} + \left(37.7\,\frac{kJ}{mol\cdot K} \right)\cdot (0 + 10^{\circ}C)\right]$ $\Delta H_{tot} = 2258.025\,kJ$

4 0

3 years ago

Two stretched copper wires both experience the same stress. The first wire has a radius of 3.9×10-3 m and is subject to a stretc

maw [93]

The stretching force acting on the second wire, given the data is 588 N

<h3>Data obtained from the question</h3>

Radius of fist wire (r₁) = 3.9×10⁻³ m
Force of first wire (F₁) = 450 N
Radius of second wire (r₂) = 5.1×10⁻³ m
Force of second wire (F₂) =?

<h3>How to determine the force of the second wire</h3>

F₁ / r₁ = F₂ / r₂

450 / 3.9×10⁻³ = F₂ / 5.1×10⁻³

cross multiply

3.9×10⁻³ × F₂ = 450 × 5.1×10⁻³

Divide both side by 3.9×10⁻³

F₂ = (450 × 5.1×10⁻³) / 3.9×10⁻³

F₂ = 588 N

Learn more about spring constant:

brainly.com/question/9199238

#SPJ1

7 0

2 years ago