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KiRa [710]
2 years ago
11

Researchers interested in studying stress gave 150 high school seniors a very difficult math exam. After the test, the researche

rs measured stress by examining physiological changes with extensive medical testing that included drawing blood samples. When the test was over, they shared the results with the students, but did not publish individual data. Explain whether or not this study conforms to each of the ethical standards: Informed Consent, Debriefing, Confidentiality, and Protection from harm.
Physics
1 answer:
belka [17]2 years ago
4 0

Answer:

<u>Informed Consent- The study didn't conform this ethical standard.</u>

<u>Debriefing- The study conforms this ethical standard.</u>

<u>Confidentiality- The study conforms this ethical standard.</u>

<u>Protection from harm- The study conforms this ethical standard.</u>

Explanation:

<u>Informed consent</u> is described as a procedure whereby researchers tend to provide details about specific research they are going to conduct, the risk & benefits involved in the study, different alternatives involved in the procedure, etc.

<u>Debriefing </u>is described as a process that is being conducted in any of the different psychological research encompassing human participants after specific research is completed. The researcher tends to describe the details of the research to the participants.

<u>Confidentiality</u> is described as one of the different code of ethics followed by health workers or psychologists. While practicing confidentiality, a psychologist tends to promise his or her participants that he or she will keep everything a secret whatever is being discussed or shared between both of them.

<u>Protection from harm</u> determines that the psychologists conducting research follows the ethics in which he or she has to protect all the participants from any kind of harm.

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Two small aluminum spheres, each of mass 0.0250 kilograms, areseparated by 80.0 centimeters.
sineoko [7]

Answer:

Total number of electrons

N = 7.25 \times 10^{24}

electrons removed from each sphere

N = 5.27 \times 10^{15}

Fraction of electrons transferred is given as

f = 7.27 \times 10^{-10}

Explanation:

As we know that moles is defined as

n =\frac{mass}{molar mass}

n = \frac{0.0250}{0.026982}

n = 0.93

so number of atoms of Al in each sphere is given as

N = 0.93(6.02 \times 10^{23})

N = 5.58 \times 10^{23}

Now number of electrons in each atom is given as

atomic number = number of electrons in each atom = 13

total number of electrons in each sphere is

N = 13 \times (5.58 \times 10^{23})

N = 7.25 \times 10^{24}

Also we know that force of attraction between them is given as

F= \frac{kq_1q_2}{r^2}

1.00 \times 10^4 = \frac{(9\times 10^9)q^2}{0.80^2}

q = 8.4 \times 10^{-4} C

now we have

q = Ne

8.4 \times 10^{-4} = N(1.6 \times 10^{-19}

N = \frac{(8.4 \times 10^{-4})}{1.6 \times 10^{-19}}

N = 5.27 \times 10^{15}

Fraction of electrons transferred is given as

f = \frac{5.27 \times 10^{15}}{7.25 \times 10^{24}}

f = 7.27 \times 10^{-10}

6 0
2 years ago
A helicopter flies 250 km on a straight path in a direction 60° south of east. The east component of the helicopter’s displaceme
GaryK [48]

Given that,

Distance in south-west direction = 250 km

Projected angle to east = 60°

East component = ?

since,

cos ∅ = base/hypotenuse

base= hyp * cos ∅

East component = 250 * cos 60°

East component = 125 km

8 0
2 years ago
Read 2 more answers
A block of mass 3m is placed on a frictionless horizontal surface, and a second block of mass m is placed on top of the first bl
tatuchka [14]

By Newton's second law, assuming <em>F</em> is horizontal,

• the net <u>horizontal</u> force on the <u>larger</u> block is

<em>F</em> - <em>µmg</em> = 3<em>mA</em>

where <em>µmg</em> is the magnitude of friction felt by the larger block due to rubbing with the smaller one, <em>µ</em> is the coefficient of static friction between the two blocks, and <em>A</em> is the block's acceleration;

• the net <u>vertical</u> force on the <u>larger</u> block is

4<em>mg</em> - 3<em>mg</em> - <em>mg</em> = 0

where 4<em>mg</em> is the mag. of the normal force of the surface pushing up on the combined mass of the two blocks, 3<em>mg</em> is the weight of the larger block, and <em>mg</em> is the weight of the smaller block;

• the net <u>horizontal</u> force on the <u>smaller</u> block is

<em>µmg</em> = <em>ma</em>

where <em>µmg</em> is again the friction between the two blocks, but notice that this points in the same direction as <em>F</em>. It is the only force acting on the smaller block in the horizontal direction, so (b) static friction is causing the smaller block to accelerate;

• the net <u>vertical</u> force on the <u>smaller</u> block is

<em>mg</em> - <em>mg</em> = 0

where <em>mg</em> is the magnitude of both the normal force of the larger block pushing up on the smaller one, and the weight of the smaller block.

(You should be able to draw your own FBD's based on the forces mentioned above.)

(c) Solve the equations above for <em>A</em> and <em>a</em> :

<em>A</em> = (<em>F</em> - <em>µmg</em>) / (3<em>m</em>)

<em>a</em> = <em>µg</em>

5 0
1 year ago
Which phrases describe NASA's goals in the coming years? Check all that apply.
Serggg [28]

Answer:

1, 3 and 4

Explanation:

Did the test and saw the answers.

3 0
2 years ago
Read 2 more answers
Two flat conductors are placed with their inner faces separated by 6.0 mm. If the surface charge density on one of the inner fac
dangina [55]

Explanation:

Relation between electric field and charge density is as follows.

           E = \frac{\sigma}{2 \epsilon}

where,    \sigma = charge density

              \epsilon = permittivity of free space = 8.85 \times 10^{-12}

So,  E_{\text{outside}} = 0

      E_{inside} = \frac{+\sigma}{2 \epsilon} - \frac{-\sigma}{2 \epsilon}

or,     E_{inside} = \frac{\sigma}{\epsilon}

Now, formula to calculate the potential difference of two conductors is as follows.

         V_{1} - V_{2} = \frac{\sigma \times d}{\epsilon}

It is given that,

           d = 6.0 mm = 6 \times 10^{-3} m

        \sigma = 40 \times 10^{-12} C/m^{2}

Hence, we will calculate the magnitude of the electric potential differences between the two conductors as follows.

        V_{1} - V_{2} = \frac{\sigma \times d}{\epsilon}

                     = \frac{40 \times 10^{-12} \times 6 \times 10^{-3}}{8.85 \times 10^{-12}}      

                     = 0.0271 volts

thus, we can conclude that value of the magnitude of the electric potential differences between the two conductors is 0.0271 volts.

7 0
2 years ago
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