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Gwar [14]
3 years ago
12

What is not an ethical concern that psychological researchers must consider?

Physics
1 answer:
natita [175]3 years ago
4 0

the answer is A privacy + confidentiality of personal info for up to 10 years.

Explanation:

i took the quiz

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PLEASE HELPPPP ME WITH THIS
lozanna [386]
I am not so sure about this it is too difficult
4 0
3 years ago
Based on the law of conversation of energy how can we reasonably improve a machines ability to do work?
tensa zangetsu [6.8K]
MARK ME BRAINLIEST!!

your answer should be “C”.
4 0
3 years ago
RHOOLIOTTO<br> How much mass would be needed to produce 2.7 x 1016 J?
Radda [10]
E = mc^2
m = e/c^2
m = 2.7*10^16/(300000^2)
m = 300000
8 0
3 years ago
Which of these atoms is the most electronegative? select one: a. si b. cl c. p d. f e. c
katrin [286]

The atom that is the most electronegative is fluorine (F).

<h3>What is electronegative?</h3>

Electronegativity, is the tendency for an atom of a given chemical element to attract shared electrons when forming a chemical bond.

Electronegativity increases across the groups from left to right of the periodic table and decreases down the group.

Examples of electronegative elements arranged in decreasing order;

  • fluorine,
  • oxygen,
  • nitrogen,
  • chlorine,
  • bromine,
  • iodine,
  • sulfur,
  • carbon, and
  • hydrogen.

Thus,  the atom that is the most electronegative is fluorine (F).

Learn more about electronegativity here: brainly.com/question/24977425

#SPJ1

4 0
2 years ago
When you irradiate a metal with light of wavelength 433 nm in an investigation of the photoelectric effect, you discover that a
sergij07 [2.7K]

Answer:

The right solution is:

(a) 2.87 eV

(b) 1.4375 eV

Explanation:

Given:

Wavelength,

= 433 nm

Potential difference,

= 1.43 V

Now,

(a)

The energy of photon will be:

E = \frac{6.626\times 10^{-34}\times 3\times 10^8}{433\times 10^{-9}}

  = 4.59\times 10^{-19} \ J

or,

  = \frac{4.59\times 10^{-19}}{1.6\times 10^{-19}}

  = 2.87 \ eV

(b)

As we know,

⇒ Vq=\frac{hc}{\lambda}-\Phi_0

By substituting the values, we get

⇒ 1.43\times 1.6\times 10^{19}=\frac{6.626\times 10^{-34}\times 3\times 10^8}{433\times 10^{-9}}-\Phi_0

⇒                       \Phi_0=2.3\times 10^{-19} \ J

or,

⇒                            =\frac{2.3\times 10^{-19}}{1.6\times 10^{-19}}

⇒                            =1.4375 \ eV

5 0
3 years ago
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