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dimulka [17.4K]
3 years ago
6

Which method is the typical first step when scientists communicate their scientific results? placing the results on an Internet

site to maximize readership distributing the results in press releases to make them well established submitting the results to a scientific journal for a peer review declaring the results in a TV interview for public feedback
Physics
2 answers:
Talja [164]3 years ago
6 0

Answer:

Option C (submitting the results to a scientific journal for a peer review) is the correct choice.

Explanation:

  • If a researcher or a scientist decides to transmit his conclusions depends on a variety of considerations. At either the conclusion of the day, the scientist should concentrate on the final use of an individual's findings before settling on how to transmit it, either through science papers, announcements at professional journals, or personally with the public.
  • Scientists often prefer to submit their articles to professional conferences rather than just going via scientific publications.

The other solutions offered are not relevant to the situation in question. But the latter is the best answer.

tangare [24]3 years ago
3 0

Answer:

it is letter c

Explanation:

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A thundercloud has an electric charge of 48.8 C near the top of the cloud and –41.7 C near the bottom of the cloud. The magnitud
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Answer: 1.51 km

Explanation:

<u>Coulomb's Law:</u> The electrostatic force between two charge particles Q: and Q2 is directly proportional to product of magnitude of charges and inversely proportional to square of separation distance between them.

Or,   \vec{F}=k \frac{Q_{1} Q_{2}}{r^{2}}

Where Q1 and Q2 are magnitude of two charges and r is distance between them:

<u>Given:</u>

Q1 = Charge near top of cloud = 48.8 C

Q2 = Charge near the bottom of cloud = -41.7 C

Force between charge at top and bottom of cloud (i.e. between Q: and Q2) (F) = 7.98 x 10^6N

k = 8.99 x 109Nm^2/C^2

<u>So,</u>

\begin{aligned}&7.98 \times 10^{6}=\left(8.99 \times 10^{9} \mathrm{Nm}^{2} / \mathrm{C}^{2}\right) \frac{48.8 \mathrm{C} \times 41.7 \mathrm{C}}{\mathrm{r}^{2}} \\&r=\sqrt{\frac{1.8294 \times 10^{13}}{7.98 \times 10^{6}}}=1.514  \times 10^{3} \mathrm{~m}=1.51 \mathrm{~km}\end{aligned}

Therefore, the separation between the two charges (r) = 1.51 km

3 0
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A mass of 1.00×10−6 μkg is the same as
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Answer:

<h3>1.01 s</h3>

Explanation:

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u is the initial velocity of the chocolate = 0m/s

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t = √1.019

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<em>Hence it will take 1.01 secs for me to catch the chocolate bar</em>

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