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Angelina_Jolie [31]
3 years ago
6

Which source is considered a primary source of scientific information? an original research article in a scientific journal a re

port about a new scientific discovery in the newspaper a book that compiles important new research findings a public lecture at a museum given by the scientist who did the research
Physics
2 answers:
Nitella [24]3 years ago
6 0
The primary source would be the original article published in a scientific journal. All other choices would be based on information from the original article.

Newspapers would only pick up the information from the journal itself, or from the authors. Books follow after the original article, after it has gained momentum among the research community. The public lecture at a museum would be based on work from the journal article.
laila [671]3 years ago
5 0

The primary source would be the original article published in a scientific journal.

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Calculate the separation between the two lowest levels for an O2 molecule in a one-dimensional container of length 5.0 cm. At wh
MariettaO [177]

Answer:

The separation between the two lowest levels =  1.24 * 10^{-39}J

The values of n where the energy of molecule reaches 1/2 kT at 300K = 2.2 * 10^{9}

The separation at this level = 1.8 * 10^{-30}J

Explanation:

Knowing the formula

En = \frac{n^{2} h^{2}  }{8 mL^{2} }

Mass of oxygen molecule

m (O2) = 32 amu * \frac{1.6605 * 10^{-27 kg} }{1 amu}

So the energy diference between the two lowest levels:

E2 - E1 = \frac{3h^{2} }{8mL^{2} }

E2 - E1 =  \frac{3 * (6.626 * 10^{-34} Js)^{2} } {8 * 32 amu * (\frac{1.6605 * 10^{-27 kg} }{1 amu})* (5*10^{-2})^{2}   } = 1.24 * 10^{-39}J

Now we should find n where the energy of molecule reaches 1/2 kT

En = \frac{n^{2} h^{2}  }{8 mL^{2} } = \frac{1}{2}kT

\frac{h^{2}  }{8 mL^{2} } = 4.13 * 10^{-14}J

n^{2} *  (4.13 * 10^{-14}J) = \frac{1}{2} (1.38 * 10^{-23}JK^{-1}) * 300K

n = 2.2 * 10^{9}

by the end is necessary to calculate the separation of the level

En - En-1 = (n^{2} - (n - 1)^{2}) * \frac{h^{2}  }{8 mL^{2} }

              = 1.8 * 10^{-30}J

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3 years ago
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The length of a 100 mm bar of metal increases by 0.3 mm when subjected to a temperature rise of 100°C. The coefficient of linear
Juli2301 [7.4K]

Answer:

α = 3×10^-5 K^-1

Explanation:

let ΔL be the change in length of the bar of metal, ΔT be the change in temperature, L be the original length of the metal bar and let α be the coefficient of linear expansion.

then, the coefficient of linear expansion is given by:

α = ΔL/(ΔT×L)

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   = 3×10^-5 K^-1

Therefore, the coefficient of linear expansion is 3×10^-5 K^-1

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3 years ago
Boyles law<br> squeezing a balloon is one way to burst it. Why? ...?
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According to Boyle's Law,  volume is inversely proportional to pressure. It means if the volume of a gas goes up the pressure goes down and if the volume of the gas goes up the pressure goes down. When the pressure of air inside the inflated balloon is more than the atmospheric pressure outside the balloon. And also when the density inside is greater than the density outside. The molecules inside the balloon move and bang around the inner walls which produces force, which provides the pressure of an enclosed air.
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(a) Calculate the buoyant force on a 2.00-L helium balloon.
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Answer:

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The net vertical force is 0.0057879 N

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