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Lera25 [3.4K]
3 years ago
8

A jet plane leaves the stratosphere and returns to the ground. How does the temperature outside the jet change during its descen

t through the troposphere?   A. It decreases and then increases.   B. It decreases steadily.   C. It increases steadily.   D. It increases and then decreases.

Physics
2 answers:
finlep [7]3 years ago
8 0

Answer : A. It decreases and then increases.

Explanation :  Troposphere is the lowermost layer of atmosphere.

Stratosphere is next layer up to the troposphere. As the jet descends from stratosphere towards the troposphere, the temperature initially decreases and then at troposphere is roughly constant and then steadily increases.

So, option (A) is correct.

inn [45]3 years ago
3 0

Answer:

The answer is not A It decreases and then increases. The correct answer is C It increases steadily.

Explanation: I just got the answer wrong by going with It decreases and then increases.

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Which of the following describes the principle of conservation of charge?
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Answer: The statement "The charge cannot be created or destroyed describes the principle of the conservation of charge".

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According to the conservation of charge, the charge can neither be created nor destroyed. It can be transferred from one system to another.

In an isolated system, the total electric charge remains constant. The net quantity of electric  charge is always conserved in the universe.

Therefore, "the charge cannot be created or destroyed" describes the principle of the conservation of charge.

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Given the isotope 2Fes, which has an actual mass of 55.934939 u: a) b) Determine the mass defect of the nucleus in atomic mass u
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Answer:

Mass defect of each iron-56 nuclei:

The binding energy per nucleon of Iron-56 is approximately 8.6 MeV.

Explanation:

According to the physics constants table on Chemistry Libretexts:

  • Proton rest mass: \rm 1.0072765\;amu;
  • Neutron rest mass: \rm 1.0086649\; amu.
  • Speed of light in vacuum: \rm 2.99792458\times 10^{8}\;m\cdot s^{-1}.
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<h3>a)</h3>

The mass defect of a nucleus is equal to the sum of the mass of its parts (protons and, in most cases, neutrons) minus the mass of the nucleus.

The atomic number of iron is 26. There are 26 protons in each iron-56 nucleus. The mass number 56 indicates that there are 56 nucleons (neutrons and protons) in each iron-56 nucleus. The other 56 - 26 = 30 particles are neutrons.

The mass of protons and neutrons in each iron-56 nucleus will be:

\rm 26 \times 1.0072765 + 30 \times 1.0086649 = 56.464736\;amu.

According to this question, the mass of an iron-56 nucleus is equal to 55.934939 amu. The mass defect will be

\rm 56.464736 - 55.934939 = 0.514197\;amu.

<h3>b)</h3>

By the mass-energy equivalence,

E = m\cdot c^{2}.

Refer to this equation, the speed of light in vacuum c^{2} is the conversion factor between mass m and energy E. The value of c is usually given only in SI units \rm m\cdot s^{-1}. Accordingly, the value of c^{2} will be in the SI unit \rm m^{2}\cdot s^{-2} = J\cdot kg^{-1}.

Convert million electron-volts to joules.

One electron-volt is equal to the electrical work done moving an electron across a potential difference of one volt.  

\begin{aligned}\rm 1 MeV&= \rm 10^{6}\; eV\\ &= \rm (10^{6}\times 1.6021765\times 10^{-19}\;C)\times 1\; V\\&=\rm 1.6021765\times 10^{-19}\;J\end{aligned}.

Convert the unit of c^{2} from \rm m^{2}\cdot s^{-2} = J\cdot kg^{-1} to the desired \rm MeV \cdot amu^{-1}:

\begin{aligned}c^{2} &= \rm {\left(2.99792458\times 10^{8}\;m\cdot s^{-1}\right)}^{2}\\&=\rm 8.987551787\times 10^{16}\; m^{2}\cdot s^{-2}\\ &= \rm 8.987551787\times 10^{16}\; J\cdot kg^{-1}\\&= \rm 8.987551787\times 10^{16}\; J\cdot kg^{-1}\times \frac{1\;MeV}{1.6021765\times 10^{-13}\;J}\times \frac{1\times 10^{-3}\;kg}{6.022142\times 10^{23}\;amu}\\&\approx \rm 931.602164\;MeV\cdot amu^{-1}\end{aligned}.

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\begin{aligned}E &= m\cdot c^{2}\\&= \rm 0.514197\;amu \times 9.31602164\;MeV\cdot amu^{-1} \\&=\rm 479.027038\; MeV \end{aligned}.

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\displaystyle \rm \frac{479.027038\; MeV}{56} \approx 8.6\; MeV.

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