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2 years ago

What is the average distance between Miranda and uranus

Physics

2 answers:

r-ruslan [8.4K]2 years ago

7 0

<span>471.6 km i hope this helps you</span>

Lesechka [4]2 years ago

4 0

106234 miles..........

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For a molecule of O2 at room temperature (300 K), calculate the average angular velocity for rotations about the x or y axes. Th

Ksivusya [100]

Answer: 6.65 rad/s

Explanation:

Firstly, we need to know that according the kinetic theory of gases, the average kinetic energy $KE$ of a molecule with $i$ degrees of freedom is:

$KE=\frac{i}{2}kT$ (1)

Where:

$i=5$ because oxigen is a diatomic molecule and has 5 degrees of freedom

$k=1.38064852 (10)^{-23} \frac{m^{2}kg}{s^{2}K}$ is the Boltzman constant

$T=300 K$ is the temperature

Then:

$KE=\frac{5}{2}(1.38064852 (10)^{-23} \frac{m^{2}kg}{s^{2}K})(300 K)$ (2)

$KE=1.035 (10)^{-20} \frac{m^{2}kg}{s^{2}}$ (3) With this value of average kinetic energy we can find the average angular velocity, with the following equation:

$KE=\frac{1}{2}I \omega^{2}$ (4)

Where:

$I$ is the moment of inertia

$\omega$ is the angular velocity

Now, $I=m R^{2}$ (5)

Being $m=31.98 g/mol=0.03198 kg/mol$ the molar mass of Oxigen molecule and $R=0.121(10)^{-9}m$ the distance between atoms

$I=(0.03198 kg/mol)(0.121(10)^{-9}m)^{2}$ (6)

$I=4.682(10)^{-22} \frac{kg}{mol} m^{2}$ (7)

Substituting (7) and (3) in (4):

$1.035 (10)^{-20} \frac{m^{2}kg}{s^{2}}=\frac{1}{2} (4.682(10)^{-22} \frac{kg}{mol} m^{2}) \omega^{2}$ (8)

Finding $\omega$ :

$\omega=6.65 rad/s$ (9) This is the average angular velocity for a molecule of O2

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A roller coaster uses the track in this picture. Where will the roller coaster car have the most potential energy?

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Explanation:

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In a real hemoglobin molecule, the tendency of oxygen to bind to a heme site increases as the other three heme sites become occu

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Since there are four states, then the grand partition function of the system is

$Z = 1+2 e^{(0.55eV+ \alpha )/kT} +e^{(1.3eV+ 2\alpha )/kT}$

where α is the chemical potential

Then, the occupancy of the system is

$bar \ n=(e^{(0.55eV+ \alpha )/kT} +e^{(1.3eV+ 2\alpha )/kT})/Z$

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Answer:

C.) Wave B has about 2.25 times more energy than wave A

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