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Serjik [45]
2 years ago
7

What fraction of a piece of concrete will be submerged when it floats in mercury? the density of concrete is 2.3×103kg/m3 and de

nsity of mercury is 13.6×103kg/m3?
Physics
1 answer:
chubhunter [2.5K]2 years ago
3 0
<span>17% Any object that is floating will display the volume of fluid that matches the mass of the object that's floating. So if the object is 100% as dense as the fluid, it will just barely sink, if it's 50% as dense as the fluid, 50% will be submerged, etc. So let's see what percent of mercury's density is concrete. 2.3x10^3 / 13.6x10^3 = 0.169117647 = 16.9117647% Rounding to 2 significant figures gives 17%. So 17% of a piece of concrete will be submerged in mercury when it's floating in mercury.</span>
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he first excited state of the helium atom lies at an energy 19.82 eV above the ground state. If this excited state is three-fold
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Answer:

Relative population is  2.94 x 10⁻¹⁰.

Explanation:

Let N₁ and N₂ be the number of atoms at ground and first excited state of helium respectively and E₁ and E₂ be the ground and first excited state energy of helium respectively.

The ratio of population of atoms as a function of energy and temperature is known as Boltzmann Equation. The equation is:

\frac{N_{1} }{N_{2} } =  \frac{g_{1}e^{\frac{-E_{1} }{KT} }  }{g_{2}e^{\frac{-E_{2} }{KT} }}

\frac{N_{1} }{N_{2} } = \frac{g_{1}e^{\frac{-(E_{1}-E_{2})  }{KT} }  }{g_{2}}

Here g₁ and g₂ be the degeneracy at two levels, K is Boltzmann constant and T is equilibrium temperature.

Put 1 for g₁, 3 for g₂, -19.82 ev for (E₁ - E₂) and 8.6x10⁵ ev/K for K and 10000 k for T in the above equation.

\frac{N_{1} }{N_{2} } = \frac{1\times e^{\frac{-(-19.82)}{8.6\times 10^{-5}\times 10000} }  }{3}

\frac{N_{1} }{N_{2} } = 3.4 x 10⁹

\frac{N_{2} }{N_{1} } =  2.94 x 10⁻¹⁰

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