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vladimir2022 [97]
3 years ago
15

A solid weighs 237.5 g in air and 12.5 g in a liquid in which it is wholly submerged. The density of the liquid is 900 kg/m³.

Physics
1 answer:
djyliett [7]3 years ago
4 0

(i) The density of the solid is 950 kg/m³

(ii) the density of another liquid in which the same solid would float is 200 kg/m³

The given parameters;

mass of the solid in air, Ma = 237.5 g

mass of the solid in liquid, Ms = 12.5 g

density of the liquid, \rho_l = 900 \ kg/m^3

(i) The density of the solid is calculated as follows;

Weight in air - Weight in liquid = upthrust

\frac{\rho_o}{\rho_l}  = \frac{Ma}{M_a - Ms} \\\\\frac{\rho_o}{900}  = \frac{237.5}{237.5 - 12.5}\\\\\frac{\rho_o}{900}  =1.0556\\\\\rho_o = 1.0556 \times 900\\\\\rho_o = 950 \ kg/m^3

(ii) the density of another liquid in which the same solid would float with one-fifth of its volume exposed above the liquid surface.​

This is a low density liquid, because one-fifth of the solid is above the liquid surface while four-fifth is below the liquid surface.

the specific gravity of the liquid = ¹/₅ = 0.2

density of water , \rho_w = 1000 \ kg/m^3

S.G = \frac{\rho_l}{\rho_w} \\\\0.2 = \frac{\rho_l}{1000} \\\\\rho_l = 0.2 \times 1000\\\\\rho_l = 200 \ kg/m^3

Thus, the density of another liquid in which the same solid would float is 200 kg/m³

Learn more here: brainly.com/question/14400760

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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.
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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}.

Total binding energy in each iron-56 nucleus:

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

Again, the mass number 56 indicates that there are 56 nucleons in each iron-56 nucleus. The binding energy per nucleon of iron-56 \mathrm{^{56}Fe} will be:

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

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