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marta [7]
4 years ago
14

What is universal constant (G)

Physics
2 answers:
ahrayia [7]4 years ago
8 0

☄ <u>Universal</u><u> </u><u>Gravitational</u><u> </u><u>Constant</u><u>(</u><u>G</u><u>)</u>

Gravitational constant is a constant of proportionality when F(Gravitational force) is proportional to product of masses and inversely proportional to the distance between them.

  • It can be defined as:- Universal gravitational constant is the magnitude of the force between a pair of 1 kg masses that are kept 1 metre apart.
  • The SI unit of G is N m²/kg² and the CGS unit of G is Dyne cm² / g².

<h3>☄ <u>Do</u><u> </u><u>you</u><u> </u><u>know</u><u>?</u></h3>

The value of G (universal gravitational constant) was found out by the scientist Henry Cavendish by using the sensitive torsion balance.

<u>━━━━━━━━━━━━━━━━━━━━</u>

forsale [732]4 years ago
5 0

Answer:

Here is the answer. Hope this helps you!

Explanation:

Now, from Universal law of gravitation, we come to know that:

F= force of attraction

M = the object with greater mass (taken from centre of the objects)

m = object with smaller mass (taken from centre of the objects)

d = distance (taken from centre of the objects)

F ∝ Mm

F ∝ 1/d²

Therefore, we get the formula

F ∝ Mm/d²

F = GMm/d²

Here, G is the constant of proportionality. It also accepted as the Universal gravitation constant. It has a value of 6.67*10^-11 Nm^2/kg^2.

The universal gravitation constant has allowed us to use the above formula to help us calculate and come to know about

  1. the force that binds us to the earth
  2. the motion of moon around the earth
  3. the motion of planets around the sun
  4. the tides on earth caused by the moon and the sun.

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1.41 × 10³⁰ MeV

As we know, E=mc², where E is energy, m is mass and c is the speed of light(i.e. 3×10⁸ m/s).

Given mass = 2.5 kgs

∴ E = (2.5)×(3×10⁸)² J = 22.5×10¹⁶ J

As our answer is in joules so we have to convert it into mega electron volt(MeV)

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Calculate the energy (in eV/atom) for vacancy formation in some metal, M, given that the equilibrium number of vacancies at 296o
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Explanation:

The given data is as follows.

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                                            = 569 K

     Density of the metal = 8.85 g/cm^{3} = 8.85 \times 10^{-6} g/m^{3}      (as 1 cm^{3} = 10^{-6} m^{3})

     Atomic mass = 51.40 g/mol

    Vacancies = 9.19 \times 10^{23} m^{-3}

Formula to calculate the number of atomic sites is as follows.

           n = \frac{\rho \times N_{A}}{\text{atomic weight}}

              = \frac{8.85 \times 10^{-6} \times 6.022 \times 10^{23}}{51.40 g/mol}

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Now, we will calculate the energy as follows.

                E = -KT \times ln (\frac{\text{no. of vacancies}}{\text{no. of atomic sites}})

where,    K = 8.62 \times 10^{-5}

         E = -8.62 \times 10^{-5} \times 569 K \times ln (\frac{9.19 \times 10^{23}}{1.036 \times 10^{17} atom/m^{3}})

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Therefore, we can conclude that energy (in eV/atom) for vacancy formation in given metal, M, is 78.46 eV/atom.

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