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Aneli [31]
4 years ago
4

A planet has two moons with identical mass. Moon 1 is in a circular orbit of radius r. Moon 2 is in a circular orbit of radius 2

r. The magnitude of the gravitational force exerted by the planet on Moon 2 is which of the following compared with the gravitational force exerted by the planet on Moon 1?
half as large
twice as large
one-fourth as large
four times as large
the same
Physics
1 answer:
PIT_PIT [208]4 years ago
5 0

Answer: Half as large

Explanation:

Newton's law of universal gravitation for Moon 1 is:

F_{1}=G\frac{Mm}{r^{2}} (1)

And for Moon 2:

F_{2}=G\frac{Mm}{(2r)^{2}} (2)

Taking into account the mass of Moon 1 is equal to the mass of Moon 2

Where:

F_{1} is the gravitational force exerted by the planet on Moon 1

F_{2} is the gravitational force exerted by the planet on Moon 2

G=6.674(10)^{-11}\frac{m^{3}}{kgs^{2}} is the Gravitational Constant  

M is the mass of the planet

m is the mass of each Moon

r is the distance between the planet and Moon 1

2r is the distance between the planet and Moon 2

Dividing (2) by (1):

\frac{F_{2}}{F_{1}}=\frac{G\frac{Mm}{(2r)^{2}}}{G\frac{Mm}{(r)^{2}}}

\frac{F_{2}}{F_{1}}=\frac{1}{2}

Isolating F_{2}:

F_{2}=\frac{1}(2) F_{1}

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Doubly ionized lithium Li2+ (Z = 3) and triply ionized beryllium Be3+ (Z = 4) each emit a line spectrum. For a certain series of
EleoNora [17]

Answer:

tex]\lambda_{Be}[/tex] = 22.78 nm

Explanation:

Bohr's model for the hydrogen atom has been used by other atoms with a single electric charge by changing the number of charges by the charge of the new atom (atomic number)

      E_{n}= k e² / 2a₀ (1 /n²)

      ao = h'² / k m e²               h' = h/2πi

For another atom with a single electron in the last layer

      a₀ ’= h’² / k m (Ze)²  

      a₀ ’= a₀ / Z²

Therefore, when replacing in the equation

      E_{n} = - Z²  Eo/n²

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The transition occurs when the electron stops from one level to another

         E_{n} -  E_{m} = Z² E₀ (1 / n² - 1 / m²) = Z² ΔE

Let's relate this expression to the wavelength

       c = λ f

      E = h f

      E = h c /λ

      h c / λ = Z² ΔE

     λ = 1 / Z² (hc / ΔE)

     λ = 1 / Z² λ_hydrogen

Let's apply this last equation to our case

Lithium Z = 3

     E_{n} = - 9 Eo / n²

     

      40.5 10-9 = 1/9 λ_hydrogen

Beryllium Z = 4

      λ = 1/16 λ_hydrogen

Let's write our two equations is and solve

     40.5 10-9 = 1/9 λ_hydrogen

    tex]\lambda_{Be}[/tex] = 1/ 16 λ_hydrogen

      40.5 10⁻⁹ = 1/9 (16 \lambda_{Be} )

    tex]\lambda_{Be}[/tex] = 40.5 9/16

  tex]\lambda_{Be}[/tex] = 22.78 nm

6 0
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Explanation: Hopefully this helps u.  Have a great rest of your day. I hope this is the right answer

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