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Nitella [24]
3 years ago
8

7 Calculate the weight of an apple of mass 100 grams

Physics
1 answer:
Leni [432]3 years ago
4 0

Answer:

See the anwers below

Explanation:

To solve this problem we must understand that the weight of anyW = 0.1*10\\W = 1 [N]body is defined as the product of mass by gravitational acceleration. Gravitational acceleration depends on the place in space where a certain body is located.

W = m*g\\

where:

W = weight [N]

m = mass = 100 [g] = 0.1 [kg]

g = gravity acceleration [m/s²] or [N/kg]

<u>Now for the Earth</u>

<u />W =m*g\\W =0.1*10\\W = 1[N]<u />

<u>For the Moon</u>

<u />W = m*g\\W =0.1*1.6\\W = 0.16 [N]<u />

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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

     E₀ = 13,606 eV

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
3 years ago
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