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Amiraneli [1.4K]
1 year ago
13

A rightward force of 12.0 N is applied to a 2.0-kg object to accelerate it across a horizontal

Physics
1 answer:
ad-work [718]1 year ago
8 0

Answer:

below

Explanation:

Net accelerating force becomes  12-8 = 4 N

F = ma

4 = 2 * a

a = 2 m/s^2

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An object experiences an impulse, moves and attains a momentum of 200 kg·m/s. If its mass is 50 kg, what is its velocity?
Nady [450]

Answer:

4 m/s

Explanation:

Momentum is defined as:

p=mv

where

m is the mass of the object

v is its velocity

For the object in this problem, we know:

p = 200 kg m/s is the momentum

m = 50 kg is the mass

Solving for the velocity, we find:

v=\frac{p}{m}=\frac{200}{50}=4 m/s

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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

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3 years ago
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Answer:

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