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makkiz [27]
3 years ago
11

Which item is an example of a primary source? A. An academic paper analyzing John F. Kennedy's presidency B. A transcript of a p

residential speech C. A textbook about American foreign policy D. A newspaper article reflecting on the anniversary of the end of the Cold War
Physics
1 answer:
Viktor [21]3 years ago
7 0

A transcript of a presidential speech (APEX Class ;)

You might be interested in
(please help asap)
OLEGan [10]

Answer:

A. Yard

Explanation:

Should be correct lol :)

6 0
3 years ago
Read 2 more answers
8) Calculate the kinetic energy of a truck that has a mass of 6100 kg and is moving at 55 m/s.
Inessa [10]

Answer:

<h2>9,226,250 J</h2>

Explanation:

The kinetic energy of an object can be found by using the formula

k =  \frac{1}{2} m {v}^{2}  \\

v is the velocity

m is the mass

From the question we have

k =  \frac{1}{2}  \times 6100 \times  {55}^{2}  \\  = 3050 \times 3025 \\  = 9226250

We have the final answer as

<h3>9,226,250 J</h3>

Hope this helps you

7 0
3 years ago
What is the acceleration of a car that starts from rest and attains a final speed of 20 m
OlgaM077 [116]

Answer:

323.9

Explanation:

3 0
3 years ago
N2O is the chemical formula of a covalent compound used in the production of whipping cream. The elements that make up this comp
vlabodo [156]
 N2O is nitrous oxide becuase of the nitrogen and oxygen

7 0
3 years ago
Read 2 more answers
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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