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Lunna [17]
4 years ago
10

0.00432 km = _____ mm

Physics
2 answers:
damaskus [11]4 years ago
4 0
If you're talking about millimeter it's 4320. (i think)
kobusy [5.1K]4 years ago
3 0
4320 i believe is your answer
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2. A force of 50 N is applied to the object for a distance of 2.0 m. Assume that object(the mass of the object is 3kg)
Gwar [14]

Answer:

2. A force of 50 N is applied to the object for a distance of 2.0 m. Assume that object(the mass of the object is 3kg)

was at rest at the beginning, what speed did it achieved because of the work done on it? (Hint:

Calculate the works performed by the force first.)

I figured that is 8.2m/S,I am just not sure can anyone help me i much appreciate it.

4 0
3 years ago
You throw a 5.5 g coin straight down at 4.0 m/s from a 25-m-high bridge.
yuradex [85]

<u><em>Answer</em></u>

A) 1,347.5 Joules

B) 22.49 m/s


<u><em>Explanation</em></u>

<u>Part A</u>

The work done by the gravity is known as potential energy.

It is given by;

P.E = mgh

Where m is mass, g is acceleration due to gravity and h is the vertical height.

P.E = 5.5 × 9.8 ×25

= 1,347.5 Joules.

<u>Part B</u>

Using the Newton's third Law of motion,

V² = U² +2as

Where v is final velocity, u is the initial velocity, and s is the displacement of the stone.

V² = 4² + (2×9.8×25)

= 16 + 490

= 506

V = √506

= 22.49 m/s


6 0
4 years ago
Read 2 more answers
3. A car has a mass of 1000kg. What is its weight?<br><br> A) 9.81x10 ^ 3N<br> B) 1.89x10 ^ 3N
Strike441 [17]

Answer

A

Explanation:

Weight = Mass x Acceleration due to gravity

Weight = 1000 x 9.81

= 9810 or 9.81x10^ 3

6 0
3 years ago
The electrons that produce the picture in a
vredina [299]
Given:
u = 10⁵ m/s, the entrance velocity
v = 2.5 x 10⁶ m/s, the exit velocity
s = 1.6 cm = 0.016 m, distance traveled

Let a = the acceleration.
Then
u² + 2as = v²
(10⁵ m/s)² + 2*(a m/s²)*(0.016 m) = (2.5 x 10⁶ m/s)²
0.032a = 6.25 x 10¹² - 10¹⁰ = 6.24 x 10¹²
a = 1.95 x 10¹⁴ m/s²

Answer: 1.95 x 10¹⁴ m/s²

4 0
3 years ago
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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