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grigory [225]
3 years ago
14

How is a habitat different from an ecosystem?

Physics
2 answers:
Whitepunk [10]3 years ago
6 0
The answer to ur question is Habitat the natural home or environment of an animal, plant, or other organism and ecocystem is <span>a biological community of interacting organisms and their physical environment.</span>
bonufazy [111]3 years ago
4 0
The habitat can be a place where Simone with keep their food and all that and an ecosystem is where you can leave and not worry about your past there
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Which weather event is capable of destroying homes and uprooting trees due to a low pressure area at its center
Bond [772]

You're talking about a <em>tornado</em>.

It's not so much the low pressure that's so dangerous in the center of a tornado.  It's more a matter of the high winds that are <em>caused </em>by the low pressure.

7 0
4 years ago
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Select each example of a projectile
mafiozo [28]
Baseball, javelin, and maybe the clock but not sure on that... Just say baseball and javelin
4 0
3 years ago
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A body accelerate uniformly from rest at 2m/s square.Calculate its velocity after traveling 9m
Thepotemich [5.8K]
The correct answer for this question is 6m/s. I hope this helps.
5 0
3 years ago
A coin 15.0 mm in diameter is placed 15.0 cm from a spherical mirror. The coin's image is 5.0 mm in diameter and is erect. Is th
s344n2d4d5 [400]

Answer:

15 cm

Explanation:

h_{o} = Diameter of the coin = 15 mm

h_{i} = Diameter of the image of coin =  5 mm

d_{o} = distance of the coin from mirror = 15 cm

d_{i} = distance of the image of coin from mirror = ?

Using the equation

\frac{d_{i}}{d_{o}} = \frac{- h_{i}}{h_{o}}

\frac{d_{i}}{15} = \frac{- (5)}{15}

d_{i} = - 5 cm

R = radius of curvature

Using the mirror equation

\frac{1}{d_{o}} + \frac{1}{d_{i}} = \frac{2}{R}

\frac{1}{15} + \frac{1}{- 5} = \frac{2}{R}

R = - 15 cm

6 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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