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ankoles [38]
3 years ago
8

5. Did objects have to touch to interact? What causes this?

Physics
2 answers:
dimulka [17.4K]3 years ago
6 0

Answer:

★ For example, a useful analogy for explaining the Earth's gravity force is that the Earth can pull on objects without touching them just like a magnet can affect other objects without touching them. In Addition, the main notion to convey here is that forces can act at a distance with no perceivable substance in between.

Explanation:

Hope you have a great day :)

djverab [1.8K]3 years ago
5 0

Answer/Explanation

A useful analogy for explaining the Earth's gravity force is that the Earth can pull on objects without touching them just like a magnet can affect other objects without touching them. The main notion to convey here is that forces can act at a distance with no perceivable substance in between.

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The Nucleus of the Atom is in the center of the Atom, not in the outer rings & orbitals.
chubhunter [2.5K]

Answer:

true

Explanation:

this the nucleus is located at the centre and contains protons and neutrons

3 0
3 years ago
An electron and a proton are held on an x axis, with the electron at x = + 1.000 m
mixas84 [53]

Answer:

  r2 = 1 m

therefore the electron that comes with velocity does not reach the origin, it stops when it reaches the position of the electron at x = 1m

Explanation:

For this exercise we must use conservation of energy

the electric potential energy is

          U = k \frac{q_1q_2}{r_{12}}

for the proton at x = -1 m

          U₁ =- k \frac{e^2 }{r+1}

for the electron at x = 1 m

          U₂ = k \frac{e^2 }{r-1}

starting point.

        Em₀ = K + U₁ + U₂

        Em₀ = \frac{1}{2} m v^2 - k \frac{e^2}{r+1} + k \frac{e^2}{r-1}

final point

         Em_f = k e^2 ( -\frac{1}{r_2 +1} + \frac{1}{r_2 -1})

   

energy is conserved

        Em₀ = Em_f

        \frac{1}{2} m v^2 - k \frac{e^2}{r+1} + k \frac{e^2}{r-1} = k e^2 (- \frac{1}{r_2 +1} + \frac{1}{r_2 -1})              

       

        \frac{1}{2} m v^2 - k \frac{e^2}{r+1} + k \frac{e^2}{r-1} = k e²(  \frac{2}{(r_2+1)(r_2-1)} )

we substitute the values

½ 9.1 10⁻³¹ 450 + 9 10⁹ (1.6 10⁻¹⁹)² [ - \frac{1}{20+1} + \frac{1}{20-1} ) = 9 109 (1.6 10-19) ²( \frac{2}{r_2^2 -1} )

          2.0475 10⁻²⁸ + 2.304 10⁻³⁷ (5.0125 10⁻³) = 4.608 10⁻³⁷ ( \frac{1}{r_2^2 -1} )

          2.0475 10⁻²⁸ + 1.1549 10⁻³⁹ = 4.608 10⁻³⁷     \frac{1}{r_2^2 -1}

          \frac{2.0475 \ 10^{-28} }{1.1549 \ 10^{-37} } = \frac{1}{r_2^2 -1}

          r₂² -1 = (4.443 10⁸)⁻¹

           

          r2 = \sqrt{1 + 2.25 10^{-9}}

          r2 = 1 m

therefore the electron that comes with velocity does not reach the origin, it stops when it reaches the position of the electron at x = 1m

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Time Intervals in Ice Ages
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The photosynthetic wave interaction between visible light and a photosensitive part of a plant is very important t how plants use light to grow. 
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Maybe the above ???
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jasenka [17]

Answer:

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Explanation: Thanks for the points luv ^-^.

5 0
3 years ago
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