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krok68 [10]
3 years ago
14

Explain where most of the mass of an atom is located. Also, explain why some particles that make up the atom do not contribute m

uch to its mass.
Physics
2 answers:
Naddika [18.5K]3 years ago
7 0
Within an atom, there are three elementary particles: the proton, neutron, and electron. Most of the mass of an atom is situated within the nucleus, which is the central part of the atom. It is made up of protons and neutrons, which are the heaviest subatomic particles. The electrons within the atom, orbit around the nucleus at a very far distance. Electrons are also a part of the lightest group of subatomic particles called leptons. That is why these electrons don't contribute much to the majority of an atoms mass. They are very light and they orbit at very far distances.
mariarad [96]3 years ago
4 0
Most mass of the atom is concentrated in the nucleus of the atom. The nucleus is located at the center of the atom which contains the subatomic particles such as neutrons and protons. Neutrons bear no charge while protons are subatomic particles with positive charges. The electron do not make up much of the mass because they just surround the nucleus
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You toss a 0.40-kg ball at 9.0 ms/ to a 14-kg dog standing on an iced-over pond. The dog catches the ball and begins to slide on
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Answer:

a)   v_{f} = 0.25 m / s  b) u = 0.25 m / s

Explanation:

a) To solve this problem let's start with the conservation of the moment, for this we define a system formed by the ball plus the dog, in this case all the forces are internal and the moment is conserved

We will write the data

     m₁ = 0.40 kg

     v₁₀ = 9.0 m / s

     m₂ = 14 kg

     v₂₀ = 0

Initial

     po = m₁ v₁₀

Final

     p_{f} = (m₁ + m₂) vf

     po = pf

     m₁ v₁₀ = (m₁ + m₂) v_{f}

      v_{f} = v₁₀ m₁ / (m₁ + m₂)

      v_{f} = 9.0 (0.40 / (0.40 +14)

      v_{f} = 0.25 m / s

b) This is the reference frame of the center of mass of the system in this case the speed of this frame is the speed of the center of mass

      u = 0.25 m / s

In the direction of movement of the ball

c) Let's calculate the kinetic energy in both moments

Initial

     K₀ = ½ m₁ v₁₀² +0

     K₀ = ½ 0.40 9 2

     K₀ = 16.2 J

Final

     K_{f}= ½ (m₁ + m₂) v_{f}2

      K_{f} = ½ (0.4 +14) 0.25 2

    K_{f} = 0.45 J

   

    ΔK = K₀ -  K_{f}

    ΔK = 16.2-0.445

    ΔK = 1575 J

These will transform internal system energy

d) In order to find the kinetic energy, we must first find the velocities of the individual in this reference system.

      v₁₀’= v₁₀ -u

      v₁₀’= 9 -.025

      v₁₀‘= 8.75 m / s

      v₂₀ ‘= v₂₀ -u

      v₂₀‘= - 0.25 m / s

      v_{f} ‘=   v_{f} - u

      v_{f} = 0

Initial

    K₀ = ½ m₁ v₁₀‘² + ½ m₂ v₂₀‘²

    Ko = ½ 0.4 8.75² + ½ 14.0 0.25²

    Ko = 15.31 + 0.4375

    K o = 15.75 J

Final

   k_{f} = ½ (m₁ + m₂) vf’²

  k_{f} = 0

All initial kinetic energy is transformed into internal energy in this reference system

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