Scott is making a chart to show the difference between chemical reactions and nuclear reactions. Which part of his chart is inco
1 answer:
Answer:
The incorrect part of the chart is;
Line 1
Chemical Reaction
Nuclear Reaction
Line 1 Happens in the nucleus of atoms Happens between atoms
Explanation:
The given chart can be presented as follows;
Chemical Reaction Nuclear Reaction
1. Happens in the nucleus of atoms Happens Between atoms
2. Forms new compounds Forms new atoms
3. Involves small amounts of energy Involves large amounts of energy
From the above chart, line 1 is incorrect because a chemical reaction happens between the valence electrons of the reacting atoms while a nuclear reaction is the reaction that takes place between the nucleus of two or more atoms or the nucleus of an atom and a subatomic particle to produce one or more than new nuclide of a new atom with a change in the mass of the nucleus being represented by the large amount of energy in the reaction. Therefore, a nuclear reaction takes place in the nucleus of an atom, forms new atoms and involves large amounts of energy
The correct arrangement is presented as follows;
Chemical Reaction Nuclear Reaction
1. Happens Between atoms Happens in the nucleus of atoms
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To develop this problem it is necessary to apply the concepts related to the proportion of a neutron star referring to the sun and density as a function of mass and volume.
Mathematically it can be expressed as
Where
m = Mass (Neutron at this case)
V = Volume
The mass of the neutron star is 1.4times to that of the mass of the sun
The volume of a sphere is determined by the equation
Replacing at the equation we have that
Therefore the density of a neutron star is
Answer:
1) E = 0 , 2) zero, 3) E = - 2,162 10⁴ N/C , 4) directed towards the center of the sphere , 5) E = 1.412 10⁴ N / C , 6)direction coming out of the sphere
Explanation:
The electric field is a vector quantity, therefore we can calculate the field due to each charge distribution and add vector
E = E₁ + E₂
To calculate each field we can use Gauss's law, which states that the flow is equal to the charge by the Gaussian surface divided by ε₀
For this case, let's take a sphere as a Gaussian surface
Ф = ∫E dA = /ε₀
The area of a sphere is
A = 4π r²
E = 1 / 4πε₀ q_{int} / r²
1) r = 4 cm
This radius is smaller than the radius of the sphere, therefore the charge inside is zero and therefore the field is zero
E = 0
2) there is no field
3) r = 8 cm
Let's calculate each field, for the inner surface
This radius is larger than the internal radius, so the field is
σ = q_{int} / A
The area of the sphere is
V = 4 π R_in²
Rho = q_{int} / 4π R_in²
q_{int} = ρ 4π R_in²
E₁ = 1 /ε₀ ρ r_in² / r²
For the outer surface
This radius is smaller so there is no load inside the Gaussian surface and therefore the field is zero
E₂ = 0
Total E
E = E₁ + 0
E = 1 /ε₀ ρ₁ R_in² / r²
Let's calculate
E = 1 /8,854 10⁻¹² 250 10⁻⁹ (7/8)²
E = - 2,162 10⁴ N / C
4) as the electric field is negative, it is directed towards the center of the sphere
5) r = 12 cm
In this case the two surfaces contribute to the electric field,
Inner surface
Q₁ = ρ₁ 4π R_in²
E₁ = 1 /ε₀ ₁rho1 R_in² / r²
Outer surface
Q₂ = ρ₂ 4π R_out²
E₂ = 1 /ε₀ ρ₂ R_out² / r²
The total field is
E = E₁ + E₂
E = 1 /ε₀ | ρ | [- R_in² + R_out²] / r²
Let's calculate
E = 1 /8,854 10⁻¹² 250 10⁻⁹ [- 7² + 11²] / 12²
E = 1.412 10⁴ N / C
6) As the field is positive, it is directed radially with direction coming out of the sphere