<span>The Balmer series or Balmer lines in atomic physics, is the designation of one of a set of six named series describing the spectral line emissions of the hydrogen atom. The Balmer series is calculated using the Balmer formula, an empirical equation discovered by Johann Balmer in 1885. this is all I know sorry</span>
Lighting is the static electricity stored in the clouds that is disposed to the earth.
Answer:
The law of multiple proportions states that when two elements can combine in different ratios to form different compounds, the masses of the element combining with the fixed mass of another element result in whole number ratios. This shows that the law of multiple proportions is followed
Explanation:
Answer:
F_Balance = 46.6 N ,m' = 4,755 kg
Explanation:
In this exercise, when the sphere is placed on the balance, it indicates the weight of the sphere, when another sphere of opposite charge is placed, they are attracted so that the balance reading decreases, resulting in
∑ F = 0
Fe –W + F_Balance = 0
F_Balance = - Fe + W
The electric force is given by Coulomb's law
Fe = k q₁ q₂ / r₂
The weight is
W = mg
Let's replace
F_Balance = mg - k q₁q₂ / r₂
Let's reduce the magnitudes to the SI system
q₁ = + 8 μC = +8 10⁻⁶ C
q₂ = - 3 μC = - 3 10⁻⁶ C
r = 0.3 m = 0.3 m
Let's calculate
F_Balance = 5 9.8 - 8.99 10⁹ 8 10⁻⁶ 3 10⁻⁶ / (0.3)²
F_Balance = 49 - 2,397
F_Balance = 46.6 N
This is the balance reading, if it is calibrated in kg, it must be divided by the value of the gravity acceleration.
Mass reading is
m' = F_Balance / g
m' = 46.6 /9.8
m' = 4,755 kg
Answer:
q₃ = -4.81 nC
Explanation:
We can use the Gauss Law here:
∅ = q/∈₀
where,
∅ = Net Flux = - 216 N.m²/C
q = total charge enclosed inside sphere = ?
∈₀ = permittivity of free space = 8.85 x 10⁻¹² C/N.m²
Therefore,
- 216 N.m²/C = q / 8.85 x 10⁻¹² C²/N.m²
q = (-216 N.m²/C)(8.85 x 10⁻¹² C²/N.m²)
q = - 1.91 nC
So, the total charge will be sum of all three charges:
q = q₁ + q₂ + q₃
- 1.91 nC = 1.74 nC + 1.16 nC + q₃
q₃ = - 1.91 nC - 1.74 nC - 1.16 nC
<u>q₃ = -4.81 nC</u>
