The universe is made of a lot of galaxies. The likely explanation for a galaxy having more than one nucleus is that the galaxy must have swallowed several smaller galaxies that were its neighbors.
- There are some reason behind the theory that the eating up of a galaxy is by another galaxy. The galaxy is said to have a powerful past due to the fact that a lot of smaller galaxies were eaten up.
The universe is known to have a very largest galaxy called the giant elliptical galaxy that is made up of about a trillion stars.
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- According to Newton's Third Law of Motion, to every action, there is an equal and opposite reaction; action and reaction act on different bodies.
- Here, the action force is in the leftward direction, so the reaction will be in the opposite direction.
- If the action force is the swimmer pushing water in the leftward direction, then the reaction force is in the rightward direction.
- And the reaction force will be given by the water on the swimmer.
<u>Answer</u><u>:</u>
<u>The </u><u>reaction </u><u>force </u><u>is </u><u>the </u><u>water </u><u>pushing </u><u>the </u><u>swimmer </u><u>in </u><u>the </u><u>rightward </u><u>direction</u><u>.</u>
Hope you could get an idea from here.
Doubt clarification - use comment section.
voltage across 2.0μf capacitor is 5.32v
Given:
C1=2.0μf
C2=4.0μf
since two capacitors are in series there equivalent capacitance will be
[tex] \frac{1}{c} = \frac{1}{c1} + \frac{1}{c2} [/tex]


=1.33μf
As the capacitance of a capacitor is equal to the ratio of the stored charge to the potential difference across its plates, giving: C = Q/V, thus V = Q/C as Q is constant across all series connected capacitors, therefore the individual voltage drops across each capacitor is determined by its its capacitance value.
Q=CV
given,V=8v


charge on 2.0μf capacitor is


=5.32v
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Answer:
X₃₁ = 0.58 m and X₃₂ = -1.38 m
Explanation:
For this exercise we use Newton's second law where the force is the Coulomb force
F₁₃ - F₂₃ = 0
F₁₃ = F₂₃
Since all charges are of the same sign, forces are repulsive
F₁₃ = k q₁ q₃ / r₁₃²
F₂₃ = k q₂ q₃ / r₂₃²
Let's find the distances
r₁₃ = x₃- 0
r₂₃ = 2 –x₃
We substitute
k q q / x₃² = k 4q q / (2-x₃)²
q² (2 - x₃)² = 4 q² x₃²
4- 4x₃ + x₃² = 4 x₃²
5x₃² + 4 x₃ - 4 = 0
We solve the quadratic equation
x₃ = [-4 ±√(16 - 4 5 (-4)) ] / 2 5
x₃ = [-4 ± 9.80] 10
X₃₁ = 0.58 m
X₃₂ = -1.38 m
For this two distance it is given that the two forces are equal