Answer:
A light year is the distance light travels in a year. ... And an astronomical unit is the average distance between the earth and the sun. So the distance to the sun is by definition one AU. A parsec is the distance at which one astronomical unit subtends an angle of one second of arc.
Answer:
crimping tool
Explanation:
This is a tool employed in affixing a connector to the end of a network cable.
Answer: distance d = 4.73e10m
Explanation: Suppose the charge on the black hole is 5740 C which is a positive charge.
Using electric potential V formula:
V = kq / d
Where K = 9.05×10^9Nm^2/C
And e = 1.6×10^-19C
But you don't need to substitute it.
1090 V = 8.99e9N·m²/C² * 5740C /d
Make d the subject of formula
d = 4.73e10 m
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Answer:
<u><em>The aufbau principle</em></u>
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<u><em>The Pauli exclusion principle</em></u>
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<u><em>Hund's rule of maximum multiplicity</em></u>
Explanation:
<u><em>The aufbau principle:</em></u>
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The fundamental electronic configuration is achieved by placing the electrons one by one in the different orbitals available for the atom, which are arranged in increasing order of energy.
<u><em>The Pauli exclusion principle:</em></u>
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Two electrons of the same atom cannot have their four equal quantum numbers. Because each orbital is defined by the quantum numbers n, l, and m, there are only two possibilities ms = -1/2 and ms = +1/2, which physically reflects that each orbital can contain a maximum of two electrons, having opposite spins
<u><em>Hund's rule of maximum multiplicity:</em></u>
This rule says that when there are several electrons occupying degenerate orbitals, of equal energy, they will do so in different orbitals and with parallel spins, whenever this is possible. Because electrons repel each other, the minimum energy configuration is one that has electrons as far away as possible from each other, and that is why they are distributed separately before two electrons occupy the same orbital.
Answer:
Explanation:
All the displacement will be converted into vector, considering east as x axis and north as y axis.
5.3 km north
D = 5.3 j
8.3 km at 50 degree north of east
D₁= 8.3 cos 50 i + 8.3 sin 50 j.
= 5.33 i + 6.36 j
Let D₂ be the displacement which when added to D₁ gives the required displacement D
D₁ + D₂ = D
5.33 i + 6.36 j + D₂ = 5.3 j
D₂ = 5.3 j - 5.33i - 6.36j
= - 5.33i - 1.06 j
magnitude of D₂
D₂²= 5.33² + 1.06²
D₂ = 5.43 km
Angle θ
Tanθ = 1.06 / 5.33
= 0.1988
θ =11.25 ° south of due west.
