For rotational equilibrium of the door we can say that torque due to weight of the door must be counter balanced by the torque of external force

here weight will act at mid point of door so its distance is half of the total distance where force is applied
here we know that

now we will have


so our applied force is 72.5 N
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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.
Any process in which a mixture of materials separates out partially
Let's ask this question step by step:
Part A)
a x b = (3.0i + 5.0j) x (2.0i + 4.0j) = (12-10) k = 2k
ab = (3.0i + 5.0j). (2.0i + 4.0j) = 6 + 20 = 26
Part (c)
(a + b) b = [(3.0i + 5.0j) + (2.0i + 4.0j)]. (2.0i + 4.0j)
(a + b) b = (5.0i + 9.0j). (2.0i + 4.0j)
(a + b) b = 10 + 36
(a + b) b = 46
Part (d)
comp (ba) = (a.b) / lbl
a.b = (3.0i + 5.0j). (2.0i + 4.0j) = 6 + 20 = 26
lbl = root ((2.0) ^ 2 + (4.0) ^ 2) = root (20)
comp (ba) = 26 / root (20)
answer
2k
26
46
26 / root (20)
Answer:
Let No be initial no of atoms
N = N0 / 2 after 1 half-life
N = N0 / 4 after 2 half-lives
So after 2 half-lives 20 of the 80 atoms remain