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For \(\mu = 2.83\) BM: \(2.83 = \sqrt{n(n+2)}\) โ \(n = 2\)
Checking \(\text{Ni}^{2+}\): Ni (Z=28) = [Ar] 3dโธ 4sยฒ. \(\text{Ni}^{2+}\) = [Ar] 3dโธ โ 2 unpaired electrons โ \(\mu = \sqrt{2 \times 4} = \sqrt{8} \approx 2.83\) BM โ
$\frac{I_{max}}{I_{min}} = \left( \frac{a_1 + a_2}{a_1 - a_2} \right)^2 = 16 \implies \frac{a_1 + a_2}{a_1 - a_2} = 4$.
$a_1 + a_2 = 4a_1 - 4a_2 \implies 5a_2 = 3a_1 \implies \frac{a_1}{a_2} = \frac{5}{3}$.
Intensity ratio $\frac{I_1}{I_2} = \left( \frac{a_1}{a_2} \right)^2 = \frac{25}{9}$.
Of the following, which is closest to $\sqrt[3]{30}$?
Estimate by testing perfect cubes:
- $2^3 = 8$
- $3^3 = 27$
- $4^3 = 64$
Since $27 < 30 < 64$, we know $3 < \sqrt[3]{30} < 4$.
$30$ is much closer to $27$ than to $64$, so $\sqrt[3]{30}$ is closer to $3$ than to $4$.
More precisely: $\sqrt[3]{30} \approx 3.107$
Among the options $\{6, 5, 4, 3\}$, the value $3$ is closest.
A ball is dropped from rest at $t = 0$. After 6 s, another ball is thrown downward from the same point with speed $v$. Both balls meet at $t = 18\text{ s}$. Find $v$ ($g = 10\text{ m/s}^2$).
Ball 1 falls for 18 s: $h_1 = \dfrac{1}{2}(10)(18)^2 = 1620\text{ m}$
Ball 2 falls for $18 - 6 = 12\text{ s}$: $h_2 = 12v + \dfrac{1}{2}(10)(12)^2 = 12v + 720$
Setting $h_1 = h_2$: $12v + 720 = 1620 \Rightarrow 12v = 900 \Rightarrow v = \mathbf{75\text{ m/s}}$
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