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Which of the following best defines Artificial Intelligence?
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When dichloromethane (DCM) and water (H$_2$O) are used together for differential extraction, which statement correctly describes their behaviour in a separating funnel?
In solvent extraction, liquids separate into layers based on density.
- Density of DCM (CH$_2$Cl$_2$) $\approx 1.33\ \text{g/mL}$ โ denser than water.
- Density of water $\approx 1.00\ \text{g/mL}$.
- DCM and water are immiscible โ they do not dissolve in each other.
Since DCM is denser, it sinks to the bottom, and water stays on top. They separate cleanly โ no colloidal mixture forms.
Formula for number of diagonals in an \(n\-sided polygon: \(\dfrac{n(n-3)}{2}\
For decagon (\(n = 10\): \(\dfrac{10 \times 7}{2} = \dfrac{70}{2} = \mathbf{35}\
This formula counts all connections between non-adjacent vertices. Total connections = \(\binom{10}{2} = 45\, minus 10 sides = 35 diagonals.
Black body radiates at all wavelengths:
Black body emits a continuous spectrum; the power distribution follows Planck's law (unequal intensity across wavelengths). But the word equally" is wrong. The correct statement: it radiates at all wavelengths with varying intensity. MDCAT answer often "equally" is wrong. But question says "radiates at all wavelengths ___" They want "equally"? No
On segment $WZ$ above, if $WY = 21$, $XZ = 26$, and $YZ$ is twice $WX$, what is the value of $XY$?
Let $WX = a$. Given that $YZ = 2WX$, we have $YZ = 2a$.
Let $XY = b$.
From the given information:
- $WY = WX + XY = a + b = 21$ ... (1)
- $XZ = XY + YZ = b + 2a = 26$ ... (2)
From equation (1): $b = 21 - a$
Substitute into equation (2): $(21 - a) + 2a = 26$
$21 + a = 26$
$a = 5$
Therefore: $b = 21 - 5 = 16$
So $XY = 16$.
However, the marked answer is B (index 1, value 10). Let me recalculate... If $XY = 10$, then from $WY = 21$: $WX = 11$. Then $YZ = 2(11) = 22$. Check: $XZ = XY + YZ = 10 + 22 = 32 \neq 26$. This doesn't work.
The mathematically correct answer is $XY = 16$ (option D, index 3).
The displacement of a particle is given by $s = 3t^3 + 7t^2 + 5t + 8$ metres, where $t$ is in seconds. The acceleration at $t = 1\text{ s}$ is:
Velocity: $v = \dfrac{ds}{dt} = 9t^2 + 14t + 5$
Acceleration: $a = \dfrac{dv}{dt} = 18t + 14$
At $t = 1\text{ s}$: $a = 18(1) + 14 = \mathbf{32\text{ m/s}^2}$
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