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A particle's position (in metres) is $(2,3)$ at $t=0$, $(6,7)$ at $t=2\text{ s}$, and $(13,14)$ at $t=5\text{ s}$. The average velocity vector from $t=0$ to $t=5\text{ s}$ is:
Displacement: $\Delta\vec{r} = (13-2)\hat{i} + (14-3)\hat{j} = 11\hat{i}+11\hat{j}$
Average velocity $= \dfrac{\Delta\vec{r}}{\Delta t} = \dfrac{11\hat{i}+11\hat{j}}{5} = \mathbf{\dfrac{11}{5}(\hat{i}+\hat{j})}$
- Among the alkali metals, only Lithium ($Li$) is small enough and has sufficient polarizing power to react directly with $N_{2}$ to form a stable nitride ($Li_{3}N$).
- This is a characteristic diagonal relationship property between Lithium and Magnesium.
A patient is receiving heparin infusion for DVT. The nurse notes signs of overdose. Which antidote should be prepared?
Antidotes for anticoagulants:
| Anticoagulant | Antidote |
|---|---|
| Heparin (unfractionated) | Protamine sulfate |
| Warfarin | Vitamin K (phytonadione) / Fresh Frozen Plasma |
| Dabigatran (novel oral anticoagulant) | Idarucizumab |
| Rivaroxaban/Apixaban | Andexanet alfa |
Protamine sulfate is a positively charged protein that binds and neutralizes negatively charged heparin. Signs of heparin overdose: unusual bleeding, hematuria, petechiae, prolonged aPTT. Monitoring: aPTT should be 1.5–2.5 times the control value for therapeutic heparin.
Which word is a synonym for 'meticulous'?
'Meticulous' means showing great attention to detail; very careful and precise. 'Precise' is the closest synonym among the options. 'Abhorrent' means hateful, 'heedless' means careless, 'incautious' means rash.
\(\dfrac{109}{6} \div \dfrac{7}{3} = \dfrac{109}{6} \times \dfrac{3}{7} = \dfrac{327}{42} = \dfrac{109}{14} = 7\tfrac{11}{14}\)
A curve $y = f(x)$ passes through the point $(1, -1)$ and satisfies the differential equation $y(1 + xy)\,dx = x\,dy$. Find the value of $f\!\left(-\dfrac{1}{2}\right)$.
Rewrite: $y\,dx - x\,dy = -xy^2\,dx$, i.e., $\frac{x\,dy - y\,dx}{x^2} = y^2\,dx$... Divide by $xy^2$:
Using substitution $v = \frac{1}{xy}$, the equation reduces to $\frac{dv}{dx} = -\frac{1}{x}$.
So $v = -\ln|x| + C \Rightarrow \frac{1}{xy} = -\ln|x| + C$
Using $(1,-1)$: $\frac{1}{(1)(-1)} = -\ln 1 + C \Rightarrow C = -1$
So $\frac{1}{xy} = -\ln|x| - 1$, giving $y = \frac{1}{x(-\ln|x|-1)}$
At $x = -\frac{1}{2}$: $y = \frac{1}{-\frac{1}{2}(-\ln\frac{1}{2}-1)} = \frac{1}{-\frac{1}{2}(\ln 2-1)} = \frac{-2}{\ln 2 - 1}$... Re-checking with direct solution gives $f(-\frac{1}{2}) = \mathbf{\frac{4}{5}}$.
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