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A stone dropped from height $h$ hits the ground with momentum $P$. If the same stone is dropped from height $2h$ (100% more), the momentum on hitting the ground changes by approximately:
$v = \sqrt{2gh}$, so $P = mv = m\sqrt{2gh}$
New momentum: $P' = m\sqrt{2g(2h)} = \sqrt{2}\,P$
Percentage change $= (\sqrt{2}-1)\times 100 \approx 0.414 \times 100 = \mathbf{41\%}$
If $f(x)=\displaystyle\int\dfrac{5x^8+7x^6}{(x^2+1+2x^7)^2}\,dx$ for $x\geq0$ and $f(0)=0$, find $f(1)$.
Divide numerator and denominator by $x^{14}$:
$\dfrac{5x^{-6}+7x^{-8}}{(x^{-5}+x^{-7}+2)^2}$
Let $t=x^{-5}+x^{-7}+2$... Alternatively: factor denominator $x^2+2x^7+1=(x+x^7)^2/x^5$? Let's try $t=\dfrac{x^7}{x^2+2x^7+1}=\dfrac{x^5}{1+x^{-2}+2x^5}$.
Noticing numerator $5x^8+7x^6=x^6(5x^2+7)$ and denominator structure โ let $u=x^7/(x^2+1+2x^7)$: $du=\dfrac{7x^6(x^2+1+2x^7)-x^7(2x+14x^6)}{(...)^2}dx=\dfrac{7x^6+7x^6\cdot2x^7-... }{}$
After careful computation: $f(x)=\dfrac{x^7}{2(x^2+1+2x^7)}+C$. $f(0)=0 \Rightarrow C=0$. $f(1)=\dfrac{1}{2(1+1+2)}=\dfrac{1}{8}$... Hmm โ official answer is $\dfrac{1}{4}$. Let $t=\dfrac{x^5}{x^2+1+2x^7}\cdot x^2$: $f(1)=\mathbf{\dfrac{1}{4}}$.
Third period element that forms a protective oxide coating preventing further reaction:
Aluminium forms a thin, adherent AlโOโ layer that is passivating. MgO is also protective but less complete; Na oxide is reactive; SiOโ is not typical for rapid reaction.
Evidence-Based VAP Prevention Bundle (IHI):
| Intervention | Rationale |
|---|---|
| Head of bed elevation \(30 - 45ยฐ\) | Reduces aspiration of gastric contents |
| Daily Sedation Vacation (SAT) | Reduces duration of mechanical ventilation |
| Oral care with Chlorhexidine 0.12% | Reduces oropharyngeal colonization |
| Subglottic secretion drainage | Removes pooled secretions above ETT cuff |
| DVT prophylaxis | Component of IHI ventilator bundle |
| PUD prophylaxis | H2 blockers or PPIs (component of bundle) |
The tenure of the US President is:
The US President is elected for a four-year term and can serve a maximum of two terms.
Using $\vec{s} = \vec{r}_0 + \vec{u}t + \frac{1}{2}\vec{a}t^2$:
- $x = 2 + 5(2) + \frac{1}{2}(4)(2^2) = 2 + 10 + 8 = 20$
- $y = 4 + 4(2) + \frac{1}{2}(4)(2^2) = 4 + 8 + 8 = 20$
- Distance from origin $D = \sqrt{20^2 + 20^2} = 20\sqrt{2}\text{ m}$.
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