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Fill in the blank: He has been working ______ this project for two weeks.
The correct preposition is 'on' when referring to working on a task or project. 'Working on a project' is the standard phrase.
A body moves eastward at $30\text{ m/s}$. After 10 s its velocity becomes $40\text{ m/s}$ northward. The magnitude of average acceleration is:
Change in velocity: $\Delta\vec{v} = 40\hat{j} - 30\hat{i}$
$|\Delta v| = \sqrt{30^2 + 40^2} = \sqrt{900+1600} = \sqrt{2500} = 50\text{ m/s}$
Average acceleration $= \dfrac{50}{10} = \mathbf{5\text{ m/s}^2}$
Given \(E_2 = -328\) kJ/mol, so \(E_1 = -328 imes 4 = -1312\) kJ/mol.
\(E_4 = \dfrac{-1312}{16} = -82\) kJ/mol
Key point: as \(n\) increases, energy becomes less negative โ outer orbits are less tightly bound.
A water supply at a desert outpost is sufficient to last 21 days for a group of 15 people. At the same average rate of consumption per person per day, how many days would that same water supply last for a group of only 9 people?
Total water supply $=$ (people) $\times$ (days) $\times$ (rate per person per day).
Total supply $= 15 \times 21 = 315$ person-days.
For 9 people: days $= \dfrac{315}{9} = 35$ days.
Answer: 35.0 days.
The Rule of Nines is used in burn assessment. A patient has burns covering the entire right arm and anterior trunk. What percentage of TBSA (Total Body Surface Area) is burned?
The Rule of Nines for adult burn assessment:
| Body Region | TBSA % |
|---|---|
| Head and neck | 9% |
| Each upper limb (arm) | 9% |
| Anterior trunk | 18% |
| Posterior trunk | 18% |
| Each lower limb (leg) | 18% |
| Perineum/genitalia | 1% |
| Total | 100% |
Let $f:\mathbb{R}\to\mathbb{R}$ be differentiable with $|f(x)-f(y)|\le 2|x-y|^{3/2}$ for all $x,y\in\mathbb{R}$. If $f(0)=1$, evaluate $\displaystyle\int_0^1 f^2(x)\,dx$.
From the condition $|f(x)-f(y)|\le 2|x-y|^{3/2}$, divide both sides by $|x-y|$:
$\left|\frac{f(x)-f(y)}{x-y}\right|\le 2|x-y|^{1/2}$
Taking $y\to x$: $|f'(x)|\le 2\cdot0=0$, so $f'(x)=0$ for all $x$.
Hence $f$ is constant. Since $f(0)=1$, we have $f(x)=1$ for all $x$.
$\int_0^1 f^2(x)\,dx = \int_0^1 1\,dx = \mathbf{1}$
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