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A force $\vec{F} = 6t^2\,\hat{i} + 4t\,\hat{j}$ acts on a particle of mass $3\text{ kg}$, initially at rest. Find its velocity at $t = 3\text{ s}$.
Acceleration: $\vec{a} = \dfrac{\vec{F}}{m} = 2t^2\,\hat{i} + \dfrac{4t}{3}\,\hat{j}$
Velocity: $\vec{v} = \displaystyle\int_0^3 \vec{a}\,dt = \left[\dfrac{2t^3}{3}\,\hat{i} + \dfrac{2t^2}{3}\,\hat{j}\right]_0^3 = \dfrac{2(27)}{3}\,\hat{i} + \dfrac{2(9)}{3}\,\hat{j} = \mathbf{18\hat{i} + 6\hat{j}}$
A satellite orbits Earth at constant speed $v$. A mass $m$ is ejected from it so that it just barely escapes Earth's gravitational pull. What additional kinetic energy must be given to mass $m$?
The satellite is in circular orbit at some radius $r$: $v = \sqrt{GM/r}$, so $v^2 = GM/r$.
Escape speed from radius $r$: $v_e = \sqrt{2GM/r} = v\sqrt{2}$
Extra KE needed $= \frac{1}{2}mv_e^2 - \frac{1}{2}mv^2 = \frac{1}{2}m(2v^2) - \frac{1}{2}mv^2 = \frac{1}{2}mv^2$
In the figure above, each of the four squares has sides of length $x$. If $\triangle PQR$ is formed by joining the centers of three of the squares, what is the perimeter of $\triangle PQR$ in terms of $x$?
Each square has side length $x$, so the center of each square is at distance $\frac{x}{2}$ from each edge.
From the arrangement:
- $P$ is the center of the top-left square at $\left(\frac{x}{2}, \frac{x}{2}\right)$
- $Q$ is the center of the top-right square at $\left(\frac{3x}{2}, \frac{x}{2}\right)$
- $R$ is the center of the bottom-right square at $\left(\frac{3x}{2}, \frac{3x}{2}\right)$
Distances:
- $PQ = x$ (horizontal distance between adjacent centers)
- $QR = x$ (vertical distance between adjacent centers)
- $PR = \sqrt{x^2 + x^2} = x\sqrt{2}$ (diagonal)
Perimeter $= PQ + QR + PR = x + x + x\sqrt{2} = 2x + x\sqrt{2}$
A sales invoice to Jane was not recorded in any book of prime entry. Which accounts are affected?
| Option | Sales Ledger Control Account | Jane's Personal Account |
|---|---|---|
| A | No | No |
| B | No | Yes |
| C | Yes | No |
| D | Yes | Yes |
Option A is correct.
Since the invoice was never entered in any book of prime entry, no postings were made anywhere — neither the control account nor Jane's personal account is affected. This is a complete error of omission.
- Initial wavelength: $\lambda_{1} = \frac{h}{mv}$
- New mass $m' = 2m$, New velocity $v' = 3v$
- New wavelength: $\lambda_{2} = \frac{h}{(2m)(3v)} = \frac{h}{6mv}$
- Therefore, $\lambda_{2} = \frac{\lambda_{1}}{6}$
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