Physics
University Physics II
Practice University Physics II with 300 published problems. Preview a problem, explore the course outline, and work through one problem at a time.
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Course outline
6 topicsElectric field and Gauss law
Electric potential and capacitance
DC and transient circuits
Magnetic forces and sources
Faraday induction and inductance
Maxwell equations and electromagnetic waves
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Proof: make a limit precise
For each positive integer \(n\), let \(a_n=n/(n+2)\). Prove from the definition of convergence that \(a_n\to1\).
Show a hint
The definition asks: given any \(\varepsilon>0\), can you make \(|a_n-1|<\varepsilon\) for every sufficiently large \(n\)? First simplify \(|a_n-1|\).
Show worked explanation
Let \(\varepsilon>0\). For every positive integer \(n\), \[ \left|\frac{n}{n+2}-1\right|=\left|\frac{-2}{n+2}\right|=\frac{2}{n+2}. \] Choose a positive integer \(N>2/\varepsilon\). Such an integer exists because the positive integers are unbounded. If \(n\ge N\), all denominators are positive, so \[ \frac{2}{n+2}\le\frac{2}{N+2}<\frac{2}{N}<\varepsilon. \] Thus, for every positive tolerance \(\varepsilon\), this \(N\), chosen for the given \(\varepsilon\), works for every \(n\ge N\). This is exactly the definition of \(a_n\to1\).
Calculation: complete the square
Solve \(x^2-6x+5=0\) over the real numbers by completing the square. Check each solution in the original equation.
Show a hint
Move \(5\) to the other side. Half of \(-6\) is \(-3\), and its square is \(9\). Add \(9\) to both sides to form \((x-3)^2\).
Show worked explanation
The identity \( (x-3)^2=x^2-6x+9 \) explains the completed square. Each step below preserves equality: \[ x^2-6x=-5,\qquad x^2-6x+9=4,\qquad (x-3)^2=4. \] A real number whose square is \(4\) is either \(2\) or \(-2\). Therefore \(x-3=2\) or \(x-3=-2\), giving \(x=5\) or \(x=1\). Check both values: \[ 5^2-6(5)+5=25-30+5=0,\qquad 1^2-6(1)+5=1-6+5=0. \] Both satisfy the original equation. The two square-root cases cover all real possibilities, so the solution set is \(\{1,5\}\).
Physical model: distance and displacement
A cart travels along a straight line: \(18\,\mathrm{m}\) east in \(5\,\mathrm{s}\), then \(12\,\mathrm{m}\) west in \(3\,\mathrm{s}\), with no pause between the two legs. Find its average velocity and average speed for the whole trip.
Show a hint
Take east as positive. Average velocity uses signed displacement divided by total time. Average speed uses total distance traveled divided by total time.
Show worked explanation
With east positive, the displacements of the two legs are \(+18\,\mathrm{m}\) and \(-12\,\mathrm{m}\). The net displacement is \(18-12=6\,\mathrm{m}\) east, while the total distance traveled is \(18+12=30\,\mathrm{m}\). Total time is \(5+3=8\,\mathrm{s}\). \[ \text{Average velocity}=\frac{6\,\mathrm{m}}{8\,\mathrm{s}}=0.75\,\mathrm{m/s}\text{ east}, \] \[ \text{Average speed}=\frac{30\,\mathrm{m}}{8\,\mathrm{s}}=3.75\,\mathrm{m/s}. \] Check: the total distance is at least the magnitude of the displacement, so average speed must be at least the magnitude of average velocity. Here \(3.75\ge0.75\), as expected.