Study questions platform-wide or filter by specific tests with correct answers revealed.
The maximum horizontal range of a projectile is 16 km ($g = 10\text{ m/s}^2$). The minimum muzzle velocity required is:
Maximum range occurs at $45°$: $R_{\max} = \dfrac{u^2}{g}$
$u^2 = R_{\max}\times g = 16000 \times 10 = 160000$
$u = \sqrt{160000} = \mathbf{400\text{ m/s}}$
Evaluate: $\displaystyle\lim_{x \to 0} \dfrac{(1-\cos 2x)(3+\cos x)}{x\tan 4x}$
Use standard limits: $\lim_{x\to0}\dfrac{1-\cos2x}{x^2}=2$ and $\lim_{x\to0}\dfrac{\tan4x}{x}=4$.
$= \lim_{x\to0}\dfrac{(1-\cos2x)}{x^2}\cdot\dfrac{(3+\cos x)}{\tan4x/x} = \dfrac{2\cdot(3+1)}{4} = \dfrac{8}{4} = \mathbf{2}$
Compare $y = \alpha x - \beta x^2$ with $y = x \tan\theta - \frac{gx^2}{2u^2 \cos^2\theta}$.
- $\tan\theta = \alpha \implies \theta = \tan^{-1}\alpha$.
- Max height occurs at $x = \frac{\alpha}{2\beta}$ (where $dy/dx = 0$).
- $H = \alpha(\frac{\alpha}{2\beta}) - \beta(\frac{\alpha}{2\beta})^2 = \frac{\alpha^2}{2\beta} - \frac{\alpha^2}{4\beta} = \frac{\alpha^2}{4\beta}$.
The time period of a pendulum is $T = 2\pi\sqrt{\frac{L}{g}}$, so $L \propto T^2$.
- Initial length: $L = kT^2$
- New length: $L + \Delta L = k T_M^2$
- Strain: $\frac{\Delta L}{L} = \frac{T_M^2 - T^2}{T^2} = (\frac{T_M}{T})^2 - 1$
- From $Y = \frac{\text{Stress}}{\text{Strain}} = \frac{Mg/A}{\Delta L/L}$
- Therefore, $\frac{1}{Y} = \frac{\Delta L/L}{Mg/A} = [(\frac{T_M}{T})^2 - 1]\frac{A}{Mg}$
Sign in to join the conversation and share your thoughts.
Log In to Comment