半径为$R$, 密度为$1$的均匀平面薄板关于其切线的转动惯量为
A: $\frac{3\pi R^4}{4}$
B: $\frac{5\pi R^4}{4}$
C: $\frac{5\pi R^3}{4}$
D: $\frac{4\pi R^3}{3}$
A: $\frac{3\pi R^4}{4}$
B: $\frac{5\pi R^4}{4}$
C: $\frac{5\pi R^3}{4}$
D: $\frac{4\pi R^3}{3}$
举一反三
- 函数$f(x)=\sin x + \cos x,x \in [0,2 \pi]$的上凸区间为 A: $[0,\frac{\pi}{4}] \cup [\frac{5}{4} \pi,2 \pi] $ B: $[\frac{\pi}{4},\frac{5}{4} \pi]$ C: $[0,\frac{3}{4}\pi] \cup [\frac{7}{4} \pi,2 \pi] $ D: $[\frac{3}{4} \pi,\frac{7}{4} \pi] $
- 下列各组角中,可以作为向量的方向角的是(<br/>) A: $\frac{\pi }{3},\,\frac{\pi }{4},\,\frac{2\pi }{3}$ B: $-\frac{\pi }{3}\,,\frac{\pi }{4}\,,\frac{\pi }{3}$ C: $\frac{\pi }{6},\,\pi ,\,\frac{\pi }{6}$ D: $\frac{2\pi }{3},\,\frac{\pi }{3},\,\frac{\pi }{3}$
- Solve $\int_{-\frac{1}{2}}^1{1-x^2}dx=$? A: $\frac{\pi}{3}+\frac{\sqrt{3}}{8}$. B: $\frac{\pi}{2}$. C: $\frac{\pi}{6}+\frac{\sqrt{3}}{4}$. D: $\frac{\pi}{4}$.
- 如图所示，电荷\(-\)Q 均匀分布在半径为R、长为L的圆弧上，圆弧的两端有一小空隙，空隙长为\(\Delta\)L(\(\Delta\)L< A: \(\frac{-Q\Delta L}{4\pi\varepsilon_0R^2L} \vec i\), \(\frac{-Q}{4\pi\varepsilon_0R}\) B: \(\frac{-Q\Delta L}{8\pi\varepsilon_0R^3} \vec i\), \(\frac{-Q}{4\pi\varepsilon_0R}\) C: \(\frac{Q\Delta L}{4\pi\varepsilon_0R^2L} \vec i\), \(\frac{Q}{4\pi\varepsilon_0R}\) D: \(\frac{-Q\Delta L}{4\pi\varepsilon_0R^2L} \vec i\), \(\frac{-Q\Delta L}{4\pi\varepsilon_0RL}\)
- 函数\(f(x) = x^2,\; x \in [-\pi,\pi]\)的Fourier级数为 A: \(\frac{\pi^2}{3}+4\Sigma_{n=1}^{\infty} \frac{(-1)^n}{n^2} \sin nx ,\; x \in [-\pi,\pi]\) B: \(\frac{\pi^2}{3}+4\Sigma_{n=1}^{\infty} \frac{(-1)^n}{n^2} \cos nx ,\; x \in [-\pi,\pi]\) C: \(\frac{2\pi^2}{3}+4\Sigma_{n=1}^{\infty} \frac{(-1)^n}{n^2} \sin nx ,\; x \in [-\pi,\pi]\) D: \(\frac{2\pi^2}{3}+4\Sigma_{n=1}^{\infty} \frac{(-1)^n}{n^2} \cos nx ,\; x \in [-\pi,\pi]\)