优学院: 曲面(xcosz+ycosx-dfrac {pi} {2}z=dfrac {pi} {2})在点((dfrac {pi} {2},1-dfrac {pi} {2},0))处的切平面方程是（ ）
举一反三
- （接上题）（2）设经分界面反射的波的振幅和入射波的振幅相等，则反射波的波函数是 A: $y_{r}=Acos \left(2\pi \nu t+\dfrac{2\pi\nu}{u}x-\dfrac{\pi}{2} \right),0\le x\le\dfrac{3\lambda}{4}$ B: $y_{r}=Acos \left(2\pi \nu t+\dfrac{2\pi\nu}{u}x \right),0\le x\le\dfrac{3\lambda}{4}$ C: $y_{r}=Acos \left(2\pi \nu t+\dfrac{2\pi\nu}{u}x-\dfrac{\pi}{4} \right),0\le x\le\dfrac{3\lambda}{4}$ D: $y_{r}=Acos\left(2\pi \nu t-\dfrac{2\pi\nu}{u}x \right),0\le x\le\dfrac{3\lambda}{4}$
- （2）设经分界面反射的波的振幅和入射波的振幅相等，则反射波的波函数是 A: $y_{r}=Acos \left(2\pi \nu t+\dfrac{2\pi\nu}{u}x-\dfrac{\pi}{2} \right),0\le x\le\dfrac{3\lambda}{4}$ B: $y_{r}=Acos \left(2\pi \nu t+\dfrac{2\pi\nu}{u}x \right),0\le x\le\dfrac{3\lambda}{4}$ C: $y_{r}=Acos \left(2\pi \nu t+\dfrac{2\pi\nu}{u}x-\dfrac{\pi}{4} \right),0\le x\le\dfrac{3\lambda}{4}$ D: $y_{r}=Acos\left(2\pi \nu t-\dfrac{2\pi\nu}{u}x \right),0\le x\le\dfrac{3\lambda}{4}$
- $n=1$时，经典图像中电子围绕质子作半径为$a_0$的圆周运动时的总能量。(1) 总能量为： A: $E=\dfrac{e^2}{8\pi \epsilon_0 a_0}$ B: $E=-\dfrac{e^2}{4\pi \epsilon_0 a_0}$ C: $E=\dfrac{e^2}{4\pi \epsilon_0 a_0}$ D: $E=-\dfrac{e^2}{8\pi \epsilon_0 a_0}$
- 下列函数在给定区间上满足罗尔定理条件的是( ). A: $f(x)=\dfrac 1{x},\; [-2,0]$ B: $f(x)=(x-4)^2,\;[-2,4]$ C: $f(x)=\sin x,\; [-\dfrac{3\pi}{2},\dfrac{\pi}{2}]$ D: $f(x)=|x|,\; [-1,1]$
- 两根无限长的均匀带电直线平行，相距$2a$，线电荷密度分别为$+\lambda$和$-\lambda$，试求(1) 每单位长度的带电直线受的作用力 A: $E=0$ B: $E=\dfrac{\lambda^2}{4\pi \epsilon_0 a}$ C: $E=\dfrac{\lambda^2}{2\pi \epsilon_0 a}$ D: $E=\dfrac{\lambda}{4\pi \epsilon_0 a}$