\(t\sin(2t)\)的拉普拉斯变换为
A: \(\dfrac{2}{s^2+4}\)
B: \(\dfrac{s}{s^2+4}\)
C: \(\dfrac{4}{(s^2+4)^2}\)
D: \(\dfrac{4s}{(s^2+4)^2}\)
A: \(\dfrac{2}{s^2+4}\)
B: \(\dfrac{s}{s^2+4}\)
C: \(\dfrac{4}{(s^2+4)^2}\)
D: \(\dfrac{4s}{(s^2+4)^2}\)
举一反三
- 如果简单正向闭曲线L所围成区域的面积为S，那么$S = (\quad ).$ A: $\dfrac{1}{2}\oint_L {xdx - ydy} $ B: $\dfrac{1}{2}\oint_L {ydy - xdx} $ C: $\dfrac{1}{2}\oint_L {ydx - xdy} $ D: $\dfrac{1}{2}\oint_L {xdy - ydx} $
- 39号元素钇的核外电子排布式是下列排布中的（ ） A: 1 s 2 2 s 2 2 p 6 3 s 2 3 p 6 3 d 10 4 s 2 4 p 6 4 d 1 5 s 2 B: 1 s 2 2 s 2 2 p 6 3 s 2 3 p 6 3 d 10 4 s 2 4 p 6 5 s 2 5 p 1 C: 1 s 2 2 s 2 2 p 6 3 s 2 3 p 6 3 d 10 4 s 2 4 p 6 4 d 2 5 s 1 D: 1 s 2 2 s 2 2 p 6 3 s 2 3 p 6 3 d 10 4 s 2 4 p 6 5 s 2 5 p 1
- （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: $y_{r}=Acos\left(2\pi \nu t-\dfrac{2\pi\nu}{u}x \right),0\le x\le\dfrac{3\lambda}{4}$
- [tex=17.429x1.286]VoKfr1IIzGJmh8L2l120JqjixkQ1tL3XlyBgbwU9pvS9UKZRTovLmaVq8Zv4SelU6NNNx16xNXr2asYvjnme4w==[/tex][img=310x163]179b0f602527037.png[/img] 未知类型：{'options': ['0.5 s(2) 1 s(3) $2 s$(4) $4 s$', '\xa01 s(3) $2 s$(4) $4 s$', '2 s(4) $4 s$', '4 s'], 'type': 102}