last week. The band played two sold-out shows at the twenty thousand seat Madison Square Garden in New York City.
VOA: special.2010.08.13
So, at this place where it hits zero, 0 that means that the square of the wave function is also going to be zero, right.
它达到0的地方,这意味着波函数的,平方也是,如果我们看概率密度图。
Musically, just musically speaking, that was the best show that I've seen at Madison Square Garden.
只从音乐的角度讲,这是我在麦迪逊广场花园看过的最精彩的表演。
And when we take the wave function and square it, that's going to be equal to the probability density of finding an electron at some point in your atom.
当我们把波函数平方时,就等于在某处,找到一个电子的概率密度。
60 If we look at 435 times 160, take the square root of that, we will end up with 264 kilojoules per mole, which sensibly lies between these two values.
35乘以,取平方根,我们会得到264千焦每摩,很合理地落在两数值之间。
Look, part of the reason I'm flaming at you is, something like square root, it seems dumb to write specs on it. Everybody knows what this is going to do. But you want to get into that discipline of good hygiene, good style.
也是非常重要的,我在这里说的,一些理由就是像平方根,为它写规范听上去很傻,每个人都知道它是干什么的。
Usually the creativity is not with the person who's playing at Madison Square Garden.
通常,在麦迪逊广场花园表演的人是缺乏创造力的。
I'm selling tickets to a comedy show at the Times Square Art Center,
我在时代广场的艺术中心卖喜剧表演的门票,
Have you ever bought any of the artists' work at Union Square?
你在联合广场买过艺术家的作品吗?
Whenever you see a particle moving in a circle, even if it's at a constant speed, it has an acceleration, v square over r directed towards the center.
只要看到质点做圆周运动,即使是匀速圆周运动,也存在一个加速度,大小为 v^2 / r,方向指向圆心
Leicester Square is a mess at the moment
现在莱斯特广场一团糟。
So again if we look at this in terms of its physical interpretation or probability density, what we need to do is square the wave function.
如果我们从物理意义或者,概率密度的角度来看这个问题,我们需要把波函数平方。
Let's call x the thing I'm trying to find the square root of. Let's start at 1. Square it.
让我们把求平方根的对象,这个数称为x,让我们从1开始。
When we look at this angular part, we see that it's always the square root of 1 over 4 pi, it doesn't matter what the angle is, it's not dependent on the angle.
当我们看这角向部分,可以看到它总是等于1除以4pai开根号,这和是什么角度没有关系,它和角度无关。
We know it's going around in a circle because if I find the length of this vector, which is the x-square part, plus the y-square part, I just get r square at all times, because sine square plus cosine square is one.
我们之所以知道它做圆周运动,是因为我求出了这个矢量的模长,也就是 x 的平方加上 y 的平方,我就得到了它在任意时刻的模长平方,因为正弦平方加余弦平方始终等于1
I think you can tell by analogy with what I did in one dimension that the position of that object at any time t is going to be the initial position plus velocity times t plus one half a t square.
你们可以类比一下我在一维情况下的结论,这个物体在任意时刻 t 的位移,等于初始位移,加上 v ? t + 1/2 ? a ? t^2
So I guess I should go back, and let me do this correctly this way. Again, I can look at test, test and I guess test now if I want to get the element out-- angle bracket or square bracket?
但是这不是我想要的对不对?,我认为我该回到这里,让我改正这个错误,好,那么我还是对元组,那么现在我想从test中取出元素来?
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