So, it's Newtonian mechanics, and the reason for this is because Newtonian mechanics does not work on this very, very small size scale.
牛顿力学,因为牛顿力学,在这种很小的尺度下不适用,我们说过,牛顿力学。
I'm not sure if she has to work this weekend or not.
我不清楚这周末她要不要上班。
This is a sign of my economic status: I don't have to work. It is middle and lower-class women who do not get to observe purdah in this way.
但这恰好是我经济地位的标志:,我不需要工作,只有中下阶层的女人,才像这样不遵守深闺的习俗。
Our question is not was Plato overlooking something he should have thought of, is does this argument work or not.
我们的问题不是柏拉图忽略了,一些他本该想到的事情,而是这个论证说不说得通。
And then this country has to work with many other nations and listen to them, not apologize, not gravel.
然后我们需要,与其他国家合作,听取他们的建议,不用道歉,不用困惑。
It's really impossible, in fact, that you not have some opinions about this work, because it's an intimate part of our culture.
但实际上要你们对这想,是不太可能的,因为这已经成为我们文化不可或缺的一部分了。
When I flail my arms around I generate work and heat. This is not a constant volume process.
这不是一个恒容过程,但如果我是一个系统,当我做这些的时候。
So I don't know if I can--whether the teachnic this work or not.
我不知道能否让这幻灯片工作
So this is another kind of engineering, an engineering not to look more closely at how humans work but an engineering to improve their function when it's failing.
这便是另一种工程技术,它并不研究人体运作原理,而是强化某部位的功能,如果这个部位失效了的话
Y He's not returning A or B or temp, and definitely not X or Y; so he just did all of this work and yet that's it.
他不会返回A或B或temp,肯定也不是X或;,所以他刚才做了所有的工作,就是那样。
You'll also see in a little while why both of these things would work this way, but it's not what I wanted. OK?
我应该把括弧,而不是方括号放在这里,你们稍后就会,明白为什么这两个符号在这里都是行得通的?
This is why he gets the Nobel prize and Lewis did not get the Nobel prize, but Pauling's work was quantitative.
这是为什么他荣获了诺贝尔奖,而Lewis没获得,鲍林的工作是数据性的,是量化的。
I don't think they work, though I want to remark before I turn to them that in saying this I'm not necessarily criticizing Plato.
我不认为它们行得通,在此我得声明,说这些话时,我并不一定是在批判柏拉图
So in other words, if thankful is still false at this point, it means this condition did not work out very well, and so I'm going to do this block of code again.
换句话说,如果thankful在这时候还是false值,这表示这个条件没有实现地很好,所以我将再次去处理这段代码。
This is actually very representative of when you do research in the laboratory, you will find often things do not work quite exactly as they worked 20 minutes ago when you just checked it in your office, for example.
在实验室做研究时,这个实际上是非常具有代表性的,举个例子,当时只是在办公室核查它时,你经常会发现事物无法,像它们20分钟之前那样精确的工作。
If I now scroll down to the actual implementation, now does this work or not work?
如果现在我向下滚动到实际的执行中,这个是可用的还是不可用的呢?
Now, this is not the nicest way to do it but it'll work. I can look at the type of the value of base and compare it to the type of an actual float and see, are they the same?
这不是最好的办法但它确实有用,我可以得到底的值的类型然后,和一个真的浮点数的类型比比,看他们是不是一样?
Because this is not really doing any work it's just talking about doing work.
因为我并没真正做任何事而只是在,谈论该做什么。
So, it's very easy to calculate, however, the number of radial nodes, and this works not just for s orbitals, but also for p orbitals, or d orbitals, or whatever kind of work of orbitals you want to discuss.
径向节点,的数量,这不仅对s轨道适用,对p轨道,d轨道,或者任何你们想讨论的轨道,都是适用的,它就等于。
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