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The U.N.Deputy Resident and Humanitarian Coordinator in Juba,South Sudan,Lise Grande, says the number of internal conflicts in the semi-autonomous region has risen from last year's levels.
VOA: standard.2010.03.23
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And note that as Z increases, as the proton number increases the radius decreases for a given n number.
并注意到当Z不断增加，对于一个给定的n，即当质子数增加的时候,半径的n值就减小了。
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Whatever the n number is, with the exception of helium, helium is the oddity because there's only two elements in n equals one shell.
无论n是多少，除了氦之外，氦是个特例,因为只有两个元素,在n为1的这一层。
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U.N.Special Representative Radhika Coomaraswamy says during the past year, a significant number of child soldiers were released by the National Liberation Forces in Burundi and more children are to be released in the Central African Republic and the Philippines as a result of active intervention by the United Nations.
VOA: standard.2009.09.16
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to the n, every value in the 1 bit vector we looked at last time is either 0 or 1. So it's a binary n number of n bits, 2 to the n.
从2到n，我们上次看到的,位向量的每个值不是0就是,所以它是n，比特的二进制数,从2到。
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"These targets are set for 2015 and are to achieve at least 90 percent measles vaccination coverage nationally and 80 percent coverage in every district, reduce measles cases to less than five per million population, reduce measles mortality by U.N.health agencies have made huge strides in reducing the number of children dying from measles.
VOA: standard.2010.05.21
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And we can calculate that with the formula that we used, which was just n minus l minus 1 equals the number of nodes.
这个我们可以用我们以前用过的那个公式来计算，也就是节点数等于ｎ减去ｌ减去１
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So in other words, every time I merge the point that I kept emphasizing verbally there and that I'm only touching each number once, means that we have to account for the amount of time it takes to merge N which is going to be just N. Now, this is again one of these cyclical answers.
换言之，之前在做合并时,我不停地强调,对每个数字我只碰了一次，这就是说,我们要记录合并所花的时间量,也就是这里的,这又是一种循环性的答案。
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It could have ended any number of other ways, and just so this story could end a number of ways.
它本可能以N种其它方式结尾，也正是如此这个故事才能以若干方式结尾。
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You have n branches on the tree and you count the number of leaves and sum them up.
一共有n根树枝,你数了叶子的数量然后把他们加起来
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With respect to n, the number of moles.
对n的偏导数，这里的n是摩尔数。
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If you start with only one, you have two pieces of DNA, then you'll get 2 to the Nth fragments after N cycles because each cycle you're doubling the number.
如果你从仅仅一个DNA开始,你有两条DNA链,经过N次循环后,就得到二的N次方个DNA片段,因为每次循环都使其数量翻倍
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The differences between those energy levels are becoming smaller and smaller as the n number rises.
那些能级间的差异,会随着n值的增加变得越来越小。
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So, this is a constant n number.
所以这是个固定的n值。
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We didn't just need that n, not just the principle quantum number that we needed to discuss the energy, but we also need to talk about l and m, as we did in our clicker question up here.
我们不仅需要ｎ,不仅要这个可以,决定能量的主量子数，还需要ｍ和ｌ,就像我们做这道题这样。
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Remember, we saw that last time looking at the binary numbers. 2 to the n is a big number.
还记得吗，我们上次看过的二进制数,从2到n是一个很大的数。
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b The repulsive term goes as some constant lower case b divided by R to the n. N is not the quantum number.
这种斥力很想一个固定的小写字母,被R到n分开的话，N不是量子数。
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So, it turns out that n is not the only quantum number needed to describe a wave function, however. There's two more you can see come out of it.
事实上，n不是描述一个波函数需要的,唯一的量子数，你们可以看到，还需要，两个量子数。
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And then the number of pairs at month n is the number of pairs at month n - 1 plus the number of pairs at month n - 2.
我们让第一个月的兔子数是1对,第n个月的兔子对数,是第n-1个月的兔子对数。
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So if n is greater than or equal to 1 and n is less than or equal to 3, let's just judge this thing a small number arbitrarily.
所以n大于等于1，并且n小于等于,我们武断的判定它为,一个小的数字。
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So this line or these lines of code up here are arguably constant time steps to say if N is less than 2 in return, that it will always take maybe one step, maybe two steps, some number of fixed CPU cycles.
如果N小于2并返回,那么这些行所对应的代码,通常只需要执行一步，或者两步，具体数字与CPU周期有关。
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In other words, these energy states are not filling just in ascending n number.
也就是说，能态不只是,按照n的升序进行排列的。
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So, there are 2 different things that we can compare when we're comparing graphs of radial probability distribution, and the first thing we can do is think about well, how does the radius change, or the most probable radius change when we're increasing n, when we're increasing the principle quantum number here?
当比较这些径向概率分布图,的时候，我们可以比较两个东西，第一个就是考虑当我们增加ｎ,当我们增加主量子数的时候，半径怎么变，最可能半径怎么变化？
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l So, if we're talking about a 4 p orbital, and our equation is n minus 1 minus l, the principle quantum number is 1 4, 1 is 1 -- what is l for a p orbital?
我们方程是n减去1减去，主量子数是，4，1是1，--p轨道的l是多少？,学生：
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Specify the quantum number n and divide by Z.
只需说明n的具体值，并用Z去除就行了。
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And we can generalize to figure out, based on any principle quantum number n, how many orbitals we have of the same energy, n and what we can say is that for any shell n, there are n squared degenerate orbitals.
我们可以总结出来，在，主量子数为n的情况下，同一个能量上，有多少个轨道，我们可以说，对任何壳层，有n平方个简并轨道。
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You count n leaves--you count the number of leaves.
你数了n片叶子...不对
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psi So we're going to for psi, and before that, we're going to figure out that instead of n just that one quantum number n, we're going to have a few other quantum numbers that fall out of solving the Schrodinger equation for what psi is.
我们要讲到，但在这之前，我们已经知道了,主量子数，现在我们需要知道，其他一些,解psi的薛定谔方程,所需要的量子数。
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If n is greater than zero, I decided I would say, "You picked a positive number, backslash n," so put the cursor on the next line, else if n was not less than zero, I say, "You picked a negative number, backslash n."
如果n比0大，我就决定来说：,“你选择了一个整数，反斜杠n“，所以你把光标,放在下一行，另外如果n不小于0，我说：,“你选择一个负数，反斜杠n“
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And when we make these comparisons, one thing I want to point out is that we need to keep the constant principle quantum number constant, so we're talking about a certain state, so we could talk about the n equals 2 state, or the n equals 3 state.
当我们做这些比较时，我想指出的一件事是,我们需要保持常量原则,保持量子数是常数，所以我们在讨论一个确定的态时，我们可以谈论n等于2的态，或者n等于3的态。
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