It was over here. Because the square root of a quarter is not smaller than a quarter it's bigger than a quarter. Right?

答案超出了这个区间,因为0。25的平方根会比0。25大,是应该比0。25大对不对？

麻省理工公开课 - 计算机科学及编程导论课程节选

In a high-level language, square root might simply be a primitive that you can use rather than you having to go over and code it.

一个高级的语言，平方根可能仅仅,只是你能使用的一个基本要素,而不是你需要去编译它。

麻省理工公开课 - 计算机科学及编程导论课程节选

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。

麻省理工公开课 - 化学原理课程节选

Now you have to know from elementary calculus that v times dv/dt is really d by dt of v square over 2.

根据初等微积分知识你们得知道,v乘以dv/dt其实就是d/2)/dt

耶鲁公开课 - 基础物理课程节选

Times the square of the elementary Z charge times n squared over Z.

乘以元电荷的平方,乘以n的平方除以。

麻省理工公开课 - 固态化学导论课程节选

So the square root of n squared r e over r h.

这里的n值是什么呢？

麻省理工公开课 - 化学原理课程节选

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

耶鲁公开课 - 基础物理课程节选

take the derivative of this, get the velocity vector and you notice his magnitude is a constant Whichever way you do it, you can then rewrite this as v square over r.

对这个式子求一次导,就能得到速度矢量,你会发现其模长是常数,不管用什么方法,加速度也可以写成 v^2 / r

耶鲁公开课 - 基础物理课程节选

So, if we just rearrange this equation, what we find is that z effective is equal to n squared times the ionization energy, IE all over the Rydberg constant and the square root of this.

我们可以发现有效的z等于n的平凡，乘以电离能除以里德堡常数,这些所有再开方，所以等于n乘以,除以RH整体的平方根。

麻省理工公开课 - 化学原理课程节选

If you want to write it as a vector equation, you want to write it as v square over r minus-- I want to say that it's pointing in the direction toward the center.

如果要写出矢量方程,就应该写成 -v^2 / r,也就是说,方向就向圆心

耶鲁公开课 - 基础物理课程节选

The magnitude is v square over r.

模长为 v^2 / r

耶鲁公开课 - 基础物理课程节选

As long as we understand that, we can do this cancellation and this says on the left-hand side the change in the quantity v square over 2 is a times the change in the quantity x.

只要我们明白这一点,我们就可以这样做简化,在等号左边,v^/2的变化量,为a乘以x的变化量

耶鲁公开课 - 基础物理课程节选

Now, I hope you guys know that much calculus, that when you take a derivative of a function of a function, namely v square over 2 is a function of v, and v itself is a function of t, then the rule for taking the derivative is first take the v derivative of this object, then take the d by dt of t, which is this one.

我希望你们了解更多的微积分知识,当你对复合函数求导时,也就是说v^/2是关于v的函数,而v本身是关于t的函数,求导的法则应该是,第一步是这一部分对v求导,然后v再对t求导,得到这一部分

耶鲁公开课 - 基础物理课程节选