I could get R-Patz maybe, but then he'd just bite the crap out of my hand!
那我就一只手闲着好了,我应该能赢过罗伯特吧,不过回头他就咬我手腕啥的了。
The two flips out, which eats up this two, so I get mg R and then I get a theta.
这两个没用了,被抵消,所以得到mgR和θ
So, the number of nuclei, 119 if we were to sit and count these as well, is 119. So, we'll multiply that by just pi, r squared, to get that cross-section, and divide all of that by 1 .
如果你们数的话,原子核的数是,我们用它乘以πr的平方,得到横截面积,除以1。
R So let's get rid of R in this expression here.
利用这个式子,我们就可以消去。
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
If you get them backwards, logr you will integrate one over r and will get log r.
如果你逆推的话,对1/r积分得到。
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