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。
One, two, three fitting and get R-square test. Enter any matrix derived advantageous values.
说明:一,二,三次拟合,并得到R方的检验。输入任何矩阵,得出险要的值。
But now notice, I can substitute now for this v squared, I can substitute the square of this that is the orbital speed and then I get M G over R.
现在注意我能,代替这个v平方,我能代入这个平方,那是轨道速度,然后得到MG在R上。
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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