From (1) and (3)
u = Wc[tau]/p[Omega];
dl/u = g d[Omega]³p² dp/2[zeta]W³c²[tau]²;
_
/p0
t = | g d[Omega]³p² dp/2[zeta]W³c²[tau]²,
_/p1
= gd[Omega]³(p0³ - p1³)/6[zeta]W³c²[tau]². (14)
Encyclopaedia Britannica, 11th Edition, "Husband" to "Hydrolysis" Volume 14, Slice 1 · 第 115 / 207 题