推荐回答（3个）
首先想问个问题：立方体正对叠放是不是A的底面和B的顶面刚好重合？？o(∩_∩)o
1.题目的“A的体积为1.0×10-3米3”是个恰好量，开立方后会发现这个数是0.1m，也就是说这个立方体的长宽高都是0.1m，所以是正方体。
P=F/S，S是指物体接触面的底面积，即为A接触B的面积（（问题就是在这里，你没给出“立方体正对叠放是不是A的底面和B的顶面刚好重合？？”这个条件））如果确认是刚好重合，那就是0.1*0.1=0.1m2 压强PA=F/S=30/0.1=300Pa（之后认为你那个“正对”就是刚好重合的意思）
2.看了遍题目，说实话，以我的物理知识还不可以解答，我只可以告诉你B对桌面的压强一直不变，是100pa，至于表达式：pA的变化式：100pa+pA的变化式
不好意思，个人能力不足，没能帮到你！祝你好成绩啦o(∩_∩)o ！！
1：mA=pA×VA=3kg
F=G=mAg=30N
S=0.1m3(那个应该是正方体吧）
压强PA=F/S=30/0.1=300Pa
2：涉及到开三次方，我不会打
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