问题
问答题
有一个空心铝球，重4.5N，体积为0.5dm3，如果把这个铝球浸没在水中，
（1）结合计算，说明它为什么上浮？
（2）漂浮的空心铝球露出水面的体积是多少？
答案
（1）当铝球浸没在水中时，它排开水的体积等于铝球的体积，即：
V排=V球=0.5dm3=0.5×10-3m3，
此时铝球受到的浮力为：
F浮=ρ水gV排=1.0×103kg/m3×10N/kg×0.5×10-3m3=5N，
∵F浮＞G，
∴铝球上浮；
（2）当铝球漂浮时：
F浮′=G=4.5N，
V排=
=F /浮 ρ水g
=0.45×10-3m3；4.5N 1.0×103kg/m3×10N/kg
V露=V球-V排 =0.5×10-3m3-0.45×10-3m3=0.05×10-3m3=0.05dm3．
答：（1）因为F浮＞G，所以铝球上浮；
（2）漂浮的空心铝球露出水面的体积是0.05dm3．
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