> 物理 >
将质量为270g的实心铝块挂在弹簧秤的挂钩下面,并把铝块浸没在某种液体中,测力计的示数变为2N,
人气:190 ℃ 时间:2019-11-04 17:28:11
解答
m=270g=0.27kg
v=m/ρ=0.27kg/2700kg/m^3=0.0001m^3
G=mg=0.27kg×10N/kg=2.7N
F浮=G-F=2.7N-2N=0.7N
ρ液=F浮/vg=0.7N/(0.0001m^3×10N/kg)=700kg/m^3
推荐
猜你喜欢
© 2024 79432.Com All Rights Reserved.
电脑版|手机版