Question

如图，光滑长木板AB可绕O转动，质量不计，A端用竖直绳与地板上拉着，在离O点0.4 m的B处挂一密度为0.8×l0 3

Answer 1

① F _浮= G _排= 0.5N

② F _浮=ρ水gV排 = 0.5N

V = 2V _排=" 2" ×0.5×10 ^-4m ³= 1×10 ^-4m ³

G=" mg" = ρVg = 0.8×10 ³kg/m ³×1×10 ^-4m ³×10N/kg = 0.8N

③绳子拉力T _B=" G" – F _浮=" 0.8N" – 0.5N = 0.3N

设小球运动t 秒时T _A=" 0" ，则

T _B×OB = G球×Vt即 0.3N ×0.4m =" 2N" × 0.02 m/s ×t

解得 t =" 3" s

略