OE=1/√2×OB=1/√2×1cm≈0.71cm 根据F1l1=F2l2得:
,,
G1l1=G2l2
,
m1g·OC=m2g·OE m2=
,
m1·OC4g×1.4cm
=≈7.89g OE0.71cm
答:配重B的质量取值范围为4g~7.89g。 20. (1)G1=m1g=2mg=2×12kg×10N/kg=240N F1=G1=240N
S1=ab=1.2m×0.2m=0.24m2 FF1240N
p====1000Pa SS10.24m2 (2)G2=m2g=mg=12kg×10N/kg=120N W2=G2h=120N×0.2m=24J
(3)吊装k号木块克服重力做功Wk=Gk(k-1)h=mgh=12kg×10N/kg×0.2m=24J 吊装每一块木块克服重力做功相等; W=6Wk=6×24J=144J t=6min=360s W144J
P===0.4W
t360s
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