Question

一台起重机2min内把6t重的重物吊高20m高处求起重机的功率

Answer 1

T=2min=120s；m=6t=6000kg
起重机做的功=物体势能的增量
W=mgh=6000*10*20=1200000J
P=W/T=1200000/120=10000w

Answer 2

p=W/t
=Gh/t
=mgh/t
=6*10^3*9.8*20/(2*60)
=9800w
=9.8kw

Answer 3

酸溶液中，[H+]>[OH-]，PH酸=-lg[H+]酸
碱溶液中，[H+]<[OH-]，PH碱=-lg[H+]碱=-lg(Kw/[OH-])=14+lg[OH-]碱
若PH酸+PH碱<14，则-lg[H+]酸+14+lg[OH-]碱<14，得lg[OH-]碱<lg[H+]酸
所以[OH-]碱<[H+]酸
即碱溶液中[OH-]小于酸溶液中[H+]，所以c(酸）>c(碱)
当然这道题不很严密，若是强酸强碱肯定没问题，但若为弱酸弱碱，酸碱浓度与[H+]、[OH-]不相等，这样比较就不准确。

Answer 4

P=FV=F.S/t=6000*10*20/120=10000w

Answer 5

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