一到初中物理题 高分
来源:百度知道 编辑:UC知道 时间:2024/06/22 03:27:13
我算出来Sa<=36.4,36.4<Sb<=255,Sc>272
对吗?
设A距离为X1:(三声合一)
X1/340-X1/5100<0.1
X1/1700-X1/5100<0.1
X1/340-X1/1700<0.1
X1<36.5m
设B距离为X2:(听到空气从来的声音,从铁管和水传来的声音合而为一)
X2/340-X2/1700>0.1
X2/340-X2/5100>0.1
X2/1700-X2/5100<0.1
36.5m<X2<255m
设C距离为X3:
X3/340-X3/1700>0.1
X3/340-X3/5100>0.1
X3/1700-X3/5100>0.1
X3>255m
设A距离为X1:(三声合一)
X1/340-X1/5100<0.1
X1/1700-X1/5100<0.1
X1/340-X1/1700<0.1
X1<36.5m
设B距离为X2:(听到空气从来的声音,从铁管和水传来的声音合而为一)
X2/340-X2/1700>0.1
X2/340-X2/5100>0.1
X2/1700-X2/5100<0.1
36.5m<X2<255m
设C距离为X3:
X3/340-X3/1700>0.1
X3/340-X3/5100>0.1
X3/1700-X3/5100>0.1
X3>255m
分析:声音到达人耳的先后顺序(传声物质)依次是:铁、水、空气。设声音传播的距离为s。要分辨(相邻的)空气和水传来的声音,应有:S1/340m/s-S2/1700m/s>>0.1s,解得:S1≥42.5m;要分辨(相邻的)水和铁传来的声音,应有:S2/1700m/s-S2/5100m/s≥0.1s,解得:S2≥255m。
结论:C同学与敲击处的距离s≥255m时,三种物质传播的声音到达听声者的间隔时间均等于或大于0.1s,能听到三次敲击声。B