求解一道高一物理题,谢谢!!!

2024-12-03 10:33:45
推荐回答(4个)
回答1:

水平距离由水流速度和时间决定 其中时间由竖直速度和竖直位移决定
(1)船头始终正对河岸,那么竖直方向上速度始终为船速 则水平方向上速度即为水速
10min 水平位移120m
所以V水=120m/600s=0.2m/s
(2)当船头保持与河岸向上游成a角,出发后12.5min到达正对河岸
那么V船cosa=V水 V船sina=d/750s
且由第一问 V船=d/600s
联合这三个式子得 V船=1/3m/s 河宽d=200m 角度a=37度

回答2:

(1)水流速度v=120/10=12m/min
(2)船的静水速度为x,x*cos@=v=12
河宽为y,x*sin@*12.5=y
x*10=y
联立方程,可求出后面三问

回答3:

水流速度 v1=120/(10*60)=0.2m/s

静水中速度 v2 河宽 d 夹角 α
v2 sinα *12.5*60=d
v2*10*60=d
v2cosα =v1=0.2m/s

解出 v2=1m/s d=600m α=arcsin(4/5)

回答4:

(1)V水=120/10=12m/min
(2)S宽/V船=10(第一次)
S宽/根号(V船^2-V水^2)=12.5(第二次)
S宽=200m V船=20m/min
(3)S宽=200m
(4)tan@=V船/V水=4/3
@=53°

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