20182022英语周报初一答案

34.(1)【答案】ACE(5分)【解题思路】由波动图象可知波源振幅A=5cm,波长A=8m,该波的波速为=4m/,由n=A可得T=2s,)则波源做简谐运动的表达式为y=- Asin ot=-5 sin Tl(cm),A正确;由t=0时刻的波动图象可知x=372m,波源把振动形式传播到质点P所用的时间为1=2=37,所以质点P第次振动到波谷所用的时间为=41+47=1808,B错误;t=7s时质点Q运动到波峰,具有沿y轴负方向的最大加速度,受到的合外力最大,C正确;t=5.4s时,波沿x轴正方向传播的距离s=45.4m=21.6m,传播到x=12m+21.6m=33.6m处,x=33.5m处质点的振动方向沿y轴正方向,D错误;因为t=6s=37,所以t=0到t=6s时间内,质点Q经过的路程为s=3×4A=60cm,E正确。(2)【关键能力】逻辑推理能力【解题思路】(i)作出光路图,如图所示,由几何关系得B1=∠A=30°(1分)由几何关系和反射定律有03=64=90°-261=30°(1分)r=65=90°-∠C=60°(1分)=65-64=30°由折射定律得n=51=3(2分)单色光在AC边发生反射时,因为sin(90°-03)=>sin c=2所以单色光不能从AC边射出(1分)(i)单色光在玻璃工件中的传播速度为=(2分)单色光在玻璃工件中的传播路程为s=DE+2EG=元L(1分)单色光从射入工件到射出工件所用的时间为t=553L(1分)4
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