甘肃省2023届高三年级3月大联考化学试卷 答案(更新中)
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【答案】ACD【解析】规定向下为正方向,由v=v0十gt知,前1s内产品不断上升,在t=1s时产品速度为0,故t=1s时产品距地面最高,A正确;t=1s时产品速度为0,第2s,第3s,第4s内的位移之比为1:3:5,B错误;第11s内产品的平均速度与t=10.5s时产品的速度相等,则v=vo+gt=(一10+10×10.5)m/s=95m/s,C正确;产品距离地面的高度h=一10t+7gt2(m),将t=11s代入,得出h=495m,D正确。,【答案】AD【解析】xt图像切线的斜率表示速度,则t=4s时,质点A的速率为口10-64m/s=1m/s,A正确;质点A的x-t图像为抛物线,根据运动学公式x=0ot+2at,u=u,十at,联立解得v,=4m/s,a=-0.75m/s2,B、C错误:由图可知质点B做匀速直线运动,其述度大小为。-3m/s=3.25m/s,设经过时间t时A、B相遥,由图可知,此时有xA十xB=13m,解得t=2s(另一解不合题意舍去),故t=2s时,A、B相遇,D正确。
四、应用文写作One possible version:Dear Paul,I'm writing to ask you to visit Smith with me this Sat-urday morning.Smith caught a bad cold last week and has been inhospital for three days.Now I miss him very much and wantto see him as soon as possible.If you are free this Saturdaymorning,shall we go to the hospital to see him together?Ifit's convenient for you,let's meet at the school gate at 9o'clock and go there by bus.Looking forward to your early reply.Yours,Li Hua