2023四川高三九市联考(二诊)理科综合试题

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30.(除注明外,每空1分,共9分)(1)逆浓度肾小管液葡萄糖、氨基酸等(任答一种)(2)不需要,其借助于Na+从肾小管液顺浓度进入上皮细胞的势能而进入肾小管液(2分)(3)渗透压饮水不足或失水过多或饮食过咸等细胞外液渗透压抗利尿激素

17解:(1)样本数据的众数为2=70.02分X∈[25,65)的频率为0.05+0.05+0.15+0.20=0.45<0.50;X∈[25,75)的频率为0.05+0.05+0.15+0.20+0.30=0.75>0.50所以中位数在区间[65,75)上,中位数为65+10×0.50-0.4565+≈66.7.0.305分(2)平均文化水平X=30×0.05+40×0.05+50×0.15+60×0.20+70×0.30+80×0.20+90×0.052020~2021学年度石家庄市高一第二学期第三次月考数学参考答案第2页{共4页)64.50分

16.解:(1)由题意作出粒子的运动轨迹,如图所示,在磁场中,洛伦兹力提供向心力,则有qB=m(2分)由几何关系得E=2+(R-2)(1分)解得R(1分)又由于立=k(1分)解得B记x。(1分)(2由R=aB可知粒子运动的半径为R=2(1分)临界情况为粒子从t=0时刻射入,并且轨迹恰好与aO边相切,如图所示,圆周运动的周期为72m=2(1分)由几何关系可知=2内,粒子转过的圆心角为后(分)对应运动时间为:=2=边2T(1分)应满足t1≥(1分)应满足4≥2(1分)联立可得T≤5动(1分)(3)根据题意画出粒子的运动轨迹,如图所示由题意有T0=1×2m(1分)解得T。(1分)E在电场中有gE=ma(1分)0υ往返一次用时为M=4(1分)应有△=(x+2)x。分)可解得E=(2m+1)mx(m=0.1、2-)(2分

2.C解析:“弹”离开无人机后做平抛运动,有h=ng2,x=t联立解得“弹”的水平位移x=15m,“弹”的位移s=√x+2=25m,遥控无人机应在距离目标25m处投弹C项正确。

a

第三节One possible version:Paragraph 1Steve and Zach hadnt gone far uhen there toas afamiliar bark. Brady came bounding, stopped directly infront of Steve and hit him with his head, pushing him backtowards the slope. But Steve didnt get it. Then Bradygrabbed the boy4 Jeans and started pulling. The messagewas clear, but Steve hesitated. Of course he rememberedBrady saving his unele's life when the dog was much younger.Was he still sharp enough to get them through this?Paragraph 2Brady pulled again, in spite of the boys disbelief, iurgently. OK, big guy, "seeing the fire was touchingunderbrush nearby, Steve made up his mind.Brady ledthem back down the slope and into the trees. Several timesthe big dog stopped. Often he changed directions to find theright way. The boys were so tired but the dog bullied bothboys to go on. Steve was almost numb when he heard itthe wonderful sound of rushing water! They made it back tothe river