5.物理答案

物理部分参考答案（2023年市二模）第Ⅰ卷（选择题）14．C15．D16．C17．B18．B19．CD20．CD21．CD第Ⅱ卷（非选择题）22．（6分）（1）用打点计时器测量或者用光电门测量（2分）（2）1/b（2分）k/2b（2分）23．（10分）（1）（k-1）R0（1分）5（1分）（3）a（2分）1.5（2分）10（2分）R0（Ω）（4）如图所示（2分）24．（13分）解：（1）设一级推动t时间火箭的加速度为a1，末速度为v1，二级推动t时间火箭的加速度为a2，末速度为v2，由牛顿运动定律可得：F合3mgmga2g·······························································（1）2分1mm（）分v1a1t2gt··················································································2113mgmgF合a25g····························································（3）2分2m1m22所以火箭上升的最大速度为：（）分v2v1a2t7gt····································41（2）设一级推动t时间火箭上升的高度为h1，二级推动t时间火箭上升的高度为h2，结束推动后火箭继续上升高度为h3，由匀变速直线运动规律可得：vh1tgt2··················································································（5）2分12vv9h12tgt2········································································（6）2分222失去推动后，火箭向上做匀减速运动，加速度大小为g，末速度为0，所以有：2（）分v22gh3························································································71149可得：hgt2············································································（8）1分32所以火箭上升的最大高度为：2（）分hh1h2h330gt·····························9125．（18分）解：（1）由于电子在平行金属板之间运动时水平方向做匀速运动，所以有：2dv··························································································（1）2分0T由题意可得，从tnT（n=0、1、2……）时刻射入平行板的电子竖直方向的偏距最大，且为板间距离的一半，在竖直方向上电子先做匀加速再做匀减速，设其加速度为a，可得：F电a···························································································（2）1分mU0eF电·······················································································（3）1分dd1Ta()22·········································································（4）1分2222md2可得：U···········································································（5）1分0eT2（2）由题意可得，所有电子从平行板沿水平方向射出，速度大小均为v0。如图甲所示，由几何关系可得电子在圆形磁场中做圆周运动的半径和磁场圆的半径相等，即：2rd························································································（6）2分2v2设磁场的磁感应强度为B，可得：m0evB······································（7）2分r022m联立可得：B·······································································（8）2分eTr（3）设电子在磁场中运动的时间为t，偏转角为，由题意可得：t·（9）2分v0如图所示，从平行金属板正中间射出的电子在磁场中的偏转角为，从上极板边缘射出的电子1在磁场中的偏转角为，由图乙中的几何关系可得：2··························································································（10）1分1223························································································（11）1分24t2联立可得：1···········································································（12）2分t23（二）选考题（2题，每题15分）33．【物理选修3－3】（15分）（1）（5分）CDE（2）（10分）解：①设左管液面下降h时，两管液面相平。由于右端玻璃管横截面积是左端的2倍，此时右1面液面上升h，所以有：212由hhh可得：hh8cm···············································（1）1分23对于右端粗管封闭的理想气体，变化前的压强：（）分p1p0ph88cmHg······21变化前的体积为：（）分V1SL··································································311变化后的体积为：VS(Lh)······················································（4）1分2由理想气体等温过程可得：（）分p1V1pV··················································51带入数据可得：p120cmHg······························································（6）1分②由于两管液面相平，两管上端封闭的理想气体压强相等，所以对于左端细管上端的理想气体：（）分p0ShpS(hhx)·····································································72可得：x=12.4cm·················································································（8）2分34．（15分）（1）（5分）ABD（2）（10分）3解：①由图可得两列波的波长分别为：，，振幅均为：（）分a2.5mb4mA0.08m······································112机械波时刻的波动方程为：（）分bt=0ybAsin(x0.5)·······················21b带入数据可得：y0.08sin(x0.5)m··········································（3）1分b2时刻对于的质点，，（）分t=0x=0ya0.08myb0.042m························41所以可得x=0处的质点偏离平衡位置的位移为：（）分yyayb0.04(22)m·····························································51②由题意可得两列波的波速为：va2.5m/s·····································（6）1分TA设至少经过t时间x=0.5m的质点偏离平衡位置的位移为0.16m，则有：m0.5n2tab（其中m、n=0、1、2……）·····························（7）2分vv所以整理可得：5m8n3································································（8）1分当n=4时m=7，结合（7）式可得：t=7.2s···············································（9）1分4