广东省梅州市2022届高三物理总复习质检试卷（4月）试卷（PDF版附解析）

\n\n\n\n\n\n广东省梅州市2022年普通高中毕业班综合测试2022年梅州市普通高中毕业班综合测试（二）物理参考答案和评分标准一、选择题（1-7题每题4分；8-10题每题选全6分，部分选对3分，有错选0分）题号12345678910答案CDBADAAABACABD11．（6分）BC（2分，选对但不全得1分）、1.85（2分）、9.75（9.71~9.76均可）（2分）12.（10分）3（2分）100（2分）1000（2分）C（2分）电压小于1.5V时该太阳能电池不导通；电压超过1.5V时，该太阳能电池随着电压的增大电阻急剧减小（2分）（注：表述不全的话，能描述出电压比某值小时不通电，得1分；能描述出电压超过某值时，该太阳能电池随着电压的增大电阻减小，得1分。）13.（11分）【解析】（1）飞机着舰瞬间金属棒中感应电动势EBLv0·······①（2分）感应电流E·······································································②（1分）IRr解得3I1.2510A····································································③（1分）由右手定则：感应电流方向b到a。······················································④（1分）12（2）飞机若舰至停下，由动能定理WW0(Mm)v·······⑤（2分）克安克f02解得Q=W克安=3.6×107J·······························································⑥（1分）由焦耳定律知总电热为：············································⑦（1分）QQQRrQR定值电阻和金属棒产生的热量之比为R·····································⑧（1分）Qrrr7解得：QQ1.4410J·················································⑨（1分）rRr2（注：⑦⑧式没有列出，但只要有合理的过程得出⑨，均可给分。例如：由QI(Rr)t得QRR）QRr14．（15分）【详解】（1）由题知，滑块从A到B、B到C，12由动能定理得mgRmvB··································································①（1分）2-1-\n广东省梅州市2022年普通高中毕业班综合测试12mgxmv···········································································②（1分）B2R联立解得v2gR，x2R······················································③（2分）B（注：vB第二问需要用到，无论在第一问或第二问算出，均可给分。）（2）小滑块到B点的速度大于传送带速度gR，据牛顿第二定律可得mgma····························································④（1分）22vv滑块的位移Bcx=R<2R，故滑块与传送带能共速···················⑤（2分）12a（注：④⑤式没有列出，但只要有合理的公式推导得出滑块与传送带共速，均可给分。）（3）滑块与传送带共速后，以vC=gR的速度离开传送带，由机械能守恒定律得1212mvmghmv····································································⑥（1分）1c22设长木板与滑块达到共同速度v2时，位移分别为l1、l2，由动量守恒定律知：mv1(mM)v2·····················································⑦（2分）121212由动能定理知：mgl1Mv2，mgl2mv2mv1·························⑧（2分）222联立解得l12R，l28R··································································⑨（1分）滑块相对长木板的位移ll2l1l，即滑块与长木板在达到共同速度时，物块未离开滑板。1滑块滑到木板右端时，若R<L<2R，Wfmg(6.5RL)，得Wfmg(13R2L)⑩（1分）417若2R<L<5R，，得WfmgR···········⑪（1分）Wfmg(l2ll)mg(ll1)4（注：⑧⑨式没有列出，用系统能量守恒的方式列出公式，得出滑块与长木板在达到共同速度时物块未离开滑板，同样给分）15.（1）变大（2分）、吸收（2分）。（2）（8分）【解析】①罐内气体加热前，压强、体积和温度分别为P0、V0和T0=300k，VV0加热到T=500k后，等效气体等压膨胀到V，则·····························①（2分）TT05解得VV0······················································································②（1分）3mV5加热前后，罐内气体的质量之比为·········································③（2分）mV30024PV②自然降温后，气体的最后压强设为P，则pV250························④（2分）00TT0-2-\n广东省梅州市2022年普通高中毕业班综合测试55解得PP≈1.0×105pa≈6.25×104pa···············································⑤（1分）088（注：①式写成理想气体状态方程后计算答案无误，同样给分。单写理想气体状态方程，得5PP01分。⑤式答案只写8同样给分）16.[选修3-5]【答案】（1）衍射（2分）、变长（2分）（2）（8分）【详解】（1）由几何关系可知光线在AB面上的入射角为60°，由临界角公1式可知sin60°=·············································································①（2分）n23得n=···················································································②（2分）3（2）光线在BC面上的E点同时发生反射和折射，投射到光屏上的PQ两点。且由几何关3系可得线段BE=L，在E点发生反射现象时，根据几何关系光斑P点到B点的距离3BP=BEtan60°=L········································································③（1分）sin3在E点发生折射时，由n，可得sinα=····························④（2分）0sin303BE6BQ=L·········································································⑤（1分）tan3-3-