2012西城二模理综试题答案

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1、中国权威高考信息资源门户www.gaokao.com北京市西城区2012年高三二模试卷物理试题评分细则2012.5一、选择题（共8*6=48分）13．B14．D15．C16．D17．A18．B19．B20．C二、实验题21．（共18分）（1）①匀速直线···········1分远小于···········1分②0.343··············2分③A··················2分（2）①如图所示··························3分②增大··························2分增大·····························2

2、分③0.81（0.80—0.82）·········2分—+0.62（0.60—0.64）···········3分·评分说明：本题共18分。①答图第（1）题，共6分。第（2）题，共12分。在第（1）题的连线中，只有滑动变阻器的滑动触头端的两个接线柱上的连线可以互相代换位置。22．（共16分）3（1）小球所受的电场力FEqm···g·························2分43FEqmg·····························2分4（2）根据平行四边形定则,小球受到的重力和电场力的的合力225F（mg)(Eq)mg················

3、·············2分合4根据牛顿第二定律Fma·····························2分合5所以，小球的加速度ag·····························2分4（3）根据动能定理有：12mglEqlmv0················4分22gl解得：v······································2分223．（共18分）E有用（1）总效率100％····························2分E总中国权威高考信息资源门户www.gaokao.com10100％71％····

4、·······················2分14（2）能用于发电的水的总质量mV；dd能用于发电的水的平E均下E落mg高h度mghVg(HVg(H))PP22所以，能用于发电的水的最大重力势能：dEmghVg(H)···················6分P2说明：上式m中V取1000kg/m3,g取9.8m/s2或10m/s2均可。P（3）a．当输电电压为U时，输电电流I···················1分U2P2所以，损失功率PIR()R···················2分UR所以，输电电压的最小值UP············

5、·······2分Pb．提高输电电压U(最经济)··································2分增大输电导线的横截面积（或采用电阻率小的输电导线材料）(不经济)················································1分；说明：本题共18分。第（1）问4分；第（2）问6分；第（3）问8分。24．（20分）（1）对物体B受力分析如图1所示，根据共点力平衡条件T1=m3g①··················1分对物体A和木板组成的整体受力分析如图2所示根据共点力平衡条件得F1=T1②···················1分代

6、入数据，由①、②得F1=10N···················1分T1F1T1Bm3g图2图1（2）运动过程中，三个物体的加速度大小相等，设加速度大小为a。对物体B受力分析如图3所示，根据牛顿第二定律T2–m3g=m3a③··········1分T2fT2对物体A受力分析如图4所示，aB根据牛顿第二定律f–T2=m2a④············1分am3g此时，f=μm2g⑤············1分图4代入数据，由③、④、⑤式得图3a=2.0m/s2············2分F2f对木板受力分析如图5所示，根据牛顿第二定律F2–f=m1a⑤····1分a图5中国权威高考信息资源

7、门户www.gaokao.com将加速度代入⑤式得F2=60N········1分（3）由于木板与物体A之间的最大静摩擦力等于滑动摩擦力，物体A、B运动的加速度与（2）中的加速度相等。即a=2.0m/s2设物体B上升高度hB=1.0m，所用时间为t，满足12hat···········1分B2代入数据，解得t=1.0s············1分设木板减速运动过程中的加速度为a，f对木板受力分析如图6所示a´根据牛顿