内蒙古包头市2023-2024学年高三上学期期末考试理科综合物理答案

2024-01-30 · U1 上传 · 4页 · 130.2 K

物理参考答案题号1415161718192021答案CBCDAACADBD22.(每空2分)(1)BF(2)1.8022(m1m2)(h4h2)f(3)(m1m2)gh320023.(1)(3分)49(2)B(2分)(3)AC(2分)(4)(2分)3924.(1)b到a(下到上也可以给分)..............................2分(2)22vv02ax...........................................2分v2al...............................................1分BlvBl2alI..........................................2分RRBIlmg.............................................2分BIlB2l22al.....................................3分mgmgR1{#{QQABCQKAogAAAgBAAQgCUwG4CkEQkBAAACoOxAAIsAAAiBFABAA=}#}125.(1)mgrmv2......................................2分20v06m/s...............................................1分(2)mv0(mmp)v1....................................1分11mv2(mm)v2mgL............................1分202p11v12m/s..............................................1分L2.4m...............................................1分(3)mPv1mpvPmQv2.................................1分111mv2mv2mv2................................1分2P12pP2Q2vP0.4m/sv21.6m/s.............................................1分mga15m/s2.......................................1分1m(2mmQ)g1mg2a11m/s..........................1分mQ木块与木板Q均作匀减速到共速v3v1a1t1v2a2t1...................................2分t10.1s................................................1分v31.5m/s.............................................1分木块与木板Q一起做匀减速直线运动到停止(2mmQ)g2a32m/s................................1分(mmQ)2{#{QQABCQKAogAAAgBAAQgCUwG4CkEQkBAAACoOxAAIsAAAiBFABAA=}#}0v3t20.75s........................................2分a3tt1t20.85s.........................................1分33.(1)ABD(选对1个得2分,选对2个得4分选对3个得5分,错选1个扣3分,最低得0分)(2)(ⅰ)pBpA1.6p0p0V01.6p0VB..........................................2分5VV...............................................2分B80(ⅱ)511V2VVV...................................2分A08080pV1.6pV000A.........................................2分T0TA11TT...............................................2分A5033.(1)正(2分);100(3分)sin1sin37(2)(ⅰ)n1.2.........................2分sin2sin30c5vc..............................................2分n6RR11Rt.........................................2分vc5c15(ⅱ)sinC....................................2分n611hRcosCR...................................2分63{#{QQABCQKAogAAAgBAAQgCUwG4CkEQkBAAACoOxAAIsAAAiBFABAA=}#}{#{QQABCQKAogAAAgBAAQgCUwG4CkEQkBAAACoOxAAIsAAAiBFABAA=}#}

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