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2014年5/6月Alevel物理真题及答案解析

本文出处:Alevel课程辅导 发布时间:2019-08-19 15:31:48 字体大小: A+ A-

  今天A+教育小编为大家分享下2014年5/6月Alevel物理真题及答案解析,希望对各位准备Alevel的学生有所帮助。


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  答案:C

  The power of passengers can be calculated by F*V=20*60*9.81*0.75=8.8kW. So the total power is 8.8kW.


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  答案:D

  The temperature is constant when melting.


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  答案:B

  4=ρRgx, 7=ρQgy, ρQ=2ρR, so 7x/8.


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  答案:D

  The metal can go back to the original point, so the metal is still in elastic statement.


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  答案:A

  The gradient of extension-load graph is 1/spring constant. The gradient is 0.125, so the spring constant is 8N/m.


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  答案:C

  Total extension =strain*original length. Total strain=stress/E1+stress/E2. So the answer is 5.7*106Pa.


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  答案:C

  From the graph the period is 3.5ms, the amplitude is 6.7mm.


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  答案:A

  The resistance of the wire is 12/0.83=14.45Ω, the total length is 36m, so the answer is 0.04Ω.


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  答案:A

  The total potential difference is 0.4V,the total resistance is 4Ω,so the current is 0.1A.


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  答案:A

  When the intensity is increased,the resistance of LDR will decrease,so the current will increase.


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  答案:B

  When the galvanometer is 0,the potential of PS is 5mV,so the total potential of PQ is 12.5mV. According to the potential is proportional to the resistance,the answer is 795.


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  答案:B

  The potential of V4 is 2,the potential of V3 is 3.


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  答案:C

  Intensity is directly proportional to the amplitude2&frequency2.


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  答案:C

  The particle is at its maximum velocity when it is in equilibrium position.


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  答案:C

  According to the corresponding formula,the wide length of diffraction is larger than double-slit interference,but the intensity is less.


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  答案:C

  X is the bright pattern,so the path difference is λ.


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  答案:A

  Learn it from the electric field of two same charged particles and repulsive particles.


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  答案:D

  qEx=Ek, so Ek is proportional to the x.


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  答案: B

  The potential difference is proportional to the resistance in series circuit.


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  答案:A

  R=ρ1/A, A =πd2/4.

  以上2014年5/6月Alevel物理真题及答案解析啦,如果各位同学哪里还有疑问,可随时与A+国际教育的老师进行一对一的免费咨询哦。

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