大学物理实验部分绪论答案.docx

部分绪论答案 P154.2）（18.240.02）cm3）（32.100.01）g5）（10.10.1）km6. 1）t=1.500.05min=(903)s=9.00.3*104ms=(9.00.3)*107s) 2) m=1.750.01kg=1.750.01*103g=1.750.01*107mg3) h=8.540.03cm=8.540.03*101mm=8.540.03*10-2m7.1) L=R+r=0.02+0.02=0.04cm2) P=2R=2*3.14*0.02=0.13cm EP=PP=2R2R=RR=0.0223.75=0.08%3) S=2Rh=2*3.14*23.75*8.84=1318cm2 ES=RR+hh=0.0223.75+0.028.84=0.31% S=S*ES=1318*0.31%=4cm2 S=SS=(13184)cm24) Ek=RR+rr=0.0223.75+0.0216.20=0.20% V=r2h=3.14*16.20*16.20*8.84=7.28cm3 EV=2*rr+hh=2*0.0216.20+0.028.84=0.47% V=V*EV=7.28*0.47%=0.03cm3 V=VV=(7.280.03)cm3P334.(1).因为测量了200次，所以满足正态分布，故有：=1199Vi212T=1200T=TT(2).因为测量了6次，所以满足分布，故有：测量次数i ciVi=ci-cVi210.203-0.0150.0000022520.198-0.0650.0000422530.2120.0750.0000562540.2150.1050.0001102550.192-0.1250.0001562560.2070.0250.00000625平均c=0.2045Vi=0Vi2=0.00037350测量列的标准误差=15Vi212=0.008643平均值的标准误差T=16=0.003528测量结果ci=ctv,0.683T=0.2050.004相对误差Er=2.0%(3) 仪器的感量=0.01则有：=13=0.006故结果为：m=57.3450.006gEr=0.01%5. 5组数据结果如小：测量次数ViVi211.9250.00440.0000193621.919-0.00160.0000025631.9430.02240.0005017641.911-0.00960.0000921651.905-0.01560.00024336平均1.9206Vi=0Vi2=0.00085920测量列的标准误差=14Vi212=0.014656平均值的标准误差T=15=0.006554正态分布测量结果ci=cT=1.92060.0066相对误差Er=0.34%分布测量结果ci=ctv,0.683T=1.92060.0073相对误差Er=0.38%10组数据结果如下：测量次数ViVi211.9250.00110.00000121 21.919-0.00490.00002401 31.9430.01910.00036481 41.911-0.01290.00016641 51.905-0.01890.00035721 61.917-0.00690.00004761 71.923-0.00090.00000081 81.9360.01210.00014641 91.920-0.00390.00001521 101.9400.01610.00025921 平均1.9239Vi=0Vi2=0.00138290测量列的标准误差=19Vi212=0.012396平均值的标准误差T=110=0.00392正态分布测量结果ci=cT=1.92390.0039相对误差Er=0.20%分布测量结果ci=ctv,0.683T=1.