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標題:
Error And Measurement
發問:
In an experiment, the external diameter D and internal diameter d of a metal tube wew found to be 64±2mm and 47±1mm respectively. What is the meximum percentage error in the cross-sectional area of the metal tube?My... 顯示更多 In an experiment, the external diameter D and internal diameter d of a metal tube wew found to be 64±2mm and 47±1mm respectively. What is the meximum percentage error in the cross-sectional area of the metal tube? My solution: A=pi/4(D^2-d^2) (〥A)/A=(2〥D)/D+(2〥d)/d (〥A)/Ax100%=[2(2/64)+2(1/47)]x100% So, the %error=10.5% But the answer: A=pi/4(D^2-d^2)=pi/4(D-d)(D+d) (〥A)/Ax100%=±[〥(D+d)/(D+d)+〥(D-d)(D-d)]x100% (〥A)/Ax100%=±[3/(64+47)+3/(64-47)]x100% (〥A)/Ax100%=±20% What's wrong I have made? 更新: D^2-d^2 =(D+d)(D-d) =D^2+Dd-Dd+d^2 From D^2-d^2 (2〥D)/D+2(〥d)/d =2[(〥D)/D+(〥d)/d] From D^2+Dd-Dd+d^2, (2〥D)/D+(〥D)/d+(〥d)/d+(〥D)/d+(〥d)/d+(2〥d)/d =4[(〥D)/D+(〥d)/d] Why are't they equal?
對不起容許我用中文作答。 有關計算誤差的公式應為 當兩物理量相加或減時 X = Y ± Z δX = δY + δZ 若是乘或除是 X = YZ 或 X = Y/Z δX/X = δY/Y + δZ/Z 而 A = π(D^2 – d^2)/4 是兩項相減，不是乘除所以 δA/A ≠ 2(δD/D) + 2(δd/d) 因此計法便是第二種方法 A=π(D^2-d^2)=pi/4(D-d)(D+d)/4 (δA)/Ax100%=±[δ(D+d)/(D+d)+ δ(D-d)(D-d)]x100% (δA)/Ax100%=±[3/(64+47)+3/(64-47)]x100% (δA)/Ax100%=±20% 不是第一種方法
其他解答:
Error And Measurement
發問:
In an experiment, the external diameter D and internal diameter d of a metal tube wew found to be 64±2mm and 47±1mm respectively. What is the meximum percentage error in the cross-sectional area of the metal tube?My... 顯示更多 In an experiment, the external diameter D and internal diameter d of a metal tube wew found to be 64±2mm and 47±1mm respectively. What is the meximum percentage error in the cross-sectional area of the metal tube? My solution: A=pi/4(D^2-d^2) (〥A)/A=(2〥D)/D+(2〥d)/d (〥A)/Ax100%=[2(2/64)+2(1/47)]x100% So, the %error=10.5% But the answer: A=pi/4(D^2-d^2)=pi/4(D-d)(D+d) (〥A)/Ax100%=±[〥(D+d)/(D+d)+〥(D-d)(D-d)]x100% (〥A)/Ax100%=±[3/(64+47)+3/(64-47)]x100% (〥A)/Ax100%=±20% What's wrong I have made? 更新: D^2-d^2 =(D+d)(D-d) =D^2+Dd-Dd+d^2 From D^2-d^2 (2〥D)/D+2(〥d)/d =2[(〥D)/D+(〥d)/d] From D^2+Dd-Dd+d^2, (2〥D)/D+(〥D)/d+(〥d)/d+(〥D)/d+(〥d)/d+(2〥d)/d =4[(〥D)/D+(〥d)/d] Why are't they equal?
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最佳解答:對不起容許我用中文作答。 有關計算誤差的公式應為 當兩物理量相加或減時 X = Y ± Z δX = δY + δZ 若是乘或除是 X = YZ 或 X = Y/Z δX/X = δY/Y + δZ/Z 而 A = π(D^2 – d^2)/4 是兩項相減，不是乘除所以 δA/A ≠ 2(δD/D) + 2(δd/d) 因此計法便是第二種方法 A=π(D^2-d^2)=pi/4(D-d)(D+d)/4 (δA)/Ax100%=±[δ(D+d)/(D+d)+ δ(D-d)(D-d)]x100% (δA)/Ax100%=±[3/(64+47)+3/(64-47)]x100% (δA)/Ax100%=±20% 不是第一種方法
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