Question

The length of a rod as measured in an experiment was found to be 2.48m, 2.46m, 2.49m, 2.50m, and 2.48m. Find the

average length, the absolute error in each observation and the percentage error.​

Answer 1

Answer:

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Explanation:

Average length

Average length=

Average length=2.48+2.46+2.49+2.50+2.48

Average length=2.48+2.46+2.49+2.50+2.485

Average length=2.48+2.46+2.49+2.50+2.485

Average length=2.48+2.46+2.49+2.50+2.485 =

Average length=2.48+2.46+2.49+2.50+2.485 =12.41

Average length=2.48+2.46+2.49+2.50+2.485 =12.415

Average length=2.48+2.46+2.49+2.50+2.485 =12.415

Average length=2.48+2.46+2.49+2.50+2.485 =12.415 m

Average length=2.48+2.46+2.49+2.50+2.485 =12.415 m=2.482m=2.48m(rounded off to 2 places)

Average length=2.48+2.46+2.49+2.50+2.485 =12.415 m=2.482m=2.48m(rounded off to 2 places)Absolute errors id different measurements are ,

Average length=2.48+2.46+2.49+2.50+2.485 =12.415 m=2.482m=2.48m(rounded off to 2 places)Absolute errors id different measurements are ,Δl1=2.48-2.48=0.00m

Average length=2.48+2.46+2.49+2.50+2.485 =12.415 m=2.482m=2.48m(rounded off to 2 places)Absolute errors id different measurements are ,Δl1=2.48-2.48=0.00mΔl2=2.48-2.46=0.02m

Average length=2.48+2.46+2.49+2.50+2.485 =12.415 m=2.482m=2.48m(rounded off to 2 places)Absolute errors id different measurements are ,Δl1=2.48-2.48=0.00mΔl2=2.48-2.46=0.02mΔl3=2.48-2.49=0.01m

Average length=2.48+2.46+2.49+2.50+2.485 =12.415 m=2.482m=2.48m(rounded off to 2 places)Absolute errors id different measurements are ,Δl1=2.48-2.48=0.00mΔl2=2.48-2.46=0.02mΔl3=2.48-2.49=0.01mΔl4=2.48-2.50=0.02m

Average length=2.48+2.46+2.49+2.50+2.485 =12.415 m=2.482m=2.48m(rounded off to 2 places)Absolute errors id different measurements are ,Δl1=2.48-2.48=0.00mΔl2=2.48-2.46=0.02mΔl3=2.48-2.49=0.01mΔl4=2.48-2.50=0.02mΔl5=2.48-2.48=0.00m

Average length=2.48+2.46+2.49+2.50+2.485 =12.415 m=2.482m=2.48m(rounded off to 2 places)Absolute errors id different measurements are ,Δl1=2.48-2.48=0.00mΔl2=2.48-2.46=0.02mΔl3=2.48-2.49=0.01mΔl4=2.48-2.50=0.02mΔl5=2.48-2.48=0.00mMean absoulte error =

Average length=2.48+2.46+2.49+2.50+2.485 =12.415 m=2.482m=2.48m(rounded off to 2 places)Absolute errors id different measurements are ,Δl1=2.48-2.48=0.00mΔl2=2.48-2.46=0.02mΔl3=2.48-2.49=0.01mΔl4=2.48-2.50=0.02mΔl5=2.48-2.48=0.00mMean absoulte error =Σ|ΔL|

Average length=2.48+2.46+2.49+2.50+2.485 =12.415 m=2.482m=2.48m(rounded off to 2 places)Absolute errors id different measurements are ,Δl1=2.48-2.48=0.00mΔl2=2.48-2.46=0.02mΔl3=2.48-2.49=0.01mΔl4=2.48-2.50=0.02mΔl5=2.48-2.48=0.00mMean absoulte error =Σ|ΔL|5

Average length=2.48+2.46+2.49+2.50+2.485 =12.415 m=2.482m=2.48m(rounded off to 2 places)Absolute errors id different measurements are ,Δl1=2.48-2.48=0.00mΔl2=2.48-2.46=0.02mΔl3=2.48-2.49=0.01mΔl4=2.48-2.50=0.02mΔl5=2.48-2.48=0.00mMean absoulte error =Σ|ΔL|5

Average length=2.48+2.46+2.49+2.50+2.485 =12.415 m=2.482m=2.48m(rounded off to 2 places)Absolute errors id different measurements are ,Δl1=2.48-2.48=0.00mΔl2=2.48-2.46=0.02mΔl3=2.48-2.49=0.01mΔl4=2.48-2.50=0.02mΔl5=2.48-2.48=0.00mMean absoulte error =Σ|ΔL|5 =

Average length=2.48+2.46+2.49+2.50+2.485 =12.415 m=2.482m=2.48m(rounded off to 2 places)Absolute errors id different measurements are ,Δl1=2.48-2.48=0.00mΔl2=2.48-2.46=0.02mΔl3=2.48-2.49=0.01mΔl4=2.48-2.50=0.02mΔl5=2.48-2.48=0.00mMean absoulte error =Σ|ΔL|5 =0.00+0.02+0.01+0.02+0.00

Average length=2.48+2.46+2.49+2.50+2.485 =12.415 m=2.482m=2.48m(rounded off to 2 places)Absolute errors id different measurements are ,Δl1=2.48-2.48=0.00mΔl2=2.48-2.46=0.02mΔl3=2.48-2.49=0.01mΔl4=2.48-2.50=0.02mΔl5=2.48-2.48=0.00mMean absoulte error =Σ|ΔL|5 =0.00+0.02+0.01+0.02+0.005

Average length=2.48+2.46+2.49+2.50+2.485 =12.415 m=2.482m=2.48m(rounded off to 2 places)Absolute errors id different measurements are ,Δl1=2.48-2.48=0.00mΔl2=2.48-2.46=0.02mΔl3=2.48-2.49=0.01mΔl4=2.48-2.50=0.02mΔl5=2.48-2.48=0.00mMean absoulte error =Σ|ΔL|5 =0.00+0.02+0.01+0.02+0.005

Average length=2.48+2.46+2.49+2.50+2.485 =12.415 m=2.482m=2.48m(rounded off to 2 places)Absolute errors id different measurements are ,Δl1=2.48-2.48=0.00mΔl2=2.48-2.46=0.02mΔl3=2.48-2.49=0.01mΔl4=2.48-2.50=0.02mΔl5=2.48-2.48=0.00mMean absoulte error =Σ|ΔL|5 =0.00+0.02+0.01+0.02+0.005 =0.01m

Average length=2.48+2.46+2.49+2.50+2.485 =12.415 m=2.482m=2.48m(rounded off to 2 places)Absolute errors id different measurements are ,Δl1=2.48-2.48=0.00mΔl2=2.48-2.46=0.02mΔl3=2.48-2.49=0.01mΔl4=2.48-2.50=0.02mΔl5=2.48-2.48=0.00mMean absoulte error =Σ|ΔL|5 =0.00+0.02+0.01+0.02+0.005 =0.01m∴Correct length=(2.48±0.01)m

Answer 2

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Explanation:

Answer: Explanation: average length= (2.48 + 2.46 + 2.49 + 2.50 + 2.48)/5 =2.482.