Physics, asked by vishnugupta4332, 30 days ago

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.

Answers

Answered by rakshithram32

Answer:

please keep me brainlisest

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 =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 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 =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 =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

Answered by rajusubha

make me as a Brainlist

Explanation:

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

Attachments:

Previous Question

Next Question

Similar questions

Math, 15 days ago

$\huge \displaystyle\sf\lim\limits_{x\to\infty}\sqrt[x]{\dfrac{x!}{x^{x}}}$ Or $\displaystyle\sf\lim\limits_{x\to\infty}\left(\dfrac{x!}{x}\right)^{\left(\dfrac{1}{x}\right)}$ ...

Chemistry, 15 days ago

strong john teller distortion can not be observed in which octahedral complexes...

Math, 15 days ago

$\huge \displaystyle\sf\lim\limits_{x\to\infty}\sqrt[x]{\dfrac{x!}{x^{x}}}$ Or $\displaystyle\sf\lim\limits_{x\to\infty}\left(\dfrac{x!}{x}\right)^{\left(\dfrac{1}{x}\right)}$ ...

Science, 30 days ago

hey there please send the pic of your science notes class 10 chapter life processess please don't SPAM ...

Geography, 30 days ago

(2) २००० साली कोणत्या देशामध्ये अधिक परदेशी पर्यटक आले? (3) कोणत्या वर्षापासून भारतामधील परदेशी पर्यटकांची संख्या वाढलेली दिसते? (4) २०१५ साली दोन्ही देशांतील परदेशी पर्यटकांची संख्या किती होती? (5) कोणत्या कालावधीत ब्राझीलमधील परदेशी पर्यटकांची संख्या कमी झाल्याचे आढळते (6) २०१० नंतर भारतातील परदेशी पर्यटक मोठ्या प्रमाणावर वाढण्यामागचे कारण काय 3 1, 7. पुढील प्रश्नांची सविस्तर उत्तरे लिहा :...

Science, 8 months ago

write keywords meaning and explanation i carbohydrates 2) Fibres 3) Balanced diet 4) proteins 5) Pats Constipation 7 vitamins 8) Minerals 9) Roughages...

CBSE BOARD X, 8 months ago

what were the administrative reformes brought by the nepolien ? FOR CLASS 10 HISTORY THE RISE OF NATIONALISM IN EUROPE...

Physics, 8 months ago

Plywood is transparent or translucent or opaque object...

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.​

Answers

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

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.