The error in measuring time by Suki's watch is 0.0005%,and by Alan's watch is 0.5s per day.
Q: Find the maximum absolute erroe in Suki's watch per day.
Q: Find the relative error in measuring time by Alan's watch.
Q: Find the percentage error in measuring time by Alan's watch.(correct your answer to 3 significant figuers.)
中一數學題一問 (救命!!) -10點!!
2008-06-07 2:18 am
回答 (2)
2008-06-07 3:13 am
✔ 最佳答案
maximum absolute error by Suki's watch= (24x60x60) x 0.0005%
= 0.432second per day
relative error by Alan's watch
= 0.5/(24x60x60)
= 0.5/86400
= 1/172800
percentage error by Alan's watch
= 1/172800 x 100%
= 0.0005787037%
= 0.000579% (3sig. fig.)
2008-06-07 2:49 am
Q.1
Do not quite understand, please check if correct. Is 0.0005% the percentage error?
Q.2 & 3
Error= 0.5s per day.
Actual = 24 x 60 x 60 = 86400s per day.
Therefore, relative error = 0.5/86400.
Therefore, %error = 0.5/86400 x 100% = 0.000579%.
Do not quite understand, please check if correct. Is 0.0005% the percentage error?
Q.2 & 3
Error= 0.5s per day.
Actual = 24 x 60 x 60 = 86400s per day.
Therefore, relative error = 0.5/86400.
Therefore, %error = 0.5/86400 x 100% = 0.000579%.
收錄日期: 2021-04-13 15:39:55
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