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The percentage error in the mass and speed are 2% and 3% respectively. What is the maximum error in kinetic energy calculated using these quantities? |
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Answer» SOLUTION :`(Delta m)/(m) XX 100= 2%, (Delta v)/(v) xx 100= 3%` KINETIC energy `E= (1)/(2) mv^(2)` Maximum error in kinetic energy `(Delta E)/(E )xx 100= (Delta m)/(m) xx 100 + 2 (Delta v)/(v) xx 100` Maximum error in kinetic energy `=2% + 2 xx 3% = 2 % + 6% = 8%` |
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