[1995]
Solution:
Kinetic energy \(K = \frac{1}{2}mv^2\). Percentage error in \(K = (\Delta m/m) + 2(\Delta v/v)\). Given \(Delta m/m \times\) 100% = 2% and \(\Delta v/v \times 100\)% = 3%. So, Percentage error in K = 2% + 2(3%) = 2% + 6% = 8%.
[1995]
Kinetic energy \(K = \frac{1}{2}mv^2\). Percentage error in \(K = (\Delta m/m) + 2(\Delta v/v)\). Given \(Delta m/m \times\) 100% = 2% and \(\Delta v/v \times 100\)% = 3%. So, Percentage error in K = 2% + 2(3%) = 2% + 6% = 8%.
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