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(A) Let $y = AB$. The relative error in the product is given by $\frac{\Delta y}{y} = \frac{\Delta A}{A} + \frac{\Delta B}{B}$. First,calculate the me

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$(0.25 \pm 0.08) \, m$

Vedclass · Answer

(A) Let $y = AB$. The relative error in the product is given by $\frac{\Delta y}{y} = \frac{\Delta A}{A} + \frac{\Delta B}{B}$. First,calculate the mean value: $y = 2.5 	imes 0.10 = 0.25 \, m$. Now,calculate the absolute error $\Delta y$: $\frac{\Delta y}{0.25} = \frac{0.5}{2.5} + \frac{0.01}{0.10}$ $\frac{\Delta y}{0.25} = 0.2 + 0.1 = 0.3$ $\Delta y = 0.25 	imes 0.3 = 0.075$. Rounding to the appropriate significant figures (consistent with the given data),we get $\Delta y \approx 0.08$. Thus,$AB = (0.25 \pm 0.08) \, m$.

The measurements of two quantities along with the precision of their respective measuring instruments are given as:
$A = 2.5 \, m/s \pm 0.5 \, m/s$
$B = 0.10 \, s \pm 0.01 \, s$
The value of $AB$ will be:

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The measurements of two quantities along with the precision of their respective measuring instruments are given as: $A = 2.5 \, m/s \pm 0.5 \, m/s$ $B = 0.10 \, s \pm 0.01 \, s$ The value of $AB$ will be: