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(A) According to Heisenberg's uncertainty principle,$\Delta x \cdot \Delta p \geq \frac{h}{4\pi}$.Since $\Delta p = m \cdot \Delta v$,the formula beco

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$2.86 	imes 10^{-28} \, m$

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(A) According to Heisenberg's uncertainty principle,$\Delta x \cdot \Delta p \geq \frac{h}{4\pi}$.Since $\Delta p = m \cdot \Delta v$,the formula becomes $\Delta x = \frac{h}{4\pi m \Delta v}$.Given: $m = 0.02 \, kg$,$\Delta v = 9.218 	imes 10^{-6} \, m/s$,and $h = 6.626 	imes 10^{-34} \, J \cdot s$.Substituting the values: $\Delta x = \frac{6.626 	imes 10^{-34}}{4 	imes 3.14159 	imes 0.02 	imes 9.218 	imes 10^{-6}}$.$\Delta x = \frac{6.626 	imes 10^{-34}}{2.313 	imes 10^{-6}} \approx 2.86 	imes 10^{-28} \, m$.

For a particle of mass $0.02 \, kg$,the uncertainty in its velocity is $9.218 \times 10^{-6} \, m/s$. The uncertainty in its position is .......

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