If you are driving a car of mass 1050 , kg at | vedclass

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

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$\geq 2 	imes 10^{-37} \, m$

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(C) According to Heisenberg's uncertainty principle,$\Delta x \cdot \Delta p \geq \frac{h}{4\pi}$.Since $\Delta p = m \cdot \Delta v$,the equation becomes $\Delta x \cdot m \cdot \Delta v \geq \frac{h}{4\pi}$.Given: $m = 1050 \, kg$,$v = 90 \, km/h = 90 	imes \frac{5}{18} = 25 \, m/s$.Uncertainty in velocity $\Delta v = 1 \, \% 	ext{ of } 25 \, m/s = 0.25 \, m/s$.Using $h = 6.626 	imes 10^{-34} \, J \cdot s$:$\Delta x \geq \frac{6.626 	imes 10^{-34}}{4 	imes 3.1416 	imes 1050 	imes 0.25}$.$\Delta x \geq \frac{6.626 	imes 10^{-34}}{3298.68} \approx 2.008 	imes 10^{-37} \, m$.Thus,$\Delta x \geq 2 	imes 10^{-37} \, m$.

If you are driving a car of mass $1050 \, kg$ at a speed of $90 \, km/h$ and the uncertainty in the velocity of the car is $1 \, \%$,what will be the uncertainty in its position?

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