A bullet of mass 2,g has a charge of 2,mu C. | vedclass

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Question

(B) Given: Mass $m = 2\,g = 2 	imes 10^{-3}\,kg$, Charge $q = 2\,\mu C = 2 	imes 10^{-6}\,C$, Final velocity $v = 10\,m/s$, Initial velocity $u = 0\

Vedclass · Accepted Answer

$50\,kV$

Vedclass · Answer

(B) Given: Mass $m = 2\,g = 2 	imes 10^{-3}\,kg$, Charge $q = 2\,\mu C = 2 	imes 10^{-6}\,C$, Final velocity $v = 10\,m/s$, Initial velocity $u = 0\,m/s$.According to the work-energy theorem, the work done by the electric field is equal to the change in kinetic energy:$W = \Delta K$$qV = \frac{1}{2}mv^2 - \frac{1}{2}mu^2$Since $u = 0$, we have:$qV = \frac{1}{2}mv^2$$V = \frac{mv^2}{2q}$Substituting the values:$V = \frac{(2 	imes 10^{-3}\,kg) 	imes (10\,m/s)^2}{2 	imes (2 	imes 10^{-6}\,C)}$$V = \frac{2 	imes 10^{-3} 	imes 100}{4 	imes 10^{-6}}$$V = \frac{2 	imes 10^{-1}}{4 	imes 10^{-6}} = 0.5 	imes 10^5\,V = 50,000\,V = 50\,kV$.Therefore, the required potential difference is $50\,kV$.

$A$ bullet of mass $2\,g$ has a charge of $2\,\mu C$. Through what potential difference must it be accelerated, starting from rest, to acquire a speed of $10\,m/s$?

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