A) 22.8 cm/s
B) 228 cm/s
C) 16.8 m/s
D) 168 m/s
Correct Answer: A
Solution :
By using \[\frac{1}{2}m(v_{1}^{2}-v_{2}^{2})=QV\] Þ \[\frac{1}{2}\times {{10}^{-3}}\{v_{1}^{2}-{{(0.2)}^{2}}\}={{10}^{-8}}(600-0)\] Þ \[{{v}_{1}}=22.8\,cm/s\]You need to login to perform this action.
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