Listen if you have to cheat for the dba thing not worth even doing that dba tbh this is very easy as I just did it in like 5 minutes it gives you everything you need even the formulas so use your f .u .c .k. 1 .n g brain you monkey...
Answer: It is the same amount of weight as the girl is putting on the pogo stick. When you are pushing something downward then gravity will push back with the equal amount of force.
Explanation:
Answer:
v = 1.15*10^{7} m/s
Explanation:
given data:
charge/ unit area
plate seperation = 1.69*10^{-2} m
we know that
electric field btwn the plates is
force acting on charge is F = q E
Work done by charge q id
this work done is converted into kinectic enerrgy

solving for v



v = 1.15*10^{7} m/s
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