Answer:
Yes, when \[\text{p}=0\], \[\text{K}=\frac{{{p}^{2}}}{2m}=0\]
But \[\text{E}=\text{K}+\text{V}=\text{V}\left(
\text{Pot}.\text{ energy} \right)\], which may or may not be zero.
You need to login to perform this action.
You will be redirected in
3 sec