Answer 15.8b
From Mechanics
(Difference between revisions)
(New page: <math>15\textrm{.}1\ \text{m}{{\text{s}}^{-1}}</math>) |
(New page: <math>15\textrm{.}1\ \text{m}{{\text{s}}^{-1}}</math>) |
Current revision
\displaystyle 15\textrm{.}1\ \text{m}{{\text{s}}^{-1}}
