Answer 16.1
From Mechanics
(Difference between revisions)
(New page: 12 <math>\text{m}{{\text{s}}^{-1}}</math>) |
(New page: 12 <math>\text{m}{{\text{s}}^{-1}}</math>) |
Current revision
12 \displaystyle \text{m}{{\text{s}}^{-1}}