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- 18:18, 23 September 2010 (hist) (diff) Answer 16.7 (New page: a) <math> 4\textrm{.}72 \text{ m}{{\text{s}}^{-1}} </math> b) <math>58\textrm{.}0{}^\circ</math>) (top)
- 17:43, 23 September 2010 (hist) (diff) Answer 16.6 (New page: <math>(1\textrm{.}8\mathbf{i}+4\textrm{.}4\mathbf{j})\text{ m}{{\text{s}}^{\text{-1}}}</math>) (top)
- 17:37, 23 September 2010 (hist) (diff) Solution 16.6 (New page: Using <math>{{m}_{A}}{{\mathbf{v}}_{A}}+{{m}_{B}}{{\mathbf{v}}_{B}}={{m}_{A}}{{\mathbf{u}}_{A}}+{{m}_{B}}{{\mathbf{u}}_{B}}</math> and <math>{{\mathbf{u}}_{A}}={{\mathbf{u}}_{B}}</math>...)
- 17:09, 23 September 2010 (hist) (diff) Answer 16.5
- 17:07, 23 September 2010 (hist) (diff) Solution 16.5 (New page: Using <math>{{m}_{A}}{{v}_{A}}+{{m}_{B}}{{v}_{B}}={{m}_{A}}{{u}_{A}}+{{m}_{B}}{{u}_{B}}</math> Assume the 2 kg particle is moving in the positive direction, <math>\begin{align} & 2\time...)
- 16:44, 23 September 2010 (hist) (diff) 16. Exercises
- 16:43, 23 September 2010 (hist) (diff) Solution 16.4 (New page: Using <math>{{m}_{A}}{{v}_{A}}+{{m}_{B}}{{v}_{B}}={{m}_{A}}{{u}_{A}}+{{m}_{B}}{{u}_{B}}</math> and <math>{{u}_{A}}={{u}_{B}}</math> <math>\begin{align} & 0\textrm{.}2\times 1\textrm{....) (top)
- 16:25, 23 September 2010 (hist) (diff) Solution 16.2 (New page: Using <math>{{m}_{A}}{{v}_{A}}+{{m}_{B}}{{v}_{B}}={{m}_{A}}{{u}_{A}}+{{m}_{B}}{{u}_{B}}</math> and <math>{{v}_{A}}={{v}_{B}}=0</math> <math>\begin{align} & 0=40\times 3+2v \\ & v=-...) (top)
- 16:21, 23 September 2010 (hist) (diff) Solution 16.3 (New page: Using <math>{{m}_{A}}{{v}_{A}}+{{m}_{B}}{{v}_{B}}={{m}_{A}}{{u}_{A}}+{{m}_{B}}{{u}_{B}}</math> <math>\begin{align} & 3\times 4+5\times (-2)=3\times 0+5v \\ & 12-10=5v \\ & v=\frac{2}{5...) (top)
- 16:17, 23 September 2010 (hist) (diff) 16. Exercises
- 16:06, 23 September 2010 (hist) (diff) Solution 16.1 (New page: Using <math>{{m}_{A}}{{v}_{A}}+{{m}_{B}}{{v}_{B}}={{m}_{A}}{{u}_{A}}+{{m}_{B}}{{u}_{B}}</math> and <math>{{u}_{A}}={{u}_{B}}</math> <math>\begin{align} & 1000\times 20+4000\times 10=(1...) (top)
- 14:43, 17 September 2010 (hist) (diff) Image:15.9.gif (top)
- 14:27, 17 September 2010 (hist) (diff) Solution 15.9b (top)
- 14:26, 17 September 2010 (hist) (diff) Solution 15.9b (New page: <math>\begin{align} & \tan \alpha =\frac{2}{4} \\ & \alpha =26.6{}^\circ \\ \end{align}</math>)
- 14:20, 17 September 2010 (hist) (diff) Solution 15.9a (New page: <math>\mathbf{u}=\text{ 4}u\mathbf{i},\quad \mathbf{v}=\text{ 2}u\mathbf{j},\quad I=kmu</math> <math>\begin{align} & \mathbf{I}=m\mathbf{v}-m\mathbf{u} \\ & =m(2u\mathbf{j})-m(4u\mathbf{...) (top)
- 14:10, 17 September 2010 (hist) (diff) 15. Exercises (top)
- 14:07, 17 September 2010 (hist) (diff) Answer 15.9 (New page: a) <math>k=2\sqrt{5}</math> b) <math>153\textrm{.}4{}^\circ </math>) (top)
- 14:01, 17 September 2010 (hist) (diff) Solution 15.8b (New page: <math>\mathbf{F}=\text{ 24}00\mathbf{j}\ \text{N},\quad m=\text{125}0\ \text{kg},\quad \mathbf{u}=\text{ 15}\mathbf{i}\ \text{m}{{\text{s}}^{-1}}</math> <math>\begin{align} & \mathbf{I}=m...) (top)
- 13:54, 17 September 2010 (hist) (diff) Solution 15.8c (top)
- 13:53, 17 September 2010 (hist) (diff) Image:15.8.gif (top)
- 13:37, 17 September 2010 (hist) (diff) Solution 15.8c (New page: <math>\begin{align} & \tan \alpha =\frac{1\textrm{.}92}{15} \\ & \\ & \alpha =7\textrm{.}29{}^\circ \\ \end{align}</math>)
- 13:35, 17 September 2010 (hist) (diff) Solution 15.8a (New page: <math>F=\text{ 8}00\mathbf{j}\ \text{N},\quad t=\text{ 3}\ \text{s}</math> <math>\begin{align} & I=Ft \\ & =800\mathbf{j}\times 3 \\ & =2400\mathbf{j} \end{align}</math> <math>\text{...) (top)
- 13:11, 17 September 2010 (hist) (diff) Answer 15.8c (New page: <math>{{7\textrm{.}29}^{\circ }}</math>) (top)
- 13:09, 17 September 2010 (hist) (diff) Answer 15.8b (New page: <math>15\textrm{.}1\ \text{m}{{\text{s}}^{-1}}</math>) (top)
- 13:07, 17 September 2010 (hist) (diff) Answer 15.8a (New page: 2400 Ns) (top)
- 11:52, 17 September 2010 (hist) (diff) Solution 15.7b (top)
- 11:51, 17 September 2010 (hist) (diff) Image:15.7temp.gif (top)
- 11:02, 17 September 2010 (hist) (diff) Solution 15.7b (New page: First find the angle, <math>\alpha </math>, between the impulse and the initial velocity. <math>\begin{align} & \tan \alpha =\frac{12}{24} \\ &\\ & \alpha =26\textrm{.}6{}^\circ \e...)
- 10:33, 17 September 2010 (hist) (diff) Solution 15.7a (New page: <math>m=\text{ 2}\text{ kg}</math>, <math>\quad \mathbf{u}=12\mathbf{i}\ \text{ m}{{\text{s}}^{-1}}</math>, <math>\quad \mathbf{v}=6\mathbf{j}\ \text{ m}{{\text{s}}^{-1}}</math> <math>\...) (top)
- 10:16, 17 September 2010 (hist) (diff) Answer 15.7 (New page: a) <math>\text{26}\text{.8 Ns}</math> b) <math>63 \textrm{.}4{}^\circ</math>) (top)
- 09:55, 17 September 2010 (hist) (diff) Solution 15.6b (New page: <math>I=\text{96}00\ \text{Ns},\quad t=\text{ 3}\ \text{s}</math> <math>\begin{align} & I=Ft \\ & 9600=F\times 3 \\ & F=3200\text{ N} \end{align}</math>) (top)
- 09:45, 17 September 2010 (hist) (diff) Solution 15.6a (New page: <math>m=\text{ 12}00\ \text{kg}, \quad u=\text{ 8}\ \text{m}{{\text{s}}^{-1}}, \quad v=\text{ }0\ \text{m}{{\text{s}}^{-1}}</math> <math>\begin{align} & I=mv-mu \\ & =1200\times 0-1200\t...)
- 09:38, 17 September 2010 (hist) (diff) Answer 15.6 (New page: a) 9600 Ns b) 3200 N) (top)
- 09:37, 17 September 2010 (hist) (diff) 15. Exercises
- 18:34, 15 September 2010 (hist) (diff) Solution 15.5c (top)
- 18:32, 15 September 2010 (hist) (diff) Solution 15.5b (top)
- 17:25, 15 September 2010 (hist) (diff) Solution 15.5c (New page: <math>F=\text{ 1}0\ \text{N},\quad I=\text{ }0\textrm{.}\text{64}\ \text{kg}\,\text{m}{{\text{s}}^{-1}}</math> <math>\begin{align} & I=Ft \\ & 0\textrm{.}64=10t \\ & t=0\textrm{.}064\te...)
- 17:19, 15 September 2010 (hist) (diff) Solution 15.5b (New page: <math>I=\text{ }0\textrm{.}\text{64}\ \text{kg}\,\text{m}{{\text{s}}^{-1}},\quad t=\text{ }0\textrm{.}\text{2}\ \text{s}</math> <math>\begin{align} & \\ & I=Ft \\ & 0\textrm{.}64=F\tim...)
- 17:01, 15 September 2010 (hist) (diff) Solution 15.5a (New page: <math>\begin{align} & m=\text{ }0\textrm{.}0\text{4 }\ \text{kg},\quad u=\text{ 9 }\ \text{m}{{\text{s}}^{-1}}\text{ },\quad v=\text{ 7}\ \text{ m}{{\text{s}}^{-1}} \\ & \\ & I=mv-mu \\...)
- 16:52, 15 September 2010 (hist) (diff) Answer 15.5c (New page: 0.064 s) (top)
- 16:52, 15 September 2010 (hist) (diff) Answer 15.5b (New page: 3.2 N) (top)
- 16:51, 15 September 2010 (hist) (diff) Answer 15.5a (New page: 0.64 Ns) (top)
- 16:50, 15 September 2010 (hist) (diff) 15. Exercises
- 16:44, 15 September 2010 (hist) (diff) 15. Exercises
- 18:32, 14 September 2010 (hist) (diff) Solution 15.4a (top)
- 18:26, 14 September 2010 (hist) (diff) Solution 15.4b (top)
- 18:24, 14 September 2010 (hist) (diff) Solution 15.4c (top)
- 18:22, 14 September 2010 (hist) (diff) Solution 15.4c (New page: <math>\begin{align} & m=\text{ }0\textrm{.}0\text{8},\quad u=\text{ 6}\textrm{.}\text{26},\quad v=\text{ }-\text{5}\textrm{.}\text{24} \\ & \\ & I=mv-mu \\ & =0\textrm{.}08\times (-5\t...)
- 18:08, 14 September 2010 (hist) (diff) Solution 15.4b (New page: <math>\begin{align} & v=0,\quad a=\text{ }-\text{9}\textrm{.}\text{8},\quad s=\text{ 1}\textrm{.}\text{4} \\ & \\ & {{v}^{2}}={{u}^{2}}+2as \\ & {{0}^{2}}={{u}^{2}}+2\times (-9\textrm{...)
- 18:04, 14 September 2010 (hist) (diff) Solution 15.4a (New page: <math>\begin{align} & u=0,\quad a=\text{9}\textrm{.}\text{8},\quad s=\text{2} \\ & {{v}^{2}}={{u}^{2}}+2as \\ & {{v}^{2}}={{0}^{2}}+2\times 9\textrm{.}8\times 2 \\ & v=6\textrm{.}26\tex...)
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