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From Mechanics
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- 12:58, 2 October 2010 (hist) (diff) Image:17.7.gif (top)
- 12:21, 2 October 2010 (hist) (diff) 17. Exercises
- 12:20, 2 October 2010 (hist) (diff) Solution 17.6c (top)
- 12:19, 2 October 2010 (hist) (diff) Image:17.6b.gif (top)
- 12:17, 2 October 2010 (hist) (diff) Solution 17.6c
- 12:14, 2 October 2010 (hist) (diff) Image:17.6a.gif (uploaded a new version of "Image:17.6a.gif") (top)
- 12:07, 2 October 2010 (hist) (diff) Image:17.6a.gif
- 11:57, 2 October 2010 (hist) (diff) Solution 17.6c (New page: Both at the start and in the final position the kinetic energy is zero so it is the loss in potential energy which is to be detemined. The initial potential energy has been calcukated in ...)
- 11:52, 2 October 2010 (hist) (diff) Solution 17.6b (New page: The loss in potential energy is equal to the gain in kinetic energy. As the potential energy is zero when the particle is at its lowest position the kinetic energy is from part a) <math...) (top)
- 11:48, 2 October 2010 (hist) (diff) Solution 17.6a (New page: Image:17.6.gif From the figure, <math>\begin{align} & PE=5\times 9\textrm{.}8(8-8\cos 30{}^\circ ) \\ & =52\textrm{.}5\text{ J} \end{align}</math>) (top)
- 11:45, 2 October 2010 (hist) (diff) Image:17.6.gif (top)
- 11:43, 2 October 2010 (hist) (diff) Answer 17.6c (New page: 6.60 J) (top)
- 11:43, 2 October 2010 (hist) (diff) Answer 17.6b (New page: <math>4\textrm{.}58\ \text{ m}{{\text{s}}^{-1}}</math>) (top)
- 11:42, 2 October 2010 (hist) (diff) Answer 17.6a (New page: 52.5 J) (top)
- 11:04, 2 October 2010 (hist) (diff) 17. Conservation of energy
- 18:04, 1 October 2010 (hist) (diff) Solution 17.5c (New page: (Loss of) potential Energy = <math>mgh</math> gives from part b) <math>\begin{align} & 3\times 9\textrm{.}8h=324 \\ & h=\frac{324}{3\times 9\textrm{.}8}=11\textrm{.}0\text{ m} \\ \end{a...) (top)
- 18:00, 1 October 2010 (hist) (diff) Solution 17.5b (New page: PE Lost = KE Gained = 324 J using the result of part a).) (top)
- 17:58, 1 October 2010 (hist) (diff) Solution 17.5a (New page: <math>\begin{align} & \text{Gain in KE }=\frac{1}{2}\times 3\times {{15}^{2}}-\frac{1}{2}\times 3\times {{3}^{2}} \\ & =324\text{ J} \end{align}</math>) (top)
- 17:57, 1 October 2010 (hist) (diff) Answer 17.5c (New page: 11.0 m) (top)
- 17:56, 1 October 2010 (hist) (diff) Answer 17.5b (New page: 324 J) (top)
- 17:56, 1 October 2010 (hist) (diff) Answer 17.5a (New page: 324 J) (top)
- 16:30, 1 October 2010 (hist) (diff) Answer 16.10
- 16:29, 1 October 2010 (hist) (diff) Solution 16.10b (New page: From part a) <math>(5m+2\lambda m-3m-3\lambda m)\mathbf{i}+(mV-\lambda mV+2m+2\lambda m)\mathbf{j}=0\mathbf{i}+0\mathbf{j}</math> Consider the <math>\mathbf{j}</math> component. That is...)
- 16:23, 1 October 2010 (hist) (diff) Solution 16.10a
- 16:09, 1 October 2010 (hist) (diff) 16. Exercises (top)
- 09:46, 30 September 2010 (hist) (diff) Solution 17.4b (New page: As the brick has no speed at the start <math>\text {KE = KE gained}</math>. Using Energy lost due to friction = Work done against friction we get, <math>\begin{align} & \text{PE lost - ...)
- 09:22, 30 September 2010 (hist) (diff) Solution 17.4a (New page: As the brick has no speed at the start <math>\text {KE = KE gained}</math>. <math>\begin{align} & \text {KE gained=PE }\text{ lost }=\text{2}\times \text{9}\text{.8}\times \text{3}\text{...) (top)
- 09:11, 30 September 2010 (hist) (diff) Answer 17.4b (top)
- 09:10, 30 September 2010 (hist) (diff) Answer 17.4a (top)
- 09:07, 30 September 2010 (hist) (diff) 17. Conservation of energy
- 08:56, 30 September 2010 (hist) (diff) 17. Conservation of energy
- 08:44, 30 September 2010 (hist) (diff) Solution 17.3c (New page: Using Conservation of Energy which gives that the energy at the start is the same aas the energy at the highest point. Assuming the potential energy at the start is zero, <math>\begin{al...) (top)
- 08:40, 30 September 2010 (hist) (diff) Solution 17.3b (New page: <math>KE=\frac{1}{2}\times 0\textrm{.}125\times {{4}^{2}}=1\text{ J}</math>) (top)
- 08:39, 30 September 2010 (hist) (diff) Solution 17.3a (New page: <math>KE=\frac{1}{2}\times 0\textrm{.}125\times {{12}^{2}}=9\text{ J}</math>) (top)
- 08:34, 30 September 2010 (hist) (diff) Answer 17.3c (New page: 8 J) (top)
- 08:34, 30 September 2010 (hist) (diff) Answer 17.3b (New page: 1 J) (top)
- 08:33, 30 September 2010 (hist) (diff) Answer 17.3a (New page: 9 J) (top)
- 08:32, 30 September 2010 (hist) (diff) Solution 17.2 (New page: <math>KE=\frac{1}{2}\times 0\textrm{.}25\times {{9}^{2}}=10\textrm{.}125\text{ J}</math>) (top)
- 08:32, 30 September 2010 (hist) (diff) Answer 17.2 (New page: <math>10\textrm{.}125\text{ J}</math>) (top)
- 08:28, 30 September 2010 (hist) (diff) Solution 17.1 (New page: <math>KE=\frac{1}{2}\times 1200\times {{30}^{2}}=540000\text{ J}</math>) (top)
- 08:27, 30 September 2010 (hist) (diff) Answer 17.1 (New page: 540000 J) (top)
- 08:25, 30 September 2010 (hist) (diff) 17. Conservation of energy
- 17:39, 29 September 2010 (hist) (diff) m Solution 16.10a (New page: <math>\begin{align} & m(4\mathbf{i}+V\mathbf{j})+\lambda m(2\mathbf{i}-V\mathbf{j})=(m+\lambda m)(3\mathbf{i}+2\mathbf{j}) \\ & (4m+2\lambda m-3m)\mathbf{i}+(mV-\lambda mV-2\lambda m)\mat...)
- 17:23, 29 September 2010 (hist) (diff) Answer 16.10
- 17:19, 29 September 2010 (hist) (diff) Solution 16.9 (New page: <math>\begin{align} & 2m(8\mathbf{i}+7\mathbf{j})+m(U\mathbf{i}+V\mathbf{j})=2m(5\mathbf{i})+m(2\mathbf{j}) \\ & U\mathbf{i}+V\mathbf{j}=(10-16)\mathbf{i}+(2-14)\mathbf{j} \\ & =-6\math...) (top)
- 17:16, 29 September 2010 (hist) (diff) Answer 16.9 (New page: <math>U=-6\ \text{ m}{{\text{s}}^{-1}},\ V=-12\ \text{ m}{{\text{s}}^{-1}}</math>) (top)
- 17:13, 29 September 2010 (hist) (diff) Solution 16.8 (New page: <math>\begin{align} & {{\mathbf{u}}_{W}}=\mathbf{i},\ {{\mathbf{v}}_{W}}=0\textrm{.}8\cos 60{}^\circ \mathbf{i}+0.\textrm{.}8\sin 60{}^\circ \mathbf{j} \\ & m(\mathbf{i})=m(0\textrm{.}8\c...) (top)
- 17:01, 29 September 2010 (hist) (diff) Answer 16.8 (New page: <math> 0\textrm{.}916 \ \text{ m}{{\text{s}}^{-1}}</math>) (top)
- 18:57, 23 September 2010 (hist) (diff) Solution 16.7b (New page: From part a), <math>\mathbf{v}=2\textrm{.}5\mathbf{i}+4\mathbf{j}</math> which gives, <math>\begin{align} & \tan \alpha =\frac{4}{2\textrm{.}5} \\ & \alpha =58\textrm{.}0{}^\circ \\ \...)
- 18:24, 23 September 2010 (hist) (diff) Solution 16.7a (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}}</mat...) (top)
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