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- 17:46, 6 October 2010 (hist) (diff) Solution 18.4b (New page: <math>\begin{align} & v(6)=36\times 6-3\times {{6}^{2}} \\ & =108\text{ m}{{\text{s}}^{\text{-1}}} \end{align}</math>) (top)
- 17:45, 6 October 2010 (hist) (diff) Solution 18.4a (New page: <math>\begin{align} & v=\frac{ds}{dt}=36t-3{{t}^{2}} \\ & a=\frac{dv}{dt}=36-6t \\ & 36-6t=0 \\ & t=\frac{36}{6}=6\text{ s} \\ \end{align}</math>) (top)
- 17:44, 6 October 2010 (hist) (diff) Answer 18.4c (New page: <math>\text{36 m}{{\text{s}}^{-\text{2}}}</math>) (top)
- 17:42, 6 October 2010 (hist) (diff) Answer 18.4b (New page: <math>108\text{ m}{{\text{s}}^{\text{-1}}}</math>) (top)
- 17:41, 6 October 2010 (hist) (diff) Answer 18.4a (New page: 6 s) (top)
- 17:41, 6 October 2010 (hist) (diff) Answer 18.3c (top)
- 17:38, 6 October 2010 (hist) (diff) Solution 18.3d
- 17:37, 6 October 2010 (hist) (diff) Answer 18.3d (New page: 2.5 m)
- 17:35, 6 October 2010 (hist) (diff) Solution 18.3c (top)
- 17:34, 6 October 2010 (hist) (diff) Answer 18.3c (New page: t=5 s)
- 17:33, 6 October 2010 (hist) (diff) Solution 18.3b (New page: <math>s=\frac{3\times {{10}^{2}}}{25}-\frac{{{10}^{3}}}{250}=12-4=8\text{ m}</math>)
- 17:31, 6 October 2010 (hist) (diff) Solution 18.3a (top)
- 17:30, 6 October 2010 (hist) (diff) Solution 18.3a (New page: <math>v=\frac{dh}{dt}=\frac{3t}{25}-\frac{3{{t}^{2}}}{250}</math> When <math>t=\text{1}0</math> <math>v=\frac{3\times 10}{25}-\frac{3\times {{10}^{2}}}{250}=1 \textrm{.}2-1 \textrm{.}2...)
- 17:24, 6 October 2010 (hist) (diff) Solution 18.3d (New page: 2.5 m)
- 17:23, 6 October 2010 (hist) (diff) Solution 18.3c (New page: t=5 s)
- 17:23, 6 October 2010 (hist) (diff) Answer 18.3b (New page: 8 m)
- 15:47, 4 October 2010 (hist) (diff) Solution 18.2c (New page: Using Newton's second law <math>F=ma</math> and the result from part a) <math>F=4\times 10=40\text{ N}</math>) (top)
- 15:42, 4 October 2010 (hist) (diff) Solution 18.2b (New page: <math>\begin{align} & 10t-6=0 \\ & t=0\textrm{.}6\text{ s} \\ \end{align}</math>) (top)
- 15:40, 4 October 2010 (hist) (diff) Solution 18.2a (New page: <math>\begin{align} & v=\frac{ds}{dt}=10t-6 \\ & a=\frac{dv}{dt}=10\ \text{m}{{\text{s}}^{-2}} \\ \end{align}</math>) (top)
- 15:39, 4 October 2010 (hist) (diff) Answer 18.2c (New page: 40 N) (top)
- 15:38, 4 October 2010 (hist) (diff) Answer 18.2b (New page: 0.6 s) (top)
- 15:36, 4 October 2010 (hist) (diff) Answer 18.2a (New page: <math>v=10t-6,\quad a=10\ \text{m}{{\text{s}}^{-2}}</math>) (top)
- 15:30, 4 October 2010 (hist) (diff) Solution 18.1c
- 15:29, 4 October 2010 (hist) (diff) Solution 18.1c (New page: Using the result of part b) with <math>a=0</math> gives <math>\begin{align} & 2t-\frac{3{{t}^{2}}}{20}=0 \\ & t=0\ \text{s }\ \text{or }\ t=\frac{40}{3}=13.3\ \text{s} \\ \end{align}<...)
- 15:25, 4 October 2010 (hist) (diff) Solution 18.1b (New page: <math>a=\frac{dv}{dt}=2t-\frac{3{{t}^{2}}}{20}</math>)
- 15:24, 4 October 2010 (hist) (diff) Solution 18.1a (New page: <math>v=\frac{ds}{dt}={{t}^{2}}-\frac{{{t}^{3}}}{20}</math>)
- 15:23, 4 October 2010 (hist) (diff) Answer 18.1c (New page: <math>t=0\text{ }\ \text{ or } \text{ } \ t=\frac{40}{3}=13\textrm{.}3</math>)
- 15:19, 4 October 2010 (hist) (diff) Answer 18.1b (New page: <math>a=2t-\frac{3{{t}^{2}}}{20}</math>)
- 15:17, 4 October 2010 (hist) (diff) Answer 18.1a (New page: <math>v={{t}^{2}}-\frac{{{t}^{3}}}{20}</math>)
- 12:12, 4 October 2010 (hist) (diff) Solution 17.9b (New page: In this case there is resistance to the motion. Energy at start = Energy lost + Energy and finish. <math>\begin{align} & 40\times 9 \textrm{.}8\times 12=\text{Energy Lost }+\frac{1}{2}\t...) (top)
- 12:09, 4 October 2010 (hist) (diff) Solution 17.9a (New page: Gain in kinetic energy = Loss in potential energy. <math>\begin{align} & 40\times 9 \textrm{.}8\times 12=\frac{1}{2}\times 40\times {{v}^{2}} \\ & v=15 \textrm{.}3\text{ m}{{\text{s}}^{\...) (top)
- 12:03, 4 October 2010 (hist) (diff) Answer 17.9b (New page: 3242 J) (top)
- 12:03, 4 October 2010 (hist) (diff) Answer 17.9a (New page: <math>15 \textrm{.}3\text{ m}{{\text{s}}^{\text{-1}}}</math>) (top)
- 12:00, 4 October 2010 (hist) (diff) 17. Exercises
- 11:59, 4 October 2010 (hist) (diff) Solution 17.8a (New page: The energy is constant. Thus the energy in the beginning is equal to the energy at the end. <math>\begin{align} & \frac{1}{2}\times m\times {{18}^{2}}=\frac{1}{2}\times m\times {{12}^{2}}...) (top)
- 11:56, 4 October 2010 (hist) (diff) Answer 17.8a (New page: 9.18 m) (top)
- 11:56, 4 October 2010 (hist) (diff) 17. Exercises
- 11:55, 4 October 2010 (hist) (diff) Answer 17.7a
- 11:54, 4 October 2010 (hist) (diff) Answer 17.8b (New page: The mass does not need to be known as it cancels out in the equation.) (top)
- 11:53, 4 October 2010 (hist) (diff) Answer 17.7a
- 18:00, 2 October 2010 (hist) (diff) Solution 17.7b (top)
- 17:54, 2 October 2010 (hist) (diff) Solution 17.7b
- 17:53, 2 October 2010 (hist) (diff) Image:17.7d.gif (top)
- 17:50, 2 October 2010 (hist) (diff) Image:17.7c.gif (top)
- 17:44, 2 October 2010 (hist) (diff) Image:17.7b.gif (top)
- 16:53, 2 October 2010 (hist) (diff) Solution 17.7b
- 16:50, 2 October 2010 (hist) (diff) Solution 17.7b (New page: Image:17.7.gif)
- 16:48, 2 October 2010 (hist) (diff) Solution 17.7a (New page: <math>KE=\frac{1}{2}\times 0\textrm{.}05\times {{1\textrm{.}2}^{2}}=0\textrm{.}036\text{ J}</math>) (top)
- 16:46, 2 October 2010 (hist) (diff) Answer 17.7b (New page: <math>\text{24}\text{.}{{\text{7}}^{\circ }}</math>) (top)
- 16:44, 2 October 2010 (hist) (diff) Answer 17.7a (New page: 0.036 J)
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