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- 09:56, 20 May 2010 (hist) (diff) Solution 10.1a
- 09:03, 20 May 2010 (hist) (diff) Solution 10.1a
- 08:53, 20 May 2010 (hist) (diff) 10. Newton’s second law
- 08:43, 20 May 2010 (hist) (diff) Image:10.1.gif (uploaded a new version of "Image:10.1.gif") (top)
- 08:39, 20 May 2010 (hist) (diff) Solution 10.1a (New page: Image:10.1.gif)
- 08:39, 20 May 2010 (hist) (diff) Image:10.1.gif
- 17:48, 18 May 2010 (hist) (diff) Solution 9.6b (top)
- 17:33, 18 May 2010 (hist) (diff) Solution 9.6b (New page: Image:9.6.gif)
- 17:29, 18 May 2010 (hist) (diff) Solution 9.6a (top)
- 17:08, 18 May 2010 (hist) (diff) Image:9.6.gif (top)
- 17:08, 18 May 2010 (hist) (diff) Solution 9.6a (New page: Image:9.6.gif)
- 12:42, 18 May 2010 (hist) (diff) Solution 9.5
- 12:23, 18 May 2010 (hist) (diff) Image:9.5.gif (uploaded a new version of "Image:9.5.gif") (top)
- 12:09, 18 May 2010 (hist) (diff) Solution 9.5 (New page: Image:9.5.gif In the figure <math>R1</math> is the air resistance. Resolving up the plane <math>\begin{align} & F1+F-mg\sin {{30}^{\circ }}=0 \\ & \\ & F1=mg-F\sin {{30}^{\circ ...)
- 12:06, 18 May 2010 (hist) (diff) Image:9.5.gif
- 09:48, 18 May 2010 (hist) (diff) 9. Newton’s first law
- 11:11, 17 May 2010 (hist) (diff) Solution 9.4 (top)
- 11:04, 17 May 2010 (hist) (diff) Solution 9.4 (New page: Image:9.4.gif As the block is moving relative to the surface the friction force must satisfy <math>F = \mu R</math> Also as the block has conmstant velocity the forces on it must be...)
- 10:44, 17 May 2010 (hist) (diff) Image:9.4.gif (top)
- 08:26, 17 May 2010 (hist) (diff) Solution 9.3 (New page: The bicycle and cyclist have constant velocity, thus the two horisontal forces must cancel out, giving the resistive force is 28 N.) (top)
- 08:14, 17 May 2010 (hist) (diff) Answer 7.3c (top)
- 18:03, 16 May 2010 (hist) (diff) Solution 9.2c (top)
- 17:41, 16 May 2010 (hist) (diff) Solution 9.2c (New page: The sum of the forces is zero as the body has constant velocity. This means the sum of the forces in all directions is zero. Resolving horisontally, <math>\begin{align} & \to :\quad 42+...)
- 17:25, 16 May 2010 (hist) (diff) Solution 9.2b (New page: The sum of the forces is zero as the body has constant velocity. This means the sum of the forces in all directions is zero. Resolving horisontally, <math>\begin{align} & \to :\quad 42\c...) (top)
- 17:17, 16 May 2010 (hist) (diff) Solution9.2a (top)
- 17:13, 16 May 2010 (hist) (diff) Solution9.2a (New page: The sum of the forces is zero as the body has constant velocity. This means the sum of the forces in all directions is zero. Horisontally, <math>\begin{align} & \to :\quad 12\ \text{N}-P...)
- 16:09, 16 May 2010 (hist) (diff) Solution 9.1 (New page: Image:9.1.gif As the parachutist is falling vertically (in other words in a straight line) and at constant speed, her velocity is constant. This means the forces acting on her must ha...) (top)
- 16:04, 16 May 2010 (hist) (diff) Image:9.1.gif (uploaded a new version of "Image:9.1.gif") (top)
- 15:59, 16 May 2010 (hist) (diff) Image:9.1.gif
- 15:45, 16 May 2010 (hist) (diff) 9. Newton’s first law
- 15:37, 16 May 2010 (hist) (diff) 9. Newton’s first law
- 17:53, 14 May 2010 (hist) (diff) Image:7.3.gif (top)
- 17:53, 14 May 2010 (hist) (diff) Answer 7.3b (New page: Image:7.3.gif) (top)
- 17:16, 14 May 2010 (hist) (diff) Image:7.2.gif (top)
- 17:16, 14 May 2010 (hist) (diff) Solution 7.2 (New page: Image:7.2.gif) (top)
- 15:00, 14 May 2010 (hist) (diff) 7. Exercises (top)
- 14:56, 14 May 2010 (hist) (diff) Solution 7.4c (top)
- 14:42, 14 May 2010 (hist) (diff) Solution 7.4c
- 14:36, 14 May 2010 (hist) (diff) Solution 7.4a (top)
- 12:45, 14 May 2010 (hist) (diff) Solution 8.10b (top)
- 18:49, 11 May 2010 (hist) (diff) Solution 6.8c (top)
- 18:45, 11 May 2010 (hist) (diff) Answer 8.10b (top)
- 18:26, 11 May 2010 (hist) (diff) Answer 8.7c (top)
- 18:08, 11 May 2010 (hist) (diff) Answer 6.8c (top)
- 18:07, 11 May 2010 (hist) (diff) Solution 6.8c
- 17:28, 11 May 2010 (hist) (diff) Answer 6.8b
- 17:27, 11 May 2010 (hist) (diff) Solution 6.8b (top)
- 17:19, 11 May 2010 (hist) (diff) Answer 6.8a (top)
- 12:31, 20 April 2010 (hist) (diff) Solution 8.10c (top)
- 12:25, 20 April 2010 (hist) (diff) m Solution 8.10c (New page: We use the result from part a). <math>\mathbf{r}=(8\ \mathbf{i}+8 \ \mathbf{ j})t+\frac{1}{2}(0\textrm{.}01\mathbf{i}+0\textrm{.}02\mathbf{j}+0\textrm{.}1\mathbf{k}){{t}^{\ 2}}+10\mathbf{...)
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