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From Mechanics
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- 17:33, 18 January 2011 (hist) (diff) Solution to Test Paper 1
- 17:22, 18 January 2011 (hist) (diff) Image:Test1ans2.gif (top)
- 17:19, 18 January 2011 (hist) (diff) Solution to Test Paper 1
- 17:07, 18 January 2011 (hist) (diff) Image:Test1ans1.gif (top)
- 18:09, 17 January 2011 (hist) (diff) Solution to Test Paper 1
- 14:49, 17 January 2011 (hist) (diff) Solution to Test Paper 1
- 14:42, 17 January 2011 (hist) (diff) Solution to Test Paper 1
- 19:55, 13 January 2011 (hist) (diff) Test (New page: {| border="1" |1 (a) || <math>\begin{align} & 44.1=\frac{1}{2}\times 9.8{{t}^{2}} \\ & t=\sqrt{\frac{44.1}{4.9}}=3\text{ s} \\ \end{align}</math> || 3 marks |- | Cell B | Cell C | C...) (top)
- 19:43, 13 January 2011 (hist) (diff) Solution to Test Paper 2 (New page: {| class="wikitable" style="text-align: center; width: 200px; height: 200px;" |- | 1 || 2 || 3 |- | 2 || 4 || 6 |- | 3 || 6 || 9 |- | 4 || 8 || 12 |- | 5 || 10 || 15 |})
- 19:15, 13 January 2011 (hist) (diff) Solution to Test Paper 1
- 18:11, 13 January 2011 (hist) (diff) Solution to Test Paper 1 (New page: <table class="wikitable" style="text-align: left;"> <th>×</th> <th>1</th> <th>2</th> <th>3</th> <tr> <td>1</td> <td>1</td> <td>2</td> <td>3</td> </tr> <tr> <td>2</td> <td>2</td> <td>4</td...)
- 16:04, 12 January 2011 (hist) (diff) Practice Test Paper 2
- 16:03, 12 January 2011 (hist) (diff) Image:Test2.4.gif (top)
- 15:44, 12 January 2011 (hist) (diff) Practice Test Paper 2
- 15:24, 12 January 2011 (hist) (diff) Practice Test Paper 2
- 14:55, 12 January 2011 (hist) (diff) Practice Test Paper 2 (New page: <math>\large {\underline{ {\textbf {Practice Test Paper (1 hour)}}}}</math> '''Note that the Mechanics Test counts for <math> \bf 50\%</math> of the final mechanics mark.''' 1. A ball ...)
- 14:31, 12 January 2011 (hist) (diff) Practice Test Paper 1 (top)
- 14:25, 12 January 2011 (hist) (diff) Image:Test1.4a.gif (top)
- 18:26, 11 January 2011 (hist) (diff) Practice Test Paper 1
- 18:10, 11 January 2011 (hist) (diff) Practice Test Paper 1
- 18:00, 11 January 2011 (hist) (diff) Practice Test Paper 1
- 17:59, 11 January 2011 (hist) (diff) Image:Test1.4.gif (top)
- 18:54, 10 January 2011 (hist) (diff) Practice Test Paper 1
- 18:39, 10 January 2011 (hist) (diff) Practice Test Paper 1
- 17:38, 10 January 2011 (hist) (diff) Practice Test Paper 1
- 16:43, 10 January 2011 (hist) (diff) Practice Test Paper 1
- 16:41, 10 January 2011 (hist) (diff) Practice Test Paper 1
- 12:50, 10 January 2011 (hist) (diff) Practice Test Paper 1 (Replacing page with ' test')
- 12:50, 10 January 2011 (hist) (diff) m Practice Test Paper 1 (New page: __NOTOC__ {| border="0" cellspacing="0" cellpadding="0" height="30" width="100%" | style="border-bottom:1px solid #797979" width="5px" | {{Selected tab|[[2. Introduction to force an...)
- 18:34, 14 October 2010 (hist) (diff) Answer 11.5 (New page: Image:11.5.gif) (top)
- 18:33, 14 October 2010 (hist) (diff) Image:11.5.gif (top)
- 18:25, 14 October 2010 (hist) (diff) 11. Exercises (top)
- 15:12, 12 October 2010 (hist) (diff) 20. Circular Motion
- 15:01, 12 October 2010 (hist) (diff) Solution 20.6b (New page: <math>F=\mu R=0\textrm{.}8\times 1200\times 9\textrm{.}8=9408\text{ N}</math>) (top)
- 14:59, 12 October 2010 (hist) (diff) Solution 20.6a (New page: The maximum value for the speed <math>v</math> is determined by the maximum value which the frictional force <math>F</math>can assume which is: <math>F=\mu R</math> gives <math>\begin{a...) (top)
- 14:51, 12 October 2010 (hist) (diff) Answer 20.6b (New page: 9408 N) (top)
- 14:51, 12 October 2010 (hist) (diff) Answer 20.6a (New page: <math>33\textrm{.}1\text{ m}{{\text{s}}^{\text{-1}}}</math>) (top)
- 14:49, 12 October 2010 (hist) (diff) Solution 20.5b (New page: <math>\begin{align} & F\le \mu R \\ & R=mg=1500\times 9\textrm{.}8 \\ & F=m\frac{{{v}^{2}}}{r}=1500\times \frac{{{15}^{2}}}{80} \\ & 1500\times \frac{{{15}^{2}}}{80}\le \mu \times 1500\...) (top)
- 14:42, 12 October 2010 (hist) (diff) Solution 20.5a (New page: <math>F=m\frac{{{v}^{2}}}{r}=1500\times \frac{{{15}^{2}}}{80}=4129\text{ N}</math>) (top)
- 14:41, 12 October 2010 (hist) (diff) Answer 20.5b (New page: 0.287) (top)
- 14:40, 12 October 2010 (hist) (diff) Answer 20.5a (New page: 4129 N) (top)
- 14:15, 12 October 2010 (hist) (diff) Solution 20.4c (New page: No change because the mass cancels out in all of the equations. For example in part b): <math>\begin{align} & mr{{\omega }^{2}}=\mu mg \\ & r=\frac{\mu g}{{{\omega }^{2}}} \end{align}<...) (top)
- 14:14, 12 October 2010 (hist) (diff) Solution 20.4b (New page: If <math>F=\mu R</math> then the coin can remain on the surface. This is the maximum value that <math>F</math> can take. That is <math>mr{{\omega }^{2}}=\mu R</math>. This gives <math>\...) (top)
- 14:02, 12 October 2010 (hist) (diff) Solution 20.4a (New page: Converting the angular speed to standard units. <math>\omega =30\text{ rpm }=\frac{30\times 2\pi }{60}=\pi \text{ rad}{{\text{s}}^{\text{-1}}}</math> For circular motion: <math>\begin{a...) (top)
- 13:51, 12 October 2010 (hist) (diff) Answer 20.4c (New page: No change.) (top)
- 13:51, 12 October 2010 (hist) (diff) Answer 20.4b (New page: 59.6 cm) (top)
- 13:50, 12 October 2010 (hist) (diff) Answer 20.4a (New page: Coin does not slide.) (top)
- 13:20, 12 October 2010 (hist) (diff) Solution 20.3 (top)
- 13:18, 12 October 2010 (hist) (diff) Solution 20.3 (New page: We start by converting the speed to SI units. <math>60\text{ km/h }=\frac{60\times 1000}{60\times 60}=\frac{50}{3}=\text{16}\textrm{.}\text{7 m}{{\text{s}}^{\text{-1}}}</math> The diagra...)
- 13:15, 12 October 2010 (hist) (diff) Image:20.3.gif (top)
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