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- 11:45, 10 April 2010 (hist) (diff) Solution 6.8a (New page: Using <math>v=u+at\ \</math> gives)
- 16:28, 8 April 2010 (hist) (diff) Solution 6.7b (New page: Using <math>s=\frac{1}{2}(u+v)t\ \</math> gives <math>s=\frac{1}{2}\times \left( 6+19 \right)\times 20=250\ \text{m}</math>) (top)
- 16:23, 8 April 2010 (hist) (diff) Solution 6.7a (New page: Using <math>v=u+at\ \</math> gives <math>\begin{align} & 19-6=a\times 20 \\ & \\ & a=\frac{13}{20}=0\textrm{.}65\ \text{m}{{\text{s}}^{-2}} \\ \end{align}</math>) (top)
- 16:17, 8 April 2010 (hist) (diff) Solution 6.6b (New page: Using <math>v=u+at\ \</math> gives <math>\begin{align} & 24-9=0\textrm{.}5t \\ & t=30\ \text{s} \\ \end{align}</math> Here we have used the result for the acceleration obtained in the...) (top)
- 16:12, 8 April 2010 (hist) (diff) Solution 6.6a (New page: Using <math>{{v}^{\ 2}}={{u}^{\ 2}}+2as\ \</math> gives <math>\begin{align} & {{24}^{2}}-{{9}^{2}}=2\times a\times 495 \\ & 576-81=990 \times a \\ & a=\frac{495}{990}=0\textrm{.}5\ \tex...) (top)
- 15:57, 8 April 2010 (hist) (diff) Solution 6.5b (New page: Using <math>v=u+at\ \</math> gives <math>v=5+0\textrm{.}2\times 5=6\text{ m}{{\text{s}}^{-1}}</math> The distance travelled during this stage using <math>s=ut+\frac{1}{2}a{{t}^{\ 2}}\ \<...) (top)
- 15:43, 8 April 2010 (hist) (diff) Solution 6.5a (New page: Using <math>s=ut+\frac{1}{2}a{{t}^{\ 2}}\ \</math> gives <math>s=0+\frac{1}{2}0\textrm{.}5\times {{10}^{2}}=25\ \text{m}</math> Using <math>v=u+at\ \</math> gives <math>v=0+0\textrm{.}...) (top)
- 10:38, 7 April 2010 (hist) (diff) Solution 6.4b (New page: Using <math>v=u+at\ \</math> gives <math>\begin{align} & 15=5+0\textrm{.}2\times t \\ & \\ \end{align}</math> where we have used the acceleration calculated in part a). This gives <...) (top)
- 10:29, 7 April 2010 (hist) (diff) Solution 6.4a (top)
- 10:02, 7 April 2010 (hist) (diff) Solution 6.4a (New page: Using <math>{{v}^{\ 2}}={{u}^{\ 2}}+2as</math> gives)
- 09:51, 7 April 2010 (hist) (diff) Solution 6.3c (New page: This question can be solved by examining the curve. The lift has a speed of <math>1 \text{ m}{{\text{s}}^{\text{-1}}}</math> halfway in the first part and halfway in the third part. Insp...) (top)
- 09:34, 7 April 2010 (hist) (diff) Solution 6.3b (New page: Acceleration at any point is given by the gradient in a velocity-time graph. On the final stage of the motion the gradient is given by <math>\frac{\left( 9-7 \right)\text{ m}{{\text{s}}...) (top)
- 09:21, 7 April 2010 (hist) (diff) Solution 6.3a (New page: The area under the velocity-time graph gives the distance travelled. This area can be calculated by breaking up the area into three parts, two triangles and a rectangle This method gives...) (top)
- 08:58, 7 April 2010 (hist) (diff) Solution 6.2b (New page: Acceleration at any point is given by the gradient in a velocity-time graph. On the first stage of the motion the gradient is given by <math>\frac{24 \text{ m}{{\text{s}}^{\text{-1}}}}{...) (top)
- 08:49, 7 April 2010 (hist) (diff) Solution 6.2a (New page: The area under the velocity-time graph gives the distance travelled. This area can be calculated using the formula for the area of a trapedium as the graph in the figure defines a trapezi...) (top)
- 08:43, 7 April 2010 (hist) (diff) Solution 6.1 (New page: The area under the velocity-time graph gives the distance travelled. This area can be calculated either by breaking up the area into three parts, two triangles and a rectangle or by using...)
- 16:32, 6 April 2010 (hist) (diff) Solution 5.9 (New page: Image:5.9.gif As the lorry and thus all parts of the lorry have constant velocity the forces on the load must be in equilibrium. Resolving horisontally to the right <math>\begin{al...) (top)
- 15:31, 6 April 2010 (hist) (diff) Image:5.9.gif (top)
- 15:15, 6 April 2010 (hist) (diff) Solution 5.8 (top)
- 14:44, 6 April 2010 (hist) (diff) Image:5.8.gif (top)
- 14:43, 6 April 2010 (hist) (diff) Solution 5.8 (New page: Image:5.8.gif)
- 11:33, 6 April 2010 (hist) (diff) Solution 5.7b (New page: Resolving vertically upwards, <math>T\cos {{30}^{\circ }}+S-mg=0</math> where <math>S</math> is the spring force '''which we assume is upwards'''. This gives <math>S=mg-\ T\cos {{30}...) (top)
- 10:24, 6 April 2010 (hist) (diff) Solution 5.7a (New page: Resolving to the right, <math>45\ \text{N}-T\cos {{60}^{\circ }}=0</math> giving <math>T=\frac{45}{0\textrm{.}5}=90\ \text{N}</math>) (top)
- 17:47, 5 April 2010 (hist) (diff) m Solution 5.6c (New page: First the maximum friction must be calculated. To do this we need the normal reaction force <math>R</math>. Resolving in the normal direction <math>\begin{align} & R-mg\cos {{20}^{\circ...) (top)
- 16:49, 5 April 2010 (hist) (diff) Answer 5.6b (top)
- 16:39, 5 April 2010 (hist) (diff) Image:5.6b.gif (top)
- 16:33, 5 April 2010 (hist) (diff) Answer 5.6b (New page: Image:5.6.gif)
- 16:32, 5 April 2010 (hist) (diff) Image:5.6.gif (top)
- 16:56, 4 April 2010 (hist) (diff) Solution 5.5d (New page: The vertical components of the forces on the tank cancel out. The resultant of all the forces on the tank is thus the resultant of the horisontal components of the forces. Maximim fricti...)
- 16:47, 4 April 2010 (hist) (diff) Solution 5.5c (New page: Image:5.5.gif The horisontal component <math>H</math> of the 5000 N force is <math>H=5000\cos {{25}^{\circ }}=4531\ \text{N}</math> The maximum friction is <math>\mu R=0.3\times...) (top)
- 16:38, 4 April 2010 (hist) (diff) Solution 5.5b
- 16:14, 4 April 2010 (hist) (diff) Solution 5.5b (New page: Image:5.5.gif)
- 16:14, 4 April 2010 (hist) (diff) Image:5.5.gif (top)
- 15:31, 4 April 2010 (hist) (diff) Solution 5.5a (New page: <math>\begin{align} & \mathbf{F}= F\cos \alpha \mathbf{i}+F\sin \alpha \mathbf{j} =\left( 5000\times \cos {{25}^{\circ }} \right)\mathbf{i}+\left( 5000\times \sin {{25}^{\circ }} \right...) (top)
- 12:44, 4 April 2010 (hist) (diff) Solution 5.4c (New page: <math>\begin{align} & F<\mu R \\ & \\ & 98<\mu 170 \\ & \\ \end{align}</math> or <math>\mu >\frac{98}{170}=0\textrm{.}58</math> correct to two significant figures) (top)
- 12:34, 4 April 2010 (hist) (diff) Solution 5.4b (New page: <math>mg=20\times 9\textrm{.}8=196\ \text{N}</math> Resolving up the plane <math>\begin{align} & F-mg\sin {{30}^{\circ }}=0 \\ & F=196\times 0\textrm{.}5=98\ \text{N} \\ \end{align}<...) (top)
- 12:26, 4 April 2010 (hist) (diff) Answer 5.4a (top)
- 12:26, 4 April 2010 (hist) (diff) Image:5.4a.gif (top)
- 12:12, 4 April 2010 (hist) (diff) Solution 5.3b (top)
- 12:02, 4 April 2010 (hist) (diff) Solution 5.3b (New page: From the previous part we have that <math>\mathbf{F}4=-\mathbf{F}1-\mathbf{F}2-\mathbf{F}3=-292\mathbf{i}-62\mathbf{j}\ \text{N}</math> Thus the magnitude <math>Q</math> is given by <ma...)
- 11:04, 4 April 2010 (hist) (diff) Solution 5.3a (top)
- 10:46, 4 April 2010 (hist) (diff) Solution 5.3a (New page: From the figure the three known forces can be calculated using, <math>\begin{align} \mathbf{F} =F\cos \alpha \mathbf{i}+F\sin \alpha \mathbf{j} \end{align}</math> We get <math>\begin...)
- 10:21, 4 April 2010 (hist) (diff) 5. Exercises
- 16:15, 2 April 2010 (hist) (diff) Image:E5.2.GIF (uploaded a new version of "Image:E5.2.GIF") (top)
- 16:11, 2 April 2010 (hist) (diff) Image:E5.2.GIF (uploaded a new version of "Image:E5.2.GIF")
- 16:01, 2 April 2010 (hist) (diff) Solution 5.2a (top)
- 16:00, 2 April 2010 (hist) (diff) Solution 5.2b (New page: Resolving horisontally to the right <math>\begin{align} & {{T}_{2}}-{{T}_{1}}\cos {{40}^{\circ }}=0 \\ & \\ & {{T}_{2}}={{T}_{1}}\cos {{40}^{\circ }}=762\times 0\textrm{.}766=584\ \te...) (top)
- 15:54, 2 April 2010 (hist) (diff) Solution 5.2a (New page: Resolving vertically <math>\begin{align} & {{T}_{1}}\cos {{50}^{\circ }}-mg=0 \\ & \\ & {{T}_{1}}=\frac{mg}{\cos {{50}^{\circ }}}=\frac{490}{0\textrm{.}643}=762\ \text{N} \\ \end{ali...)
- 15:43, 2 April 2010 (hist) (diff) 5. Exercises
- 15:42, 2 April 2010 (hist) (diff) Answer 5.1aNew (New page: Image:5.1aNew.gif) (top)
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