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 #3
avatar+26364 
+1

We have a right triangle triangle ABC where the legs AB and BC have lengths 6 and 3sqrt3, respectively.

Medians AM and CN meet at point P. What is the length of CP?

 

\(\begin{array}{rcll} \text{Let } \vec{a} &=& \vec{CB} \\ \text{Let } \vec{b} &=& \vec{AC} \\ \text{Let } \vec{c} &=& \vec{AB} = \vec{a} + \vec{b} \\ \text{Let } \vec{u} &=& \vec{CN} = \frac12\vec{c} - \vec{b} = \frac12(\vec{a} + \vec{b}) - \vec{b} = \mathbf{ \frac12 \vec{a} - \frac12 \vec{b} } \\ \text{Let } \vec{v} &=& \vec{AM} = \mathbf{ \vec{b} + \frac12 \vec{a} } \\ \end{array} \)

 

\(\begin{array}{|lrcll|} \hline \mathbf{ \vec{b}+\lambda \vec{u} - \mu \vec{v} } & \mathbf{=} & \mathbf{0} \\\\ & \vec{b}+\lambda \vec{u} - \mu \vec{v} &=& 0 \quad & | \quad \mathbf{ \vec{u} = \frac12 \vec{a} - \frac12 \vec{b} } \\ & \vec{b}+\lambda (\frac12 \vec{a} - \frac12 \vec{b}) - \mu \vec{v} &=& 0 \quad & | \quad \mathbf{ \vec{v} = \vec{b} + \frac12 \vec{a} } \\ & \vec{b}+\lambda (\frac12 \vec{a} - \frac12 \vec{b}) - \mu (\vec{b} + \frac12 \vec{a}) &=& 0 \\ & \vec{b}+ \frac{\lambda}{2} \vec{a} - \frac{\lambda}{2} \vec{b} - \mu \vec{b} - \frac{\mu}{2} \vec{a} &=& 0 \\ & \frac{\vec{a}}{2} \cdot (\underbrace{\lambda -\mu}_{=0}) &=& \vec{b} \cdot ( \underbrace{\frac{\lambda}{2} + \mu - 1}_{=0} ) \\\\ (1) & \lambda -\mu &=& 0 \\ & \mu &=& \lambda \\\\ (2) & \frac{\lambda}{2} + \mu - 1 &= & 0 \\ & \frac{\lambda}{2} + \lambda - 1 &= & 0 \\ & \frac{3}{2}\lambda &= & 1 \\ & \lambda &= & \frac{2}{3} \\ \hline \end{array} \)

 

\(\begin{array}{|rcll|} \hline \overline{CP} &=& \lambda \cdot \overline{CN} \quad & | \quad \overline{CN}^2 = (3\cdot \sqrt{3})^2 + (\frac{6}{2})^2 = 36 \\ \overline{CP} &=& \frac{2}{3} \cdot 6 \quad & | \quad \overline{CN} = \sqrt{36} = 6 \\ \overline{CP} &=& 2\cdot 2 \\ \mathbf{\overline{CP}} & \mathbf{=} & \mathbf{4} \\ \hline \end{array} \)

 

 

laugh

13.07.2017
 #2
avatar+26364 
0

How far away would light have to be from a black hole with the mass of the Earth (6 x 1024 kg)

in order to be able to escape from its gravitational pull?

(Note: use the formula for escape speed again: vesc = 1.15 × 10-5 MR−−√MR )

 

Let speed of light \(c = 3\cdot 10^8\ \frac{m}{s}\)

Let gravitational constant \( = G \) \( \)

Let Radius = R

 

\(\begin{array}{|rcll|} \hline v_{esc} &=& \sqrt{\frac{2 G M_{\text{Earth}} }{R} } \\ v_{esc} &=& \sqrt{2 G} \times \sqrt{\frac{ M_{\text{Earth}} }{R} } \quad & | \quad \sqrt{2 G} \approx 1.15 \cdot 10^{-5} \\ v_{esc} &=& 1.15 \cdot 10^{-5} \times \sqrt{\frac{ M_{\text{Earth}} }{R} } \quad & | \quad v_{esc} = c \\ c &=& 1.15 \cdot 10^{-5} \times \sqrt{\frac{ M_{\text{Earth}} }{R} } \\ c^2 &=& 1.15^2 \cdot (10^{-5})^2 \cdot \frac{ M_{\text{Earth}} }{R} \\ R &=& 1.15^2 \cdot (10^{-5})^2 \cdot \frac{ M_{\text{Earth}} }{c^2} \quad & | \quad M_{\text{Earth}} = 6 \cdot 10^{24}\ kg \qquad c = 3\cdot 10^8\ \frac{m}{s} \\ &=& 1.15^2 \cdot (10^{-5})^2 \cdot \frac{ 6 \cdot 10^{24} }{(3\cdot 10^8)^2} \\ &=& 1.15^2 \cdot 10^{-10} \cdot \frac{ 6 \cdot 10^{24} }{ 9\cdot 10^{16} } \\ &=& 1.15^2 \cdot 10^{-10} \cdot \frac{2}{3} \cdot 10^{(24-16)} \\ &=& 1.15^2 \cdot \frac{2}{3} \cdot 10^{(24-16-10)} \\ &=& 0.88167 \cdot 10^{-2} \ m \\ &=& 8.8167 \cdot 10^{-3} \ m \\ &=& 8.817 \ mm \\\\ \mathbf{R} & \mathbf{ \approx }& \mathbf{ 8.8 \ mm } \\ \hline \end{array} \)

 

The light would have to be more than \(\mathbf{8.8 \ mm}\) from the black hole with the mass of the Earth

in order to be able to escape from its gravitational pull

 

laugh

07.07.2017