Mitglied: heureka

If the 0.145 mm diameter tungsten filament in a light bulb is to have a resistance of 0.161 Ω at 20.0°C, how long should it be? The resistivity of tungsten at 20.0°C is 5.60 10-8 Ω · m

diameteter (d) = 0.145 mm,

resistance (R) = 0.161 Ω,

resistivity (ρ) = 5.60 *10-8 Ω · m

$\rho = \dfrac{RA}{L} \qquad \rm{A} = \dfrac{\pi\ d^2}{4} \\\\ \text{Length (L): } \qquad \boxed{L=\dfrac{\pi\ d^2 }{4} \dfrac{R}{\rho}}$

$\small{\text{ $ L= \left( \dfrac{ \pi * 0.145^2 *0.161 } { 4 * 5.60 \cdot 10^{-8} } \right) \left( \dfrac{mm^2\ \Omega}{\Omega\ m} \right) $ }} \qquad \boxed{ 1\, \dfrac{\Omega\ mm^2}{m} = 10^{-6}\ \Omega\ m \qquad 1\ \Omega\ m = \dfrac{10^6\ mm^2\ \Omega}{m} }\\\\ \small{\text{ $ L= \left( \dfrac{ \pi * 0.145^2 *0.161 } { 4 * 5.60 \cdot 10^{-8} } \right) \left( \dfrac{mm^2\ \Omega}{\dfrac{10^6\ mm^2\ \Omega}{m}} \right) $ }}\\\\\\ \small{\text{ $ L= \left( \dfrac{ \pi * 0.145^2 *0.161 } { 4 * 5.60 \cdot 10^{-8} } \right) \left( \dfrac{m}{10^6} \right) $ }}\\\\\\ \small{\text{ $ L= \left( \dfrac{ \pi * 0.145^2 *0.161 } { 4 * 5.60 \cdot 10^{-2} } \right) \ m $ }}\\\\ \small{\text{ $ L=\dfrac{ \pi * 0.145^2 *0.161 \cdot 10^2} { 4 * 5.60 } \ m $ }}\\\\ \small{\text{ $ L=0.04747486461 \ m $ }}\\ \small{\text{ $ L=4.747486461 \ cm \approx 4.75 \ cm $ }}$

heureka09.02.2015

+26396

How can i get the sum of interior angles of any polygon ?

The sum of interior angles of a 3 Point polygon = 1 ( triangle ) = $1*\pi$

The sum of interior angles of a 4 Point polygon = 2 ( triangle ) = $2*\pi$

The sum of interior angles of a 5 Point polygon = 3 ( triangle ) = $3*\pi$

The sum of interior angles of a 6 Point polygon = 4 ( triangle ) = $4*\pi$

$\dots$

The sum of interior angles of a n Point polygon = n-2 ( triangle ) = $\qquad\small{\text{$ (n-2)*\pi \quad $ or $ \quad (n-2)* 180 \ensurement{^{\circ}} $ }}$

heureka09.02.2015

+26396

Nullstellen von x+3+ln(x+3)

siehe: https://de.wikipedia.org/wiki/Fixpunktiteration

Wir setzen z= x+3

Die Gleichung muss zuerst in eine Fixpunktgleichung, also in eine Gleichung der Form

$z= \varphi(z)$

umgeformt werden.

$\\ z + \ln(z) = 0 \\ z= -\ln(z) \quad | \quad e^{()} \\ e^z = e^{ -\ln(z)} \\\\ e^z = \dfrac{1} { e^{ \ln(z)} }\\\\ e^z = \dfrac{1} { z } \\\\ z = \dfrac{1} { e^z } \\$

und erhalten die Gleichung:

$\boxed{z = \dfrac{1} { e^z } \small{\text{ mit } x = z-3}} \\ \small{\text{ Die Iterationsformel lautet nun: $ z_{i+1} = \dfrac{1} { e^{z_i} } $ }}$

Wir starten die Iteration mit

$\small{ $ z_0 = 1 \quad ( x = z_0-3 = 1-3 = -2)\\\\ z_1 = \dfrac{1}{e^1} = 0.36787944117 \\ z_2 = \dfrac{1}{e^{0.36787944117 } } =0.69220062756\\ z_3 = \dfrac{1}{e^{0.69220062756} } =0.50047350056\\ z_4 = \dfrac{1}{e^{0.50047350056} } =0.60624353509\\ z_5 = \dfrac{1}{e^{0.60624353509} } =0.54539578598\\ z_6 = \dfrac{1}{e^{0.54539578598} } =0.57961233550\\ z_7 = \dfrac{1}{e^{0.57961233550} } =0.56011546136\\ z_8 = \dfrac{1}{e^{0.56011546136} } =0.57114311508\\ z_9 = \dfrac{1}{e^{0.57114311508} } =0.56487934739\\ $ }}$

$\\ \small{ $ z_{10} = \dfrac{1}{e^{0.56487934739} } =0.56842872503\\ z_{11} = \dfrac{1}{e^{0.56842872503} } =0.56641473315\\ z_{12} = \dfrac{1}{e^{0.56641473315} } =0.56755663733\\ z_{13} = \dfrac{1}{e^{0.56755663733} } =0.56690891192\\ z_{14} = \dfrac{1}{e^{0.56690891192} } =0.56727623218\\ z_{15} = \dfrac{1}{e^{0.56727623218} } =0.56706789839\\ z_{16} = \dfrac{1}{e^{0.56706789839} } =0.56718605010\\ z_{17} = \dfrac{1}{e^{0.56718605010} } =0.56711904006\\ z_{18} = \dfrac{1}{e^{0.56711904006} } =0.56715704400\\ z_{19} = \dfrac{1}{e^{0.56715704400} } =0.56713549021\\ z_{20} = \dfrac{1}{e^{0.56713549021} } =0.56714771426\\ \dots $ }}$

$\\z = 0.56714329041 \\ x = z - 3= 0.56714329041 - 3\\ \textcolor[rgb]{1,0,0}{x = -2.43285670959}$

heureka09.02.2015

+26396

one over two plus one over four plus one over eight

$\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8} = ?\\\\ \boxed{ \dfrac{1}{2} = 2 * \left(\dfrac{1}{4} \right) \qquad \dfrac{1}{4} = 2 * \left(\dfrac{1}{8} \right) }\\\\ \dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8} = 2*2* \left(\dfrac{1}{8}\right) + 2 * \left(\dfrac{1}{8}\right) +\left(\dfrac{1}{8}\right) = \left(\dfrac{1}{8}\right) \left(4+2+1\right) = \dfrac{7}{8}$

heureka09.02.2015

+26396

Gliese 581e is an exoplanet with a mass of 1.9 Earths that orbits a red dwarf star at a distance of 5 x1010 m (0.33 au). If its orbital period is 124 days, find the mass of the star in kg. Divide your answer by the suns mass to see how much more or less massive the star is than our sun.

$1 \, au = 149\ 597\ 870\ 700 \; m$

We have:

$\small{\text{$T = 124\ days =124 * 24 * 60 * 60\ s = 1.07136*10^7\ s $}}$ and

$\small{\text{$a = 0.33 au \approx 5 * 10^{10}\ m$}}$ and

$\small{\text{ $ m_{Exoplanet}= 1.9* m_{earth} \qquad m_{earth} = 5.97219 * 10^{24}\ kg $ }}$ see: https://en.wikipedia.org/wiki/Earth_mass

$\small{\text{ $ m_{Exoplanet}= 1.13471610000* 10^{25}\ kg $ }}$

see: https://en.wikipedia.org/wiki/Kepler%27s_laws_of_planetary_motion

$\boxed{ \dfrac{T^2}{a^3} = \dfrac{ 4 \pi^2}{G(M+m)} } \qquad \boxed { M=\dfrac{ 4 \pi^2a^3}{G T^2 } - m }$

where
T is the orbital period of the orbiting body,
M is the mass of the star,
G is the universal gravitational constant and
a is the radius, i.e. the semi-major axis of the ellipse.
m is the mass of the orbiting body.

The gravitational constant is : see: https://en.wikipedia.org/wiki/Gravitational_constant

$\small{\text{ $ G = 6.67384 \cdot 10^{-11} \dfrac{m^3}{kg \cdot s^2} $ }}$

$\small{\text{ $ M=\dfrac{ 4 \pi^2 (5 * 10^{10}\ m)^3}{ 6.67384 \cdot 10^{-11} \dfrac{m^3}{kg \cdot s^2} * (1.07136*10^7\ s) ^2 } - 1.13471610000* 10^{25}\ kg $ }}\\\\\\ \small{\text{ $ M=\dfrac{ 4 \pi^2 *5^3 * 10^{30}}{ 6.67384 \cdot 10^{-11}* 1.07136^2*10^{14 }}\ kg - 1.13471610000* 10^{25}\ kg $ }}\\\\\\ \small{\text{ $ M=\dfrac{ 4 \pi^2 *5^3 * 10^{30}\cdot 10^{11}\cdot 10^{-14 } }{ 6.67384 * 1.07136^2}\ kg - 1.13471610000* 10^{25}\ kg $ }}\\\\\\ \small{\text{ $ M=\dfrac{ 4 \pi^2 *5^3 * 10^{27 } }{ 6.67384 * 1.07136^2}\ kg - 1.13471610000* 10^{25}\ kg $ }}\\\\ \small{\text{ $ M=6.44203535336 \cdot 10^{29}\ kg - 1.13471610000* 10^{25}\ kg $ }}\\ \small{\text{ $ M=10^{25} (6.44203535336 \cdot 10^{4} - 1.13471610000 )\ kg $ }}\\ \small{\text{ $ M=64419.2188175 \cdot 10^{25} \ kg $ }}\\ \small{\text{ $ M_{star}=6.44192188175 \cdot 10^{29} \ kg $ }}$

Solar mass see: https://en.wikipedia.org/wiki/Solar_mass

$\small{\text{ $ M_\odot= 1.98855 \cdot 10^{30}\ kg $ }}\\\\ \small{\text{ $ \dfrac{ M_{star} }{ M_\odot } = \dfrac{6.44192188175 \cdot 10^{29} \ kg } {1.98855 \cdot 10^{30}\ kg } $ }}\\\\ \small{\text{ $ = \dfrac{6.44192188175 \cdot 10^{-1}} {1.98855}$ }}\\\\ \small{\text{ $ = 3.23950711913 \cdot 10^{-1} $ }}\\ \small{\text{ $ = 0.323950711913 $ }}\\ \small{\text{ $ M_{star}= 0.324\ * M_\odot $ }}$

heureka09.02.2015

+26396

+10

An asteroid has a diameter of 10km and is at a distance of 10^6km, what angle in degrees does it subtend ?

$\tan{ ( \frac{ \Theta\ensurement{^{\circ}} } {2} ) } = \dfrac{ 5\ km } { 10^6\ km } = 5*10^{-7}\\\\ \frac{ \Theta\ensurement{^{\circ}} } {2} = 0.00002864789\ensurement{^{\circ}}\\\\ \Theta\ensurement{^{\circ}} =0.00005729578\ensurement{^{\circ}}\\\\ \Theta^{'}=\Theta\ensurement{^{\circ}} \times \dfrac{60^{'}} { 1\ensurement{^{\circ}} } = (0.00005729578 * 60)^{'} = 0.00343774677^{'}\\\\ \Theta^{"}=\Theta^{'}\times \dfrac{ 60^{"} } { 1^{'} } = (0.00343774677* 60)^{"} = 0.20626480625^{"}$

heureka09.02.2015

+26396

+10

$-4^{2}-(-4)^{2}-4^{2}=\small{\text{ ????????? }}$

$\\= -16-(16) - 16 \\ = -16-16-16 \\ = - 48$

heureka08.02.2015

+26396

+10

67º13'4"-45º27'32"

$\begin{array}{rrrr} & 67\ensurement{^{\circ}} & 13' & 4" \\ =& 66\ensurement{^{\circ}} & 73' & 4" \\ =& 66\ensurement{^{\circ}} & 72' &64" \\ \hline &-45\ensurement{^{\circ}} & 27' & 32" \\ \hline =&21\ensurement{^{\circ}} & 45' & 32" \end{array}\\\\ 67\ensurement{^{\circ}} \ 13' \ 4" - 45\ensurement{^{\circ}} \ 27' \ 32" = 21\ensurement{^{\circ}} \ 45' \ 32"$

heureka08.02.2015

+26396

+10

1 Deg --> ? rad

$\boxed{\\ \theta_{rad}= \theta_{deg} * \frac{ \pi }{ 180\ensurement{^{\circ}} } }\qquad \boxed{ \theta_{deg} = \theta_{rad} *\frac { 180\ensurement{^{\circ}} }{ \pi } }$

heureka08.02.2015

+26396

+10

$\dfrac{365!}{317!*365^{48} } =\prod\limits_{i=0}^{47}} \frac{318+i}{365}$

= 0.0394020271205775608037890867509822602447496243075877 $\cdots$

heureka08.02.2015

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