Question

Heyla peeps :D

Please answer to this question.

spammers stay away please :)

thanks <3!!❤️​

Answer 1

Hi Varshu

Answer:

Let

${l}^{1} = {c}^{x} {G}^{y} ( \frac{e{}^{2}}{2\pi \: Ę} ) {}^{z}$

$c = {{l \: t {}^{ - 1} }} \\ G = {m}^{ - 1} l {}^{3} {t}^{ - 2} \\ \frac{e {}^{2} }{4\pi \: Ę} = ml {}^{3} {t}^{ - 2} \\ \\ l = (l \: t {}^{ - 1} ) {}^{x} {({m}^{ - 1} l {}^{3} {t}^{ - 2})}^{y} (ml {}^{3} {t}^{ - 2} ) {}^{z} \\ \\ {m}^{0} l {}^{1} t {}^{0} = {l}^{x + 3y + 3z} \: {m}^{ - y + z} \: {t}^{ - x - 2y - 2z) }$

by comparing powers we get

x+3y+3z = 1

-y+z = 0

x+2y+2z=0

When we solve these equations, we get

x = -2

y= 1/2 and z=1/2

There your answer is

${c}^{ - 2} {G\frac{ {e}^{2} }{4\pi \: Ę} }^{ \frac{1}{2} }$

Answer 2

Answer:

$length \: \alpha \: e {}^{2} \div 4\pi \: e. \\ length(l) = (c) {}^{a} (g) {}^{b} (e {}^{2} \div 4\pi \: e.) {}^{c} \\ where \\ c = speed \: of \: light \\ g = constant \\ dimensions \: of \\ l = (l) \\ c =( lt {}^{ - } ) \\ g = (m {}^{ - 1} l {}^{3} t {}^{ - 2} ) \\ e {}^{2} \div 4\pi \: e. = (ml {}^{3} t {}^{ - 2} )$

$(m {}^{0} l {}^{1} t {}^{0} )=( lt {}^{ - } ) {}^{a} (m {}^{ - 1} l {}^{3} t {}^{ - 2}) {}^{b} (ml {}^{3} t {}^{ - 2} ) {}^{c} \\ comparing \: on \: both \: sides \\ = > a+3b+3c=1..EQ1\\=>-b+c=0\\=>b=c...EQ2\\=>-a-2b-2c=0\\=>a+2b+2c=0...EQ3\\subtracting\: EQ3\: & \:EQ1\\=>a+3b+3c-a-2b-2c=1-0\\=>b+c=1...EQ4\\=>sub \:EQ2\: in\:EQ4\\=>b+b=1\\2b=1\\=>b=1/2\\sub\:b=1/2\:in\:EQ4\\=>1/2+c=1\\=>c=1-1/2\\=>c=2-1/2\\=>c=1/2\\sub\: c \:&\:b \:in\:EQ3\\=>a+2(1/2)+2(1/2)=0\\=>a+1+1=0\\=>a=-2$