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2 years ago

12

PLEASE ANSWER! DESPERATE! WILL MARK BRAINLIEST!!

1 answer:

Virty [35]2 years ago

7 0

Answer:

$\boxed{energy} \to \: is \: the \: capacity \: for \: doing \: work. \\ \: energy \: can \: exist\: in \: vaious \: forms : such \: as \to \\ \boxed{thermal}\boxed{mechanical}\boxed{chemical}\boxed{solar}......etc.\\ its \to \: \boxed{ solar \: energy} : without \: solar \: energy \\ \: i \: dont \: think \: life \: on \: earth \: or \: \\ other \:p lanetry \: bodies \: would \: be \: possible. \\ \: solar \: energy \: from \: different \: stars \\ \: are \: responssible \: for \: warming \: of \: thier \: solar \: systems \\ \: and \:collectively \: saves\: thier \: galaxies \: from \: being \: \boxed{dark \: cold}.$

Explanation:

♨Rage♨

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2 years ago

Read 2 more answers

Help me please I can't get the final step

inna [77]

Answer:

$\displaystyle m=\frac{2}{3},\ n=\frac{4}{3}$

Explanation:

<u>Dimensional Analysis</u>

It's given the relation between quantities A, B, and C as follows:

$\displaystyle A=\frac{3}{2}B^mC^n$

and the dimensions of each variable is:

$A=L^2T^2$

$B=LT^{-1}$

$C=LT^2$

Substituting the dimensions into the relation (the coefficient is not important in dimension analysis):

$\displaystyle L^2T^2=\left(LT^{-1}\right)^m\left(LT^2\right)^n$

Operating:

$L^2T^2=\left(L^mT^{-m}\right)\left(L^nT^{2n}\right)$

$L^2T^2=L^{m+m}T^{-m+2n}$

Equating the exponents:

$m+n=2$

$-m+2n=2$

Adding both equations:

$3n=4$

Solving:

$n=4/3$

$m=2-4/3=2/3$

Answer:

$\mathbf{\displaystyle m=\frac{2}{3},\ n=\frac{4}{3}}$

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2 years ago