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y = 3/4 * root((3*x^4)/c^2,3)

 

C is the speed of light in a vacuum.  If X and Y are distances measured in meters, does this equation make sense?

 May 18, 2020
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y = 3/4 * root((3*x^4)/c^2,3)

c is the speed of light in a vacuum.  If X and Y are distances measured in meters, does this equation make sense?

 

 

Hello Guest!

 

Is your equation called that

\(y=\frac{3}{4\cdot \sqrt{\frac{3x^4}{c^{2.3}}}}\)    (according to rule point calculation operates from left to right. 2,3 means 2.3?)

or so

\(y=\frac{3}{4}\cdot \sqrt{\frac{3x^4}{c^{2.3}}}\)  ?

I assume the second equation.

\(\color{BrickRed}y=\frac{3}{4}\cdot \sqrt{\frac{3x^4}{c^{2.3}} }\)

\(y=\frac{3}{4}\cdot \sqrt{\frac{3x^4}{c^{2.3}} }\\ y^2=\frac{9}{16}\cdot \frac{3x^4}{c^{2.3}}\\ c^{2.3}=\frac{9}{16}\cdot \frac{3x^4}{y^2}\)

\(c=(\frac{9}{16}\cdot \frac{3x^4}{y^2})^{-2.3}\)

The equation doesn't make sense.
The unit of the speed of light is \(m/sec\ (300 \times 10 ^ 9\ m / sec).\)

The unit on the right is \(m^{-4.6}\), no matter what is used for x and y.

\(c\color{red}\neq\)\((\frac{9}{16}\cdot \frac{3x^4}{y^2})^{-2.3}\)

\(y\color{red}\neq\)\(\frac{3}{4}\cdot \sqrt{\frac{3x^4}{c^{2.3}}}\)

laugh  !

 May 18, 2020
edited by asinus  May 18, 2020

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