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

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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

#1
+9774
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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!

$$y=\frac{3}{4\cdot \sqrt{\frac{3x^4}{c^{2.3}}}}$$    (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.

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

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

!

May 18, 2020
edited by asinus  May 18, 2020