The quantity with units of length only is 
Explanation:
We have to combine the following constants:
-
, Planck constant, with units ![[m^2][kg][s^{-1}]](https://tex.z-dn.net/?f=%5Bm%5E2%5D%5Bkg%5D%5Bs%5E%7B-1%7D%5D)
- G, the Newton's gravitational constant, with units ![[m^3][kg^{-1}][s^{-2}]](https://tex.z-dn.net/?f=%5Bm%5E3%5D%5Bkg%5E%7B-1%7D%5D%5Bs%5E%7B-2%7D%5D)
-
, the speed of light, with units ![[m][s^{-1}]](https://tex.z-dn.net/?f=%5Bm%5D%5Bs%5E%7B-1%7D%5D)
The combination of these constant should have units of length only, so with meters (m).
First, we notice that
has [kg] in its units, while G has
in its units, so in order to make the [kg] disappear, we have to multiply them and they should have same power, so:
![hG = [m^{2+3}][kg^{1-1}][s^{-1-2}]=[m^5][s^{-3}]](https://tex.z-dn.net/?f=hG%20%3D%20%5Bm%5E%7B2%2B3%7D%5D%5Bkg%5E%7B1-1%7D%5D%5Bs%5E%7B-1-2%7D%5D%3D%5Bm%5E5%5D%5Bs%5E%7B-3%7D%5D)
Now we have to make the seconds, [s], disappear. We do that by dividing the new quantity by
, so that the new units are:
![\frac{hG}{c^3}=\frac{[m^5][s^{-3}]}{([m][s^{-1}])^3}=\frac{[m^5][s^{-3}]}{[m^3][s^{-3}]}=[m^2]](https://tex.z-dn.net/?f=%5Cfrac%7BhG%7D%7Bc%5E3%7D%3D%5Cfrac%7B%5Bm%5E5%5D%5Bs%5E%7B-3%7D%5D%7D%7B%28%5Bm%5D%5Bs%5E%7B-1%7D%5D%29%5E3%7D%3D%5Cfrac%7B%5Bm%5E5%5D%5Bs%5E%7B-3%7D%5D%7D%7B%5Bm%5E3%5D%5Bs%5E%7B-3%7D%5D%7D%3D%5Bm%5E2%5D)
We are almost done: now the quantity has units of an area, squared meters. Therefore, in order to make it have it units of length, we just take its square root:
![\sqrt{\frac{hG}{c^3}}=\sqrt{[m^2]}=[m]](https://tex.z-dn.net/?f=%5Csqrt%7B%5Cfrac%7BhG%7D%7Bc%5E3%7D%7D%3D%5Csqrt%7B%5Bm%5E2%5D%7D%3D%5Bm%5D)
Learn more about gravitational constant:
brainly.com/question/1724648
brainly.com/question/12785992
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