Physics: Astudy of how much corn was produced in a field depending on the planting density compared the yield from fields planted with 4 plants per hectare to the yield from fields planted with 6 plants per hectare. 1. what is the ratio of the distance between the corn plants in the more-densely-planted field to the distance between the corn plants in the less-densely-planted field?

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Astudy of how much corn was produced in a field

ronniethefun

15.01.2022

$\frac{d_2}{d_1} =\frac{\sqrt{6}}{\sqrt{4}}=\sqrt{\frac{6}{4}} = \frac{\sqrt{6}}{2}= 1.225$

Or in the other way:

$\frac{d_1}{d_2} =\frac{\sqrt{4}}{\sqrt{6}}=\sqrt{\frac{4}{6}} = \frac{\sqrt{6}}{3}= 0.816$

Explanation:

For this case we need to remember that hectare is a measure of area. And we have the following ratios:

$r_1 =40000 \frac{plants}{acre}$

We can convert this into plants/m^2

$r_1 = 40000 \frac{plants}{acre} *\frac{1 acre}{10000 m^2}=4 \frac{plants}{m^2}$

$r_2=60000 \frac{plants}{acre}$

We can convert this into plants/m^2

$r_2 = 60000 \frac{plants}{acre} *\frac{1 acre}{10000 m^2}=6 \frac{plants}{m^2}$

By dimensional analysis we assume that A = L^2 we can find the ratio in terms of distance like this:

$d_1 =\sqrt{4}= 2 \frac{\sqrt{plants}}{m}$

$d_2= \sqrt{6}= 2.449 \frac{\sqrt{plants}}{m}$

And if we find the ratios in terms of distance we got:

$\frac{d_2}{d_1} =\frac{\sqrt{6}}{\sqrt{4}}=\sqrt{\frac{6}{4}} = \frac{\sqrt{6}}{2}= 1.225$

Or in the other way:

$\frac{d_1}{d_2} =\frac{\sqrt{4}}{\sqrt{6}}=\sqrt{\frac{4}{6}} = \frac{\sqrt{6}}{3}= 0.816$

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