# Umhlaba nepea0817

$$r_1 = 6370km$$ umhlaba (njengebhola $$r_1 = 6370km$$ ) $$r_1 = 6370km$$ $$r_2 = 2mm$$ ) bese $$r_2 = 2mm$$ intambo phezu kwenkabazwe ukuze ilale iqinile ebusweni. Manje uzelula zombili izintambo ngemitha elilodwa lilinye. Zombili izintambo manje kufanele zilele ngokugcwele phezu kwenkabazwe futhi - azisalali ngokuphelele ebusweni, kepha sezindiza ngaphezu kwenkabazwe. Intambo intanta ngaphezu komhlaba iphakeme kangakanani, iphakeme kangakanani ngaphezu kwepea?

Izintambo ezimbili zinobude bokuqala:

$$l_1 = 2\cdot 6370 km \cdot \pi \Leftrightarrow r_1 = 6370 km = \frac{l_1}{2 \cdot \pi}$$

njengoba

$$l_2 = 2 \cdot 2mm \cdot \pi \Leftrightarrow r_2 = 2mm = \frac{l_2}{2 \cdot \pi}.$$

Kepha manje kungemva kokunwetshwa

$$r_{1 NEU} = \frac{l_1 + 1m}{2\cdot \pi}$$

njengoba

$$r_{2 NEU} = \frac{l_2 + 1m}{2\cdot \pi}.$$

Kepha manje kuyamangaza

$$r_{1 NEU} - r_1 = \frac{l_1 + 1m}{2\cdot \pi} - \frac{l_1}{2\cdot \pi} = \frac{l_1 + 1m - l_1}{2 \cdot \pi} = \frac{1m}{2 \cdot \pi} = 0.159m$$

njengoba

$$r_{2 NEU} - r_2 = \frac{l_2 + 1m}{2\cdot \pi} - \frac{l_2}{2\cdot \pi} = \frac{l_2 + 1m - l_2}{2 \cdot \pi} = \frac{1m}{2 \cdot \pi} = 0.159m.$$

Ngakho-ke ibanga ukusuka ebusweni lizimele nge $$l_1$$ noma $$l_2$$ , $$l_2$$ kuma-radii $$r_1$$ noma $$r_2$$ emikhakheni. Impendulo $$0.159m$$ : Zombili izintambo zintanta ngobude obufanayo $$0.159m$$ ) ngaphezulu kobuso.

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