m=frac{k_{B}T}{sigmaOmega} eching

$$m = \frac { k _ { B } T } { \sigma \Omega }$$

T uchun yechish

$$\left\{\begin{matrix}T=\frac{m\Omega \sigma }{k_{B}}\text{, }&k_{B}\neq 0\text{ and }\sigma \neq 0\text{ and }\Omega \neq 0\\T\in \mathrm{R}\text{, }&m=0\text{ and }k_{B}=0\text{ and }\sigma \neq 0\text{ and }\Omega \neq 0\end{matrix}\right,$$

Chiziqli tenglamalarni yechish bo'yicha qadamlar

Tenglamaning ikkala tarafini \Omega \sigma ga ko'paytirish.

$$m\Omega \sigma =k_{B}T$$

Tomonlarni almashtirib, barcha oʻzgaruvchi shartlar chap tomonga oʻtkazing.

$$k_{B}T=m\Omega \sigma$$

Ikki tarafini k_{B} ga bo‘ling.

$$\frac{k_{B}T}{k_{B}}=\frac{m\Omega \sigma }{k_{B}}$$

k_{B} ga bo'lish k_{B} ga ko'paytirishni bekor qiladi.

$$T=\frac{m\Omega \sigma }{k_{B}}$$

k_B uchun yechish

$$\left\{\begin{matrix}k_{B}=\frac{m\Omega \sigma }{T}\text{, }&T\neq 0\text{ and }\sigma \neq 0\text{ and }\Omega \neq 0\\k_{B}\in \mathrm{R}\text{, }&m=0\text{ and }T=0\text{ and }\sigma \neq 0\text{ and }\Omega \neq 0\end{matrix}\right,$$

Chiziqli tenglamalarni yechish bo'yicha qadamlar

Tenglamaning ikkala tarafini \Omega \sigma ga ko'paytirish.

$$m\Omega \sigma =k_{B}T$$

Tomonlarni almashtirib, barcha oʻzgaruvchi shartlar chap tomonga oʻtkazing.

$$k_{B}T=m\Omega \sigma$$

Tenglama standart shaklda.

$$Tk_{B}=m\Omega \sigma$$

Ikki tarafini T ga bo‘ling.

$$\frac{Tk_{B}}{T}=\frac{m\Omega \sigma }{T}$$

T ga bo'lish T ga ko'paytirishni bekor qiladi.

$$k_{B}=\frac{m\Omega \sigma }{T}$$