trial chatgpt codes

\[ \frac{F_{\text{electric}}}{F_{\text{gravity}}}\]

\[ \frac{F_{\text{electric}}}{F_{\text{gravity}}} = \frac{\frac{1}{4\pi \varepsilon_0} \cdot \frac{q_1 q_2}{r^2}}{G \cdot \frac{m_1 m_2}{r^2}} = \frac{(9 \times 10^9) \cdot \frac{(1.6 \times 10^{-19})(1.6 \times 10^{-19})}{r^2}}{(6.67 \times 10^{-11}) \cdot \frac{(1.67 \times 10^{-27})(9.11 \times 10^{-31})}{r^2}} = 2.4 \times 10^{39} \] \[ I = m_1 x^2 + m_2 (L - x)^2 \] \[ \frac{dI}{dx} = m_1 \cdot 2x + m_2 \cdot 2(L - x)(0 - 1) \] \[ \text{Minimum when } \frac{dI}{dx} = 0 \] \[ m_1 \cdot 2x = m_2 \cdot 2(L - x) \] \[ m_1 \cdot 2x = m_2 \cdot 2L - m_2 \cdot 2x \] \[ (m_1 + m_2) \cdot 2x = m_2 \cdot 2L \] \[ \Rightarrow x = \frac{m_2 L}{m_1 + m_2} \] 

\[ \frac{\Delta L}{L} = ? \] \[ \text{Given: } \text{Hypotenuse } = L + \Delta L, \quad \text{Opposite side } = x \] Using Pythagoras’ Theorem: \[ L^2 + x^2 = (L + \Delta L)^2 \] \[ L^2 + x^2 = \left( L \left(1 + \frac{\Delta L}{L} \right) \right)^2 \] \[ L^2 + x^2 = L^2 \left(1 + \frac{\Delta L}{L} \right)^2 \] Divide both sides by \( L^2 \): \[ 1 + \frac{x^2}{L^2} = \left(1 + \frac{\Delta L}{L} \right)^2 \] \[ \sqrt{1 + \frac{x^2}{L^2}} = 1 + \frac{\Delta L}{L} \] Using the approximation \( \sqrt{1 + x} \approx 1 + \frac{x}{2} \) for small \( x \), we get: \[ 1 + \frac{x^2}{2L^2} \approx 1 + \frac{\Delta L}{L} \] \[ \Rightarrow \frac{\Delta L}{L} \approx \frac{x^2}{2L^2} \] \[ \boxed{\frac{\Delta L}{L} = \frac{x^2}{2L^2}} \]

\[ \sum F = ma \]\[ 100 - \mu mg = ma \] \[ 100 - (0.5)(10)(10) = (10)a \] \[ 50 = 10\hspace{2mm}a \] \[ a = 5 \, \text{m/s}^2 \]