fixed main

content/index.md

@@ -3,7 +3,3 @@ title: Welcome to the Warehouse
 ---
 Welcome to the warehouse of the Thoughts of 0x6d617273756c746f72, Prime of the Faith, ranked Hacker on HTB, and the founder of a bunch of defunct projects on github. Average internet nerd/sysadmin/programmer. Move on, to either his cursed musings or with your day.
 
-# Socials
-
-- Matrix: [@lanre:envs.net](https://matrix.to/#/@lanre:envs.net)
-- Bluesky: [krishna.ayyalasomayajula.net](https://bsky.app/profile/krishna.ayyalasomayajula.net)

content/physics/CrazyStuff.md

@@ -104,5 +104,89 @@ $$
 
 Just as the potential energy of a raised ball does not depend on the gravitational field of the ball, the electric potential of the test charge $q$ doesn't depend on the magnitude of the charge itself. This quantity is given the unit **Volt**, or $1\text{ V}=1\;\text{J}/\mathrm{C}$ 
 
+# Circuits
+### Ohm’s Law:
+$$
+V = IR
+$$
+- Voltage $V$ = Current $I$ × Resistance $R$
+
+### Power in Circuits:
+$$
+P = IV
+$$
+- Power $P$ = Current $I$ × Voltage $V$
+
+$$
+P = I^2R
+$$
+- Power $P$ = Current squared $I^2$ × Resistance $R$
+
+$$
+P = \frac{V^2}{R}
+$$
+- Power $P$ = Voltage squared $V^2$ / Resistance $R$
+
+### Series Circuits:
+$$
+R_{\text{total}} = R_1 + R_2 + \dots + R_n
+$$
+- Total Resistance $R_{\text{total}}$ = Sum of Individual Resistances
+
+$$
+V_{\text{total}} = V_1 + V_2 + \dots + V_n
+$$
+- Total Voltage $V_{\text{total}}$ = Sum of Individual Voltages
+
+$$
+I_{\text{total}} = I_1 = I_2 = \dots = I_n
+$$
+- Current $I_{\text{total}}$ is the same across all components
+
+### Parallel Circuits:
+$$
+\frac{1}{R_{\text{total}}} = \frac{1}{R_1} + \frac{1}{R_2} + \dots + \frac{1}{R_n}
+$$
+- Total Resistance $R_{\text{total}}$ = Reciprocal Sum of Individual Resistances
+
+$$
+V_{\text{total}} = V_1 = V_2 = \dots = V_n
+$$
+- Voltage $V_{\text{total}}$ is the same across all components
+
+$$
+I_{\text{total}} = I_1 + I_2 + \dots + I_n
+$$
+- Total Current $I_{\text{total}}$ is the sum of individual currents
+
+### Capacitance:
+$$
+Q = CV
+$$
+- Charge $Q$ = Capacitance $C$ × Voltage $V$
+
+$$
+C = \frac{\epsilon_0 A}{d}
+$$
+- Capacitance $C$ of a parallel plate capacitor: $\epsilon_0$ = permittivity of free space, $A$ = area of plates, $d$ = distance between plates
+
+### Inductance:
+$$
+V = L \frac{di}{dt}
+$$
+- Voltage across an inductor $V$ = Inductance $L$ × Rate of change of current $\frac{di}{dt}$
+
+### RL Time Constant:
+$$
+\tau = \frac{L}{R}
+$$
+- Time constant $\tau$ for an RL circuit
+
+### Kirchhoff’s Laws:
+- **Kirchhoff’s Current Law (KCL):** The sum of currents entering a junction equals the sum of currents leaving.
+- **Kirchhoff’s Voltage Law (KVL):** The sum of voltages around any closed loop equals zero.
+
+
+
 ---
 #physics

content/physics/final.md

@@ -0,0 +1,101 @@
+> [!NOTE] A mass $m$ attached to a spring with constant $k$ oscillates on a frictionless surface. Derive an expression for the velocity $v$ as a function of displacement $x$.
+>
+> [!INFO]-
+> Total mechanical energy in SHM is conserved:
+> $$
+> E = \frac{1}{2}mv^2 + \frac{1}{2}kx^2 = \frac{1}{2}kA^2
+> $$
+> Solving for $v$:
+> $$
+> v = \pm \sqrt{\frac{k}{m}(A^2 - x^2)}
+> $$
+
+> [!NOTE] A wave traveling along a rope is represented by $y(x,t) = 0.02\cos(40x - 600t)$. Determine the amplitude, wavelength, frequency, and speed of the wave.
+>
+> [!INFO]-
+> The general wave form is $y = A\cos(kx - \omega t)$:
+> - Amplitude $A = 0.02\ \mathrm{m}$
+> - Wave number $k = 40 \Rightarrow \lambda = \frac{2\pi}{k} = \frac{2\pi}{40} = 0.157\ \mathrm{m}$
+> - Angular frequency $\omega = 600 \Rightarrow f = \frac{\omega}{2\pi} = \frac{600}{2\pi} \approx 95.5\ \mathrm{Hz}$
+> - Wave speed $v = f\lambda = 95.5 \times 0.157 \approx 15\ \mathrm{m/s}$
+
+> [!NOTE] In Young's double slit experiment, fringes are formed on a screen 1.2 m away using light of wavelength $600\ \text{nm}$. The slits are separated by $0.2\ \text{mm}$. Calculate the distance between adjacent bright fringes.
+>
+> [!INFO]-
+> Fringe spacing is given by:
+> $$
+> y = \frac{\lambda D}{d} = \frac{600 \times 10^{-9} \times 1.2}{0.2 \times 10^{-3}} = 3.6\ \text{mm}
+> $$
+
+> [!NOTE] A pendulum of length $0.5\ \mathrm{m}$ is displaced by a small angle. Determine its period and explain why amplitude does not affect the result.
+>
+> [!INFO]-
+> The period is:
+> $$
+> T = 2\pi\sqrt{\frac{L}{g}} = 2\pi\sqrt{\frac{0.5}{9.8}} \approx 1.41\ \mathrm{s}
+> $$
+> In the small-angle approximation ($\theta < 10^\circ$), motion is independent of amplitude.
+
+> [!NOTE] A charged particle $q$ moves through a uniform electric field $E$. Derive the expression for the work done on the charge and its change in potential energy.
+>
+> [!INFO]-
+> Work done:
+> $$
+> W = qEd
+> $$
+> Change in potential energy:
+> $$
+> \Delta U = -qEd
+> $$
+> since electric potential energy decreases when the charge moves in the direction of the field.
+
+> [!NOTE] A capacitor of $10\ \mu\mathrm{F}$ is charged to $5\ \mathrm{V}$. Calculate the energy stored in it.
+>
+> [!INFO]-
+> $$
+> E = \frac{1}{2}CV^2 = \frac{1}{2} \cdot 10 \times 10^{-6} \cdot 25 = 1.25 \times 10^{-4}\ \mathrm{J}
+> $$
+
+> [!NOTE] A coil of wire rotates in a magnetic field. Use Faraday’s law to derive the expression for the induced emf.
+>
+> [!INFO]-
+> Faraday’s law:
+> $$
+> \mathcal{E} = -\frac{d\Phi}{dt}
+> $$
+> Magnetic flux $\Phi = B A \cos(\omega t)$, so:
+> $$
+> \mathcal{E} = B A \omega \sin(\omega t)
+> $$
+
+> [!NOTE] A wire carries a current of $3\ \mathrm{A}$ through a magnetic field of $0.5\ \mathrm{T}$ perpendicular to its length. The wire is $0.4\ \mathrm{m}$ long. Find the magnetic force.
+>
+> [!INFO]-
+> $$
+> F = ILB\sin\theta = 3 \cdot 0.4 \cdot 0.5 \cdot 1 = 0.6\ \mathrm{N}
+> $$
+
+> [!NOTE] Explain the effect of damping on the amplitude-frequency graph of a driven harmonic oscillator.
+>
+> [!INFO]-
+> Damping reduces the peak amplitude and shifts the resonant frequency slightly lower. Greater damping broadens the curve and lowers the quality factor $Q$.
+
+> [!NOTE] A mass-spring oscillator experiences light damping. Write the differential equation and general solution.
+>
+> [!INFO]-
+> Equation:
+> $$
+> m\ddot{x} + b\dot{x} + kx = 0
+> $$
+> Solution:
+> $$
+> x(t) = A e^{-\gamma t} \cos(\omega' t + \phi)
+> $$
+> where $\gamma = \frac{b}{2m}$ and $\omega' = \sqrt{\omega_0^2 - \gamma^2}$.
+
+> [!NOTE] A $0.2\ \mathrm{kg}$ object experiences a force due to gravity from Earth at a distance of $6.4 \times 10^6\ \mathrm{m}$. Calculate the force.
+>
+> [!INFO]-
+> $$
+> F = G\frac{Mm}{r^2} = 6.67 \times 10^{-11} \cdot \frac{5.97 \times 10^{24} \cdot 0.2}{(6.4 \times 10^6)^2} \approx 1.96\ \mathrm{N}
+> $$