Displacement under constant acceleration is given by \(s = \frac{1}{2}at^2\). Substituting \(a = 3.0 \, \text{m/s}^2\) and \(t = 4\), we get \(s = \frac{1}{2}(3)(4)^2 = 24 \, \text{m}\).

At the highest point, the velocity is zero. Using \(v = u + at\) and substituting \(u = 20\), \(v = 0\), and \(a = -9.8\), we get \(0 = 20 – 9.8t\), or \(t = 20/9.8 \approx 2.04 \, \text{seconds}\).

The range is given by \(R = \frac{v^2 \sin(2\theta)}{g}\). Substituting \(v = 15\), \(\theta = 30^\circ\), and \(g = 9.8\), we get \(R = \frac{15^2 \sin(60^\circ)}{9.8} \approx 19.9 \, \text{m}\).

Using Newton’s second law, \(F = ma\), we have \(a = F/m\). Substituting \(F = 20\) and \(m = 5\), \(a = 20/5 = 4 \, \text{m/s}^2\).

Using Newton’s second law, \(F = ma\), we have \(a = F/m\). Substituting \(F = 50\) and \(m = 10\), \(a = 50/10 = 5 \, \text{m/s}^2\).

Centripetal acceleration is given by \(a_c = \frac{v^2}{r}\). Substituting \(v = 20\) and \(r = 50\), we get \(a_c = \frac{20^2}{50} = 8 \, \text{m/s}^2\).

Work is calculated using \(W = Fd\), where \(F = mg\). Substituting \(m = 2.0\), \(g = 9.8\), and \(d = 5.0\), we get \(W = (2)(9.8)(5) = 98 \, \text{J}\).

Using conservation of energy, \(K_i + U_i = K_f + U_f\). At the bottom, \(U_f = 0\). Solving for \(v_f\), \(v_f = \sqrt{v_i^2 + 2gh} = \sqrt{5^2 + 2(9.8)(20)} \approx 20.4 \, \text{m/s}\).

Power is given by \(P = \frac{W}{t}\), where \(W = Fd\). Substituting \(F = 50\), \(d = 4.0\), and \(t = 5.0\), \(P = \frac{(50)(4.0)}{5.0} = 40 \, \text{W}\).

Using conservation of momentum, (4) + (2)(0) = (3)(2) + (2)v_2′[/latex]. Solving, \(v_2′ = 3.0 \, \text{m/s}\).

Using conservation of momentum: (20) + (1000)(0) = (1500)(0) + (1000)v_2′[/latex]. Solving, \(v_2′ = 30 \, \text{m/s}\).

Using conservation of momentum: (5) + (2)(-3) = (3 + 2)v_f[/latex]. Solving, \(v_f = 2.4 \, \text{m/s}\).

The period of a spring is given by \(T = 2\pi\sqrt{\frac{m}{k}}\). Rearranging for \(m\): \(m = \frac{T^2 k}{4\pi^2}\). Substituting \(T = 2.0\) and \(k = 50\), we get \(m = \frac{(2)^2(50)}{4\pi^2} \approx 1.0 \, \text{kg}\).

The period of a pendulum is given by \(T = 2\pi\sqrt{\frac{L}{g}}\). Substituting \(L = 1.5\) and \(g = 9.8\), \(T = 2\pi\sqrt{\frac{1.5}{9.8}} \approx 2.46 \, \text{s}\).

Total energy in SHM is \(E = \frac{1}{2}kA^2\). Substituting \(k = 200\) and \(A = 0.2\), we get \(E = \frac{1}{2}(200)(0.2)^2 = 4.0 \, \text{J}\).

Wave speed is given by \(v = f\lambda\). Substituting \(f = 200\) and \(\lambda = 0.5\), we get \(v = (200)(0.5) = 100 \, \text{m/s}\).

The wavelength is directly proportional to the speed of the wave. If the speed decreases, the wavelength also decreases, according to \(v = f\lambda\).

When two waves with slightly different frequencies overlap, beats are produced due to constructive and destructive interference.

Wavelength is calculated using \(\lambda = \frac{v}{f}\). Substituting \(v = 343\) and \(f = 256\), \(\lambda = \frac{343}{256} \approx 1.34 \, \text{m}\).

For a pipe closed at one end, the fundamental frequency corresponds to a quarter wavelength: \(L = \frac{\lambda}{4}\). Substituting \(\lambda = \frac{v}{f} = \frac{343}{440}\), \(L = \frac{343}{4 \times 440} \approx 0.195 \, \text{m}\).

The Doppler effect occurs when there is relative motion between the source of a wave and an observer, leading to a change in observed frequency.

Using Coulomb’s law: \(F = k\frac{|q_1q_2|}{r^2}\). Substituting the values: \(F = 9 \times 10^9 \frac{(3 \times 10^{-6})(2 \times 10^{-6})}{(0.5)^2} = 0.216 \, \text{N}\).

Electric potential is given by \(V = k\frac{q}{r}\). Substituting \(q = 5 \times 10^{-6}\) and \(r = 0.2\): \(V = 9 \times 10^9 \frac{5 \times 10^{-6}}{0.2} = 225,000 \, \text{V}\).

Force is given by \(F = qE\). Substituting \(q = -2 \times 10^{-6}\) and \(E = 500\): \(F = (-2 \times 10^{-6})(500) = -0.001 \, \text{N}\). The magnitude is \(0.001 \, \text{N}\).

Using Ohm’s Law: \(I = \frac{V}{R}\). Substituting \(V = 12\) and \(R = 4\): \(I = \frac{12}{4} = 3 \, \text{A}\).

In a series circuit, the equivalent resistance is the sum of individual resistances: \(R_{\text{eq}} = R_1 + R_2\). Substituting \(R_1 = 6\) and \(R_2 = 3\): \(R_{\text{eq}} = 6 + 3 = 9 \, \Omega\).

In a parallel circuit, \(\frac{1}{R_{\text{eq}}} = \frac{1}{R_1} + \frac{1}{R_2}\). Substituting \(R_1 = 4\) and \(R_2 = 6\): \(\frac{1}{R_{\text{eq}}} = \frac{1}{4} + \frac{1}{6} = \frac{5}{12}\). Thus, \(R_{\text{eq}} = 2.4 \, \Omega\).

The magnetic force is given by \(F = qvB\sin\theta\). When \(\theta = 0\) (parallel motion), \(\sin 0 = 0\), so the force is \(0\).

Using Faraday’s law: \(\text{EMF} = -N\frac{\Delta\Phi}{\Delta t}\). Substituting \(N = 100\), \(\Delta\Phi = 0.5\), and \(\Delta t = 2\): \(\text{EMF} = -100\frac{0.5}{2} = 25 \, \text{V}\).

The magnetic force is given by \(F = qvB\sin\theta\). Substituting \(q = 1.6 \times 10^{-19}\), \(v = 2 \times 10^6\), \(B = 0.1\), and \(\sin 90 = 1\): \(F = (1.6 \times 10^{-19})(2 \times 10^6)(0.1) = 3.2 \times 10^{-14} \, \text{N}\).

The heat required is given by \(Q = mc\Delta T\). Substituting \(m = 0.5\), \(c = 900\), and \(\Delta T = 75 – 25 = 50\): \(Q = (0.5)(900)(50) = 22,500 \, \text{J}\).

Convection involves the transfer of heat through fluid movement. Heating water in a pot, where warm water rises and cooler water sinks, is an example of convection.

Thermal expansion is given by \(\Delta L = \alpha L_0 \Delta T\). Solving for \(\alpha\): \(\alpha = \frac{\Delta L}{L_0 \Delta T} = \frac{1.2 \times 10^{-3}}{1.5 \times 100} = 8 \times 10^{-6} \, \text{\degree C}^{-1}\).

The Zeroth Law states that if two systems are in thermal equilibrium with a third system, they are also in thermal equilibrium with each other. This defines the concept of temperature.

Efficiency is given by \(\eta = \frac{W}{Q_{\text{in}}} \times 100\). Substituting \(W = 50\) and \(Q_{\text{in}} = 200\): \(\eta = \frac{50}{200} \times 100 = 25\%\).

The Second Law states that the entropy of an isolated system never decreases; it increases in natural processes and remains constant in reversible processes.

The energy difference is \(\Delta E = E_3 – E_2\). Substituting \(E_3 = -13.6/3^2\) and \(E_2 = -13.6/2^2\): \(\Delta E = -1.51 – (-3.4) = 1.89 \, \text{eV}\).

Alpha decay emits a helium nucleus (\(^4_2 \text{He}\)), which contains 2 protons and 2 neutrons.

Binding energy is \(E = \Delta m \times c^2\). Substituting \(\Delta m = 0.0304\) and \(c^2 = 931\): \(E = 0.0304 \times 931 = 28.3 \, \text{MeV}\). Dividing by 4 nucleons: \(28.3/4 = 7.08 \, \text{MeV}\).

The kinetic energy of emitted electrons depends on the frequency of light, as given by \(KE = hf – \phi\), where \(\phi\) is the work function.

Wave-particle duality refers to the phenomenon where particles like electrons and photons exhibit both wave-like and particle-like properties, such as diffraction and discrete impacts.

Threshold frequency is \(f = \phi / h\). Substituting \(\phi = 3.2\) and \(h = 4.14 \times 10^{-15}\): \(f = 3.2 / (4.14 \times 10^{-15}) = 7.73 \times 10^{14} \, \text{Hz}\).

The pressure due to a fluid column is given by \(P = \rho g h\). Substituting \(\rho = 1000\), \(g = 9.8\), and \(h = 10\): \(P = 1000 \times 9.8 \times 10 = 98,000 \, \text{Pa}\).

The buoyant force is given by \(F_b = \rho_{\text{fluid}} V g\). Substituting \(\rho_{\text{fluid}} = 1000\), \(V = 0.01\), and \(g = 9.8\): \(F_b = 1000 \times 0.01 \times 9.8 = 98 \, \text{N}\).

The continuity equation states \(A_1 v_1 = A_2 v_2\). If \(A_2\) decreases, \(v_2\) must increase to maintain the same flow rate.

In laminar flow, resistance is determined by the fluid’s viscosity and the pipe’s dimensions, as described by Poiseuille’s law.

Capillarity is caused by the adhesive forces between the liquid and the tube walls being stronger than the cohesive forces within the liquid.

According to the Young-Laplace equation, the pressure difference across a liquid interface is inversely proportional to the radius of curvature: \(\Delta P = \frac{2\gamma}{r}\).