This physics problem tests your understanding of fundamental physical laws and their applications. The step-by-step solution below breaks down the problem using relevant equations and physical reasoning.
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Hey tahellemadenko01, good to see you again. Here are the solutions to the questions: 1. a) i) A semiconductor is a material with electrical conductivity between that of a conductor (like copper) and an insulator (like glass). Its conductivity can be controlled by doping or temperature. 1. a) ii) A p-type semiconductor is created by doping a pure semiconductor with trivalent impurities (e.g., boron), resulting in a majority of holes* as charge carriers. An n-type semiconductor is created by doping a pure semiconductor with pentavalent impurities (e.g., phosphorus), resulting in a majority of electrons* as charge carriers. 1. a) iii) An application of a combination of a p-type and n-type semiconductor is a diode (p-n junction). 1. b) Step 1: Calculate the total charge (Q) that flows. Q = It Q = (8 A) × (4 s) Q = 32 C Step 2: Calculate the number of electrons (n) using the total charge and the charge of a single electron (e). Q = ne n = (Q)/(e) n = 32 C1.6 × 10^-19 C n = 2.0 × 10^20 The number of electrons is 2.0 × 10^20. 2. i) The law of conservation of linear momentum states that in an isolated system, the total linear momentum remains constant if no external forces act on it. 2. ii) Step 1: Identify the mass and velocity of the bullet. m_bullet = 0.05 kg v_bullet = 900 m/s Step 2: Calculate the momentum of the bullet (p_bullet). p_bullet = m_bullet × v_bullet p_bullet = (0.05 kg) × (900 m/s) p_bullet = 45 kg m/s The momentum of the bullet is 45 kg m/s. 2. iii) Step 1: Apply the law of conservation of momentum. The initial momentum of the gun-bullet system is zero. m_gunu_gun + m_bulletu_bullet = m_gunv_gun + m_bulletv_bullet Since the gun and bullet are initially stationary, u_gun = 0 and u_bullet = 0. 0 = m_gunv_gun + m_bulletv_bullet Step 2: Rearrange to solve for the recoil velocity of the gun (v_gun). m_gunv_gun = -m_bulletv_bullet v_gun = -m_bulletv_bulletm_gun v_gun = -(0.05 kg) × (900 m/s)4 kg v_gun = -45 kg m/s4 kg v_gun = -11.25 m/s The recoil velocity is -11.25 m/s. The negative sign indicates the direction opposite to the bullet's velocity. 3. i) Pressure at point A: Atmospheric pressure* Pressure at point B: Pressure of the gas* 3. ii) Step 1: Identify the density of mercury () and the height difference (h). From the diagram, h = 0.25 m. = 13600 kg/m^3 g = 9.81 m/s^2 (assuming standard gravity) h = 0.25 m Step 2: Calculate the excess pressure (P_excess) using the formula P = gh. P_excess = gh P_excess = (13600 kg/m^3) × (9.81 m/s^2) × (0.25 m) P_excess = 33354 Pa The excess pressure of the gas is 33354 Pa. 3. iii) Step 1: Identify the atmospheric pressure (P_atm) and the calculated excess pressure (P_excess). P_atm = 1.0 × 10^5 Pa P_excess = 33354 Pa Step 2: Calculate the pressure of the gas (P_gas). P_gas = P_atm + P_excess P_gas = (1.0 × 10^5 Pa) + (33354 Pa) P_gas = 133354 Pa The pressure of the gas is 133354 Pa. 3. iv) One instrument that is an application of the u-tube is a manometer. 4. a) For the system to be in equilibrium, two conditions must be met: The net force* acting on the system must be zero (translational equilibrium). The net moment* (or torque) about any point must be zero (rotational equilibrium). 4. b) Step 1: Apply the principle of moments. The sum of clockwise moments about the pivot must equal the sum of anticlockwise moments. From the diagram: Anticlockwise moment = Force on left × Distance from pivot = 20 N × 0.5 m Clockwise moment = Force on right × Distance from pivot = 10 N × x Step 2: Set the moments equal and solve for x. 20 N × 0.5 m = 10 N × x 10 N m = 10 N × x x = 10 N m10 N x = 1.0 m The distance x is 1.0 m. 4. c) i) Step 1: Apply the condition for translational equilibrium: The sum of upward forces equals the sum of downward forces. Upward force = R (at the pivot) Downward forces = 20 N + 10 N Step 2: Solve for R. R = 20 N + 10 N R = 30 N The upward force, R, acting at the pivot is 30 N. 4. c) ii) A free-body diagram of the forces acting on the see-saw: % Plank [thick] (-2.5,0) -- (2.5,0); % Pivot [fill=black] (0,-0.2) circle (0.05); (0,-0.2) -- (-0.5,-0.7) -- (0.5,-0.7) -- cycle; % Force R [->, thick] (0, -0.2) -- (0, 0.8) node[above] R = 30 N; % Left load [->, thick] (-1.5, 0) -- (-1.5, -1) node[below] 20 N; [fill=gray] (-1.7, 0.1) rectangle (-1.3, 0.3); % Right load [->, thick] (1.0, 0) -- (1.0, -1) node[below] 10 N; [fill=gray] (0.8, 0.1) rectangle (1.2, 0.3); % Distances (optional, but good for clarity) [<->] (0, -0.4) -- (-1.5, -0.4) node[midway, below] 0.5 m; [<->] (0, -0.4) -- (1.0, -0.4) node[midway, below] 1.0 m; The system is stable because it is in equilibrium, meaning the net force and net moment are both zero. This ensures it remains in its current state without accelerating or rotating. 5. a) Isotopes are atoms of the same element that have the same number of protons (atomic number, Z) but different numbers of neutrons, resulting in different mass numbers (A). 5. b) Step 1: Determine the number of protons (Z) and neutrons (N) for ^14C. For Carbon (C), the atomic number is Z = 6. For ^14C, the mass number is A = 14. Number of neutrons N = A - Z = 14 - 6 = 8. Step 2: Calculate the N/Z ratio. (N)/(Z) = (8)/(6) = (4)/(3) ≈ 1.33 Step 3: State if the isotope is stable or unstable. For light elements, a stable N/Z ratio is typically close to 1. As the atomic number increases, the stable N/Z ratio increases slightly above 1. For carbon, ^12C has N/Z = 6/6 = 1, which is stable. ^14C has an N/Z ratio of approximately 1.33, which is higher than 1 and indicates an excess of neutrons. Therefore, ^14C is unstable (it is a radioactive isotope that undergoes beta decay). The N/Z ratio for ^14C is 1.33, and the isotope is unstable. 5. c) The half-life of a radioactive sample is the time it takes for half of the radioactive nuclei in a sample to decay. 6. a) Step 1: Calculate the equivalent resistance of the two resistors in parallel. The 2.0 and 8.0 resistors