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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Right Blessing, let's go. Question 12. b. (i): Gravitational force and initial accelerations Step 1: Identify the given values and convert units. Mass of particle 1: m_1 = 1 kg. Mass of particle 2: m_2 = 2 kg. Separation distance: r = 50 cm = 0.50 m. Gravitational constant: G = 6.674 × 10^-11 N m^2/kg^2. Step 2: Calculate the gravitational force (F) between the particles. Using Newton's Law of Universal Gravitation: F = G (m_1 m_2)/(r^2) F = (6.674 × 10^-11 N m^2/kg^2) (1 kg)(2 kg)(0.50 m)^2 F = (6.674 × 10^-11) (2)/(0.25) N F = (6.674 × 10^-11) × 8 N F = 5.3392 × 10^-10 N Step 3: Calculate the initial acceleration of each particle. Using Newton's Second Law (F = ma): For particle 1 (m_1): a_1 = (F)/(m_1) = 5.3392 × 10^-10 N1 kg a_1 = 5.3392 × 10^-10 m/s^2 For particle 2 (m_2): a_2 = (F)/(m_2) = 5.3392 × 10^-10 N2 kg a_2 = 2.6696 × 10^-10 m/s^2 The gravitational force is 5.34 × 10^-10 N. The initial acceleration of the 1 kg particle is 5.34 × 10^-10 m/s^2. The initial acceleration of the 2 kg particle is 2.67 × 10^-10 m/s^2. Question 12. b. (ii): Escape velocity from the moon Step 1: Identify the given values and convert units. Mass of the moon: M = 7.4 × 10^22 kg. Radius of the moon: R = 1740 km = 1740 × 10^3 m = 1.74 × 10^6 m. Gravitational constant: G = 6.674 × 10^-11 N m^2/kg^2. Step 2: Calculate the escape velocity (v_e). The formula for escape velocity is: v_e = sqrt((2GM)/(R)) v_e = sqrt(2 × (6.674 × 10^-11) N m^2/kg^2) × (7.4 × 10^22 kg)1.74 × 10^6 m v_e = sqrt(9.87752 × 10^12)1.74 × 10^6 m/s v_e = sqrt(5.67673 × 10^6) m/s v_e ≈ 2382.59 m/s Rounding to two significant figures: v_e ≈ 2.4 × 10^3 m/s The escape velocity from the moon is: 2.4 × 10^3 m/s Question 13. a. (i): Viscosity and coefficient of viscosity • Viscosity: A measure of a fluid's resistance to flow. It describes the internal friction within a fluid when it is in motion. A highly viscous fluid flows slowly, while a low-viscosity fluid flows easily. • Coefficient of viscosity (): A quantitative measure of a fluid's viscosity. It is defined as the ratio of the shear stress to the rate of shear strain in a fluid. SI unit: Pascal-second (Pa · s) or Newton-second per square meter* (N s/m^2). Dimensions: [ML^-1T^-1]. Question 13. a. (ii): Stokes's Law and Terminal Velocity derivation Step 1: State the mathematical expression for Stokes's Law and define symbols. Stokes's Law describes the viscous drag force (F_d) on a small spherical object moving through a viscous fluid at low Reynolds numbers: F_d = 6 a v Where: • (eta) is the coefficient of viscosity of the fluid. • a is the radius of the spherical object. • v is the velocity of the spherical object relative to the fluid. Step 2: Sketch the forces acting on the sphere. When a metallic sphere falls under gravity in a liquid, three forces act on it: 1. Gravitational force (F_g) acting downwards. 2. Buoyant force (F_b) acting upwards (due to Archimedes' principle). 3. Viscous drag force (F_d) acting upwards (due to Stokes's Law). Step 3: Show the derivation of terminal velocity. At terminal velocity, the net force on the sphere is zero, meaning the upward forces balance the downward force: F_g = F_b + F_d Substitute the expressions for each force: • Gravitational force: F_g = mg = V g, where V is the volume of the sphere. For a sphere, V = (4)/(3) a^3. F_g = ((4)/(3) a^3) g • Buoyant force: F_b = V g, where is the density of the liquid. F_b = ((4)/(3) a^3) g • Viscous drag force (at terminal velocity v): F_d = 6 a v Substitute these into the force balance equation: ((4)/(3) a^3) g = ((4)/(3) a^3) g + 6 a v Rearrange to solve for v: 6 a v = ((4)/(3) a^3) g - ((4)/(3) a^3) g 6 a v = ((4)/(3) a^3) g ( - ) Divide both sides by 6 a: v = ((4)/(3) a^3) g ( - )6 a Simplify the expression: v = (4 a^3 g ( - ))/(3 × 6 a) v = (4 a^3 g ( - ))/(18 a) Cancel and one a: v = (4 a^2 g ( - ))/(18 ) Reduce the fraction (4)/(18) to (2)/(9): v = (2a^2( - )g)/(9) Question 13. b. (i): Conditions for liquid to be ideal Two conditions for a liquid to be considered ideal are: 1. It is incompressible: Its density remains constant regardless of pressure changes. 2. It is non-viscous: It has no internal friction, meaning its coefficient of viscosity is zero. Question 13. b. (ii): Kinetic and potential energies per unit volume Step 1: Define density. The density of the liquid is = (m)/(V). Step 2: Derive the expression for kinetic energy per unit volume. The kinetic energy (KE) of a liquid of mass m and velocity v is: KE = (1)/(2)mv^2 To find the kinetic energy per unit volume, divide by the volume V: (KE)/(V) = (1)/(2)mv^2V Substitute m = V: (KE)/(V) = (1)/(2)( V)v^2V (KE)/(V) = (1)/(2) v^2 Step 3: Derive the expression for potential energy per unit volume. The potential energy (PE) of a liquid of mass m at a height h is: PE = mgh To find the potential energy per unit volume, divide by the volume V: (PE)/(V) = (mgh)/(V) Substitute m = V: (PE)/(V) = (( V)gh)/(V) (PE)/(V) = gh Question 14. a. (i): State the zeroth and first laws of thermodynamics • Zeroth Law of Thermodynamics: If two thermodynamic systems are each in thermal equilibrium with a third system, then they are in thermal equilibrium with each other. This law establishes the concept of temperature. • First Law of Thermodynamics: The change in the internal energy ( U) of a system is equal to the heat added to the system (Q) minus the work done by the system (W). It is a statement of the conservation of energy. U = Q - W Drop the next question.