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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QUESTION 4: STRESS, STRAIN AND YOUNG MODULUS (continuation) 4.3.2 The strain. Step 1: Identify the given values. Original length (L_0) = 500 mm = 0.5 m Elongation ( L) = 0.25 mm = 0.00025 m Step 2: Calculate the strain (). = ( L)/(L_0) = 0.00025 m0.5 m = 0.0005 The strain is 0.0005. 4.3.3 The Young modulus of the material. Step 1: Identify the given and calculated values. Stress () = 101.86 MPa = 101.86 × 10^6 Pa (from 4.3.1) Strain () = 0.0005 (from 4.3.2) Step 2: Calculate the Young's Modulus (E). E = ()/() E = 101.86 × 10^6 Pa0.0005 E = 203.72 × 10^9 Pa E = 203.72 GPa The Young modulus of the material is 203.72 GPa. QUESTION 5: HEAT 5.1 Name the THREE basic gas processes. • Isobaric process (constant pressure) • Isochoric process (constant volume) • Isothermal process (constant temperature) 5.2 A square metal plate with sides 1 m long is at a temperature of 31 °C. There is a hole of 600 mm in diameter in the centre. The linear coefficient of expansion of the metal is 12.5 × 10^-6 /K. 5.2.1 Calculate: The temperature of the metal plate in if it is heated until the sides are 1,005 m long. Step 1: Identify the given values. Initial length (L_0) = 1 m Final length (L_f) = 1.005 m Initial temperature (T_0) = 31 °C Linear coefficient of expansion () = 12.5 × 10^-6 /K Step 2: Calculate the change in temperature ( T). The formula for linear expansion is L = L_0 T. T = (L_f - L_0)/( L_0) T = 1.005 m - 1 m(12.5 × 10^-6 /K) × 1 m T = (0.005)/(12.5 × 10^-6) = 400 K Step 3: Calculate the final temperature (T_f). T_f = T_0 + T T_f = 31 °C + 400 °C = 431 °C The final temperature of the metal plate is 431 °C. 5.2.2 The diameter of the hole at this temperature. Step 1: Identify the given and calculated values. Initial hole diameter (D_0) = 600 mm = 0.6 m Linear coefficient of expansion () = 12.5 × 10^-6 /K Change in temperature ( T) = 400 K (from 5.2.1) Step 2: Calculate the final diameter (D_f). The hole expands by the same linear expansion formula. D_f = D_0 (1 + T) D_f = 0.6 m (1 + (12.5 × 10^-6 /K) × 400 K) D_f = 0.6 m (1 + 0.005) D_f = 0.6 m × 1.005 D_f = 0.603 m The diameter of the hole at this temperature is 0.603 m. 5.3 A cylinder has a capacity of 0,075 m^3 and is filled with oxygen at a pressure of 380 kPa and a temperature of 50 °C. Later it is found that owing to a leak, the pressure has dropped to 280 kPa and the temperature has decreased to 25 °C. 5.3.1 Calculate the mass of the oxygen that was initially in the cylinder if the gas constant is 265 J/kg.K. Step 1: Identify the initial conditions and convert units. Volume (V) = 0.075 m^3 Initial pressure (P_1) = 380 kPa = 380 × 10^3 Pa Initial temperature (T_1) = 50 °C = 50 + 273.15 = 323.15 K Gas constant (R) = 265 J/kg.K Step 2: Use the ideal gas law to calculate the initial mass (m_1). The ideal gas law is PV = mRT. m_1 = (P_1 V)/(R T_1) m_1 = (380 × 10^3 Pa) × (0.075 m^3)(265 J/kg.K) × (323.15 K) m_1 = (28500)/(85634.75) m_1 ≈ 0.3328 kg The mass of the oxygen initially in the cylinder is 0.3328 kg. Drop the next question.