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.
ScanSolve AI Answer
Here are the solutions to questions 2, 3, 4, and 5. 2. At what temperature will the Celsius and Fahrenheit temperature scales record the same reading? Step 1: Set the Celsius and Fahrenheit temperature formulas equal to each other. The formula for converting Celsius to Fahrenheit is T_F = (9)/(5)T_C + 32. We want to find the temperature x where T_C = T_F = x. x = (9)/(5)x + 32 Step 2: Solve for x. Subtract (9)/(5)x from both sides: x - (9)/(5)x = 32 (5)/(5)x - (9)/(5)x = 32 -(4)/(5)x = 32 Multiply both sides by -(5)/(4): x = 32 × (-(5)/(4)) x = 8 × (-5) x = -40 The temperature at which Celsius and Fahrenheit scales record the same reading is -40^ or -40^. 3. A malaria patient has a body temperature of 39.5^. Convert this temperature to (a) ^ (b) K. Given temperature T_C = 39.5^. a) Convert to Fahrenheit Step 1: Use the conversion formula T_F = (9)/(5)T_C + 32. T_F = (9)/(5)(39.5) + 32 Step 2: Calculate the Fahrenheit temperature. T_F = 9 × 7.9 + 32 T_F = 71.1 + 32 T_F = 103.1^ The temperature in Fahrenheit is 103.1^. b) Convert to Kelvin Step 1: Use the conversion formula T_K = T_C + 273.15. T_K = 39.5 + 273.15 Step 2: Calculate the Kelvin temperature. T_K = 312.65 K The temperature in Kelvin is 312.65 K. 4. A steel rod increases its length by 5 mm when the temperature increases by 10^. What is the initial length of the rod if the coefficient of linear expansion for steel is 1.1 × 10^-5 per ^? Step 1: Identify the given values and the formula for linear expansion. Change in length, L = 5 mm = 5 × 10^-3 m. Change in temperature, T = 10^. Coefficient of linear expansion, = 1.1 × 10^-5 per ^. The formula for linear expansion is L = L_0 T, where L_0 is the initial length. Step 2: Rearrange the formula to solve for L_0. L_0 = ( L)/( T) Step 3: Substitute the values and calculate L_0. L_0 = 5 × 10^-3 m(1.1 × 10^-5 per ^)(10^) L_0 = 5 × 10^-31.1 × 10^-4 m L_0 = (5)/(1.1 × 10^-1) m L_0 = (5)/(0.11) m L_0 ≈ 45.45 m The initial length of the rod is approximately 45.45 m. 5. How much heat must be added to a 4.0 × 10^-3 kg steel ball bearing in order to increase its temperature by 30 K if the specific heat capacity of steel is 4.49 × 10^2 J/Kg.K? And how much will the temperature of the ball increase if it were made of gold of specific heat capacity 1.29 × 10^2 J/Kg.K rather than steel? Part 1: Heat added to the steel ball Step 1: Identify the given values for the steel ball and the formula for heat transfer. Mass of steel ball, m = 4.0 × 10^-3 kg. Change in temperature, T = 30 K. Specific heat capacity of steel, c_steel = 4.49 × 10^2 J/Kg.K. The formula for heat transfer is Q = mc T. Step 2: Substitute the values and calculate the heat added Q. Q = (4.0 × 10^-3 kg)(4.49 × 10^2 J/Kg.K)(30 K) Q = (4.0 × 4.49 × 30) × 10^-3 × 10^2 J Q = (538.8) × 10^-1 J Q = 53.88 J The heat that must be added is 53.88 J. Part 2: Temperature increase for a gold ball with the same heat added Step 3: Identify the given values for the gold ball and the heat added. Heat added, Q = 53.88 J (from Part 1). Mass of gold ball, m = 4.0 × 10^-3 kg (assuming the same mass). Specific heat capacity of gold, c_gold = 1.29 × 10^2 J/Kg.K. The formula for heat transfer is Q = mc_gold T_gold. Step 4: Rearrange the formula to solve for T_gold. T_gold = (Q)/(mc_gold) Step 5: Substitute the values and calculate the temperature increase for the gold ball. T_gold = 53.88 J(4.0 × 10^-3 kg)(1.29 × 10^2 J/Kg.K) T_gold = (53.88)/((4.0 × 1.29) × 10^-3) × 10^2 K T_gold = (53.88)/(5.16 × 10^-1) K T_gold = (53.88)/(0.516) K T_gold ≈ 104.42 K The temperature of the gold ball will increase by approximately 104.42 K. Send me the next one 📸