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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Step 1: Fill the table for heat transfer properties. | Property | Conduction | Convection | Radiation | | :-------------------- | :--------- | :--------- | :-------- | | Medium needed | Yes | Yes | No | | Occurs in vacuum | No | No | Yes | | Involves fluid motion | No | Yes | No | Step 2: Calculate the heat transferred. Given: Length L = 300 cm = 3 m Area A = 2.4 m^2 Initial temperature T_1 = 30^ Final temperature T_2 = 90^ Time t = 13 s Thermal conductivity K = 200 W/mK The temperature difference is T = T_2 - T_1 = 90^ - 30^ = 60^ = 60 K. The formula for heat transfer by conduction is: (Q)/(t) = K A ( T)/(L) Rearranging to solve for Q: Q = K A ( T)/(L) t Substitute the given values: Q = (200 W/mK) × (2.4 m^2) × 60 K3 m × (13 s) Q = (200 × 2.4 × 20 × 13) J Q = (480 × 20 × 13) J Q = (9600 × 13) J Q = 124800 J Q = 124.8 kJ The heat transferred is 124.8 kJ. Step 3: Find the magnetic field inside a solenoid. Given: Number of turns per unit length n = 500 turns/m Current I = 2 A The permeability of free space _0 = 4 × 10^-7 T·m/A The formula for the magnetic field inside a solenoid is: B = _0 n I Substitute the given values: B = (4 × 10^-7 T·m/A) × (500 turns/m) × (2 A) B = (4 × 10^-7 × 1000) T B = (4 × 10^-4) T B ≈ (4 × 3.14159 × 10^-4) T B ≈ 1.2566 × 10^-3 T The magnetic field inside the solenoid is 1.26 × 10^-3 T. Step 4: Proof that the work done in an adiabatic process can be given as w = | (R)/(1-) (T_2 - T_1) |. For a reversible adiabatic process, the work done W is given by: W = _V_1^V_2 P dV For an adiabatic process, PV^ = constant = K. Thus, P = K V^-. Substitute P into the work integral: W = _V_1^V_2 K V^- dV W = K [ V^-+1-+1 ]_V_1^V_2 W = (K)/(1-) (V_2^1- - V_1^1-) Since K = P_1 V_1^ = P_2 V_2^, we can substitute K: W = (1)/(1-) (P_2 V_2^ V_2^1- - P_1 V_1^ V_1^1-) W = (1)/(1-) (P_2 V_2 - P_1 V_1) For an ideal gas, PV = nRT. For one mole (n=1), PV = RT. Substitute P_1 V_1 = RT_1 and P_2 V_2 = RT_2: W = (1)/(1-) (RT_2 - RT_1) W = (R)/(1-) (T_2 - T_1) The magnitude of the work done is w = |W|, so: w = | (R)/(1-) (T_2 - T_1) | This completes the proof. Step 5: Proof that the relationship between pressure and volume in an adiabatic change can be given as P V^ = constant. For an adiabatic process, there is no heat exchange, so dQ = 0. According to the first law of thermodynamics: dU = dQ - dW dU = -dW For an ideal gas, the change in internal energy is dU = n C_v dT, and the work done is dW = P dV. So, n C_v dT = -P dV (1) From the ideal gas law, PV = nRT. Differentiating this equation (assuming n and R are constant): P dV + V dP = nR dT Rearranging for dT: dT = (P dV + V dP)/(nR) (2) Substitute equation (2) into equation (1): n C_v ( (P dV + V dP)/(nR) ) = -P dV (C_v)/(R) (P dV + V dP) = -P dV We know that for an ideal gas, R = C_p - C_v. So, (C_v)/(C_p - C_v) (P dV + V dP) = -P dV. Divide both sides by C_v: (1)/(C_p)C_v - 1 (P dV + V dP) = -P dV Let = (C_p)/(C_v) (the adiabatic index). (1)/( - 1) (P dV + V dP) = -P dV Multiply both sides by ( - 1): P dV + V dP = -( - 1) P dV Rearrange the terms to group P dV: V dP = -( - 1) P dV - P dV V dP = - P dV Now, separate the variables P and V: (dP)/(P) = - (dV)/(V) Integrate both sides: (dP)/(P) = - (dV)/(V) P = - V + C' where C' is the integration constant. P + V = C' Using logarithm properties, V = V^: P + V^ = C' (P V^) = C' Exponentiate both sides: P V^ = C' Thus, P V^ = constant. This also implies P_1 V_1^ = P_2 V_2^. This completes the proof. Send me the next one 📸