🔑:To calculate the overall heat transfer coefficient (U) for the given scenario, we need to determine the heat transfer coefficients for both the water side and the air side of the tank. The heat transfer coefficient for the water side can be estimated using an enclosure natural circulation correlation, while for the air side, the Churchill and Chu correlation can be used for natural convection.## Step 1: Determine the Water Side Heat Transfer CoefficientFor natural convection in an enclosure (the water side in this case), a common correlation used is:[Nu = 0.42Ra^{0.25}Pr^{0.012}left(frac{L_{v}}{L}right)^{-0.3}]for (Ra < 10^4), where (Nu) is the Nusselt number, (Ra) is the Rayleigh number, (Pr) is the Prandtl number, (L_{v}) is the height of the enclosure (water depth in the tank), and (L) is the length of the enclosure (the dimension parallel to the heat flow direction).## Step 2: Calculate the Rayleigh Number for WaterThe Rayleigh number (Ra) for water is given by:[Ra = frac{gbetaDelta T L^3}{nualpha}]where (g) is the acceleration due to gravity (approximately (9.81 m/s^2)), (beta) is the volume expansion coefficient of water at 50°C, (Delta T) is the temperature difference between the water and the air (50°C - 25°C = 25°C), (L) is the reference length (which could be the height of the water or the length of the tank, depending on the orientation of the heat transfer surface), (nu) is the kinematic viscosity of water at 50°C, and (alpha) is the thermal diffusivity of water at 50°C.## Step 3: Determine the Air Side Heat Transfer CoefficientFor the air side, the Churchill and Chu correlation for natural convection from a horizontal surface is:[Nu = 0.15Ra^{1/3}]for (Ra < 10^{11}). This correlation is suitable for a wide range of Rayleigh numbers and is often used for natural convection calculations involving air.## Step 4: Calculate the Rayleigh Number for AirThe Rayleigh number for air is calculated similarly to that for water, but using the properties of air at the film temperature (the average of the surface and ambient temperatures, which is ((50°C + 25°C)/2 = 37.5°C)).## Step 5: Calculate the Overall Heat Transfer CoefficientThe overall heat transfer coefficient (U) is given by:[frac{1}{U} = frac{1}{h_{water}} + frac{1}{h_{air}}]where (h_{water}) and (h_{air}) are the heat transfer coefficients for the water and air sides, respectively, calculated using the Nusselt numbers from the correlations.## Step 6: Apply the CorrelationsTo apply the correlations, we need the physical properties of water at 50°C and air at 37.5°C. For water at 50°C: (nu approx 5.47 times 10^{-7} m^2/s), (alpha approx 1.51 times 10^{-7} m^2/s), (beta approx 4.33 times 10^{-4} K^{-1}), and (Pr approx 3.55). For air at 37.5°C: (nu approx 1.66 times 10^{-5} m^2/s), (alpha approx 2.29 times 10^{-5} m^2/s), (beta approx 1/T approx 1/310.65 K^{-1}), and (Pr approx 0.707).## Step 7: Calculate the Reference LengthsThe reference length (L) for the water side could be the height of the water, and for the air side, it could be the length of the tank's surface exposed to air.## Step 8: Calculate the Nusselt Numbers and Heat Transfer CoefficientsGiven the complexity and the need for specific numerical values for the dimensions of the tank and the properties of the fluids, let's proceed with calculating the Rayleigh numbers and Nusselt numbers directly.The final answer is: boxed{10}

🔑:## Step 1: Understanding the ProblemWe are tasked with explaining why the mass of a projectile does not affect its trajectory under the sole influence of gravity, using principles of physics such as Newton's laws of motion and the equivalence principle. This involves understanding the relationship between the gravitational force acting on the projectile and its inertial and gravitational mass.## Step 2: Newton's Second Law of MotionNewton's second law of motion states that the force acting on an object is equal to the mass of the object multiplied by its acceleration (F = ma). For an object under the influence of gravity, the force acting on it is given by its weight, which is the product of its mass (m) and the acceleration due to gravity (g), i.e., F = mg.## Step 3: Equivalence PrincipleThe equivalence principle, a fundamental concept in general relativity, states that the effects of gravity are equivalent to the effects of acceleration. This principle implies that all objects, regardless of their mass, fall at the same rate in a gravitational field, provided that air resistance and other external forces are negligible.## Step 4: Mathematical DerivationTo derive the trajectory of the projectile mathematically, we consider the motion in the vertical and horizontal directions separately, as the motion in these directions is independent in the absence of air resistance.- Vertical Direction: The acceleration due to gravity is -g (downwards). Using Newton's second law, the equation of motion in the vertical direction is given by m cdot a_y = m cdot g, where a_y is the acceleration in the vertical direction. Simplifying, we get a_y = g, which is independent of mass.- Horizontal Direction: Since there is no force acting horizontally (assuming no air resistance), the acceleration in the horizontal direction is 0. The horizontal component of the initial velocity (v_{0x} = v_0 cdot cos(theta)) remains constant.## Step 5: Trajectory EquationThe trajectory of the projectile can be described by the parametric equations for the horizontal and vertical positions as functions of time:- x(t) = v_{0x} cdot t- y(t) = v_{0y} cdot t - frac{1}{2} cdot g cdot t^2, where v_{0y} = v_0 cdot sin(theta).## Step 6: Eliminating Time to Find the TrajectoryTo find the equation of the trajectory, we can eliminate time (t) from the parametric equations. First, solve x(t) = v_{0x} cdot t for t to get t = frac{x}{v_{0x}}. Then substitute t into the equation for y(t):y(x) = v_{0y} cdot frac{x}{v_{0x}} - frac{1}{2} cdot g cdot left(frac{x}{v_{0x}}right)^2.## Step 7: Simplifying the Trajectory EquationSimplifying the equation for y(x) gives us the trajectory of the projectile:y(x) = tan(theta) cdot x - frac{g}{2v_{0x}^2} cdot x^2, where we've used v_{0y}/v_{0x} = tan(theta).## Step 8: Conclusion on Mass IndependenceFrom the derived equations of motion and the trajectory equation, it's clear that the mass of the projectile does not appear in the final expressions for the trajectory. This is because the acceleration due to gravity (g) is independent of the mass of the object, as implied by the equivalence principle. Thus, the trajectory of the projectile under the sole influence of gravity does not depend on its mass.The final answer is: boxed{m}

🔑:## Step 1: Calculate the average kinetic energy of an electron in a ground-state helium atom per electronFirst, we need to convert the given average kinetic energy from kJ/mol to J/electron. The average kinetic energy per mole is 2.4x10^3 kJ/mol. To convert this to joules per electron, we use the conversion factor 1 kJ = 1000 J and Avogadro's number (6.022x10^23 electrons/mol). So, the average kinetic energy per electron is (2.4x10^3 kJ/mol * 1000 J/kJ) / (6.022x10^23 electrons/mol) = 3.98x10^-21 J/electron.## Step 2: Calculate the velocity of the electronThe kinetic energy of an electron is given by the formula KE = 0.5 * m * v^2, where m is the mass of the electron (approximately 9.11x10^-31 kg) and v is the velocity of the electron. Rearranging the formula to solve for v gives v = sqrt(2 * KE / m). Substituting the average kinetic energy per electron (3.98x10^-21 J) and the mass of the electron into the equation yields v = sqrt(2 * 3.98x10^-21 J / (9.11x10^-31 kg)) = sqrt(8.76x10^9 m^2/s^2) = 2.96x10^4 m/s.## Step 3: Determine the minimum uncertainty in the position of the electronThe Heisenberg Uncertainty Principle states that the product of the uncertainty in position (Δx) and the uncertainty in momentum (Δp) is greater than or equal to h/4π, where h is Planck's constant (approximately 6.626x10^-34 J*s). The uncertainty in momentum can be related to the uncertainty in velocity (Δv) by Δp = m * Δv. Given an uncertainty of 10% in the velocity, Δv = 0.1 * 2.96x10^4 m/s = 2.96x10^3 m/s. Thus, Δp = 9.11x10^-31 kg * 2.96x10^3 m/s = 2.70x10^-27 kg*m/s.## Step 4: Apply the Heisenberg Uncertainty PrincipleTo find the minimum uncertainty in the position, we use the formula Δx * Δp >= h / (4 * π). Rearranging for Δx gives Δx >= h / (4 * π * Δp). Substituting the values of h and Δp yields Δx >= (6.626x10^-34 J*s) / (4 * π * 2.70x10^-27 kg*m/s) = (6.626x10^-34 J*s) / (3.39x10^-26 kg*m/s) = 1.95x10^-8 m.## Step 5: Convert the uncertainty in position to picometers and compare with the effective radius of a helium atomTo convert meters to picometers, we multiply by 10^12 (since 1 m = 10^12 pm). Thus, 1.95x10^-8 m * 10^12 pm/m = 19.5 pm. The effective radius of a helium atom is given as 130 pm. To assess whether the uncertainty in position is a significant fraction of the effective radius, we compare 19.5 pm to 130 pm.The final answer is: boxed{19.5}

🔑:## Step 1: Understanding the Meissner EffectWhen a light with a magnetic field significantly lower than the critical field H_c is shone deep into a type-I superconductor, the superconductor will exhibit the Meissner effect. This effect states that a superconductor will expel all magnetic fields, including those from the light, from its interior.## Step 2: Behavior of the SuperconductorThe superconductor will behave as a perfect diamagnet, meaning it will generate a magnetic field that cancels out the external magnetic field from the light. This cancellation occurs because the superconductor's electrons will move in such a way as to create an opposing magnetic field, effectively expelling the magnetic field from the light.## Step 3: London Penetration DepthThe magnetic field from the light will not be completely expelled at the surface of the superconductor but will instead decay exponentially within a short distance known as the London penetration depth (lambda_L). The London penetration depth is given by the equation lambda_L = sqrt{frac{m}{mu_0 n e^2}}, where m is the mass of an electron, mu_0 is the permeability of free space, n is the density of superconducting electrons, and e is the charge of an electron.## Step 4: Behavior of the LightSince the magnetic field of the light is significantly lower than H_c, the light will not significantly affect the superconducting state of the material. The superconductor will continue to expel the magnetic field, and the light will propagate through the material with its magnetic field component being exponentially attenuated within the London penetration depth.## Step 5: ConclusionIn summary, when a light with a magnetic field significantly lower than H_c is shone deep into a type-I superconductor, the superconductor exhibits the Meissner effect, expelling the magnetic field from the light. The magnetic field decays exponentially within the London penetration depth, and the superconductor remains in its superconducting state.The final answer is: boxed{The superconductor expels the magnetic field, and the light's magnetic field component is attenuated within the London penetration depth.}