🔑:## Step 1: Understand the Stefan-Boltzmann LawThe Stefan-Boltzmann law states that the total energy radiated per unit surface area of a blackbody across all wavelengths per unit time (also known as the blackbody emissive power, E) is proportional to the fourth power of the blackbody's temperature, T. The law is given by (E = sigma T^4), where (sigma) is the Stefan-Boltzmann constant ((5.670367 times 10^{-8} W/m^2K^4)).## Step 2: Calculate the Total Energy Radiated by the StarThe total energy radiated by the star per unit time (power, P) can be calculated by multiplying the energy radiated per unit surface area by the surface area of the star ((4pi a^2)), where (a) is the radius of the star. Thus, (P = 4pi a^2 sigma T^4).## Step 3: Determine the Energy Received by the ShellSince the shell is a Gaussian surface surrounding the star and is at a distance R from the center of the star, the energy radiated by the star is distributed over the surface area of a sphere with radius R. The surface area of this sphere is (4pi R^2). However, to find the energy received by the shell, we consider that the energy emitted by the star is evenly distributed over this surface area.## Step 4: Calculate the Energy Flux Through the ShellThe energy flux (energy per unit area per unit time) through the shell can be found by dividing the total power emitted by the star by the surface area of the sphere with radius R. Thus, the energy flux is (P / (4pi R^2)).## Step 5: Calculate the Total Energy Received by the Shell in an HourTo find the total energy received by the shell in an hour, we multiply the energy flux by the surface area of the shell ((4pi R^2)) and by the number of seconds in an hour ((3600)). However, since the energy flux already accounts for the distribution over (4pi R^2), we simply multiply the power emitted by the star by the number of seconds in an hour.## Step 6: Perform the CalculationGiven (P = 4pi a^2 sigma T^4), the energy received in an hour is (E_{hour} = P times 3600 = 4pi a^2 sigma T^4 times 3600).The final answer is: boxed{4pi a^2 sigma T^4 times 3600}

🔑:## Step 1: Introduction to Nuclear ModelsThe Liquid Drop Model (LDM) and the Shell Model (SM) are two fundamental models in nuclear physics used to describe the structure of atomic nuclei. The LDM treats the nucleus as a drop of liquid, emphasizing its collective properties, while the SM describes the nucleus in terms of individual nucleon orbits, focusing on the shell structure of protons and neutrons.## Step 2: Limitations of the Liquid Drop ModelThe LDM has several limitations. It does not account for the shell structure of nuclei, which is crucial for understanding the stability of certain nuclei and the occurrence of magic numbers. Additionally, it oversimplifies the nucleus's surface tension and neglects the effects of quantum mechanics on nuclear structure.## Step 3: Limitations of the Shell ModelThe SM, while successful in explaining the shell structure and magic numbers, has its own limitations. It assumes that nucleons move independently in a mean field, neglecting the complexities of nucleon-nucleon interactions and correlations. This simplification can lead to inaccuracies in predicting the properties of nuclei, especially those far from stability.## Step 4: Impact on Understanding Nuclear StructureThe limitations of these models impact our understanding of nuclear structure by providing an incomplete picture of nuclear properties and behavior. They are particularly challenged when dealing with exotic nuclei, where the interplay between collective and single-particle degrees of freedom is significant, and in predicting the properties of nuclei that are far from the valley of stability.## Step 5: First-Principles QCD ComputationsFirst-principles Quantum Chromodynamics (QCD) computations aim to describe the nucleus directly from the underlying theory of strong interactions, without relying on phenomenological models. These computations, often performed using lattice QCD, have the potential to overcome the limitations of the LDM and SM by directly calculating nuclear properties from the interactions of quarks and gluons.## Step 6: Current State of Lattice QCDThe current state of lattice QCD computations has made significant progress in recent years, with calculations of simple nuclear systems (like the deuteron and light nuclei) showing promising results. However, computing the properties of larger nuclei remains a significant challenge due to the complexity of QCD and the computational resources required.## Step 7: Challenges and Future DirectionsDespite the challenges, ongoing efforts to improve lattice QCD calculations and the development of new computational methods and algorithms are expected to enhance our ability to compute nuclear properties from first principles. This will provide a more fundamental understanding of nuclear structure, complementing and potentially surpassing the predictive power of phenomenological models like the LDM and SM.The final answer is: boxed{QCD}

🔑:## Step 1: Understand the Problem and Identify Key ComponentsWe are tasked with deriving the equation for the distance traveled by a projectile launched from the Earth's surface with an initial velocity (v_0) at an angle (theta) above the horizontal. Key components include the initial velocity (v_0), the angle of projection (theta), the gravitational acceleration (g), and the distance traveled, which we aim to express in terms of these parameters.## Step 2: Resolve Initial Velocity into Horizontal and Vertical ComponentsThe initial velocity (v_0) can be resolved into horizontal and vertical components. The horizontal component of the initial velocity is (v_{0x} = v_0 cos(theta)), and the vertical component is (v_{0y} = v_0 sin(theta)). These components are crucial for analyzing the motion.## Step 3: Analyze Horizontal MotionThe horizontal motion of the projectile is uniform since there is no acceleration in the horizontal direction (assuming negligible air resistance). The distance traveled in the horizontal direction can be calculated using the formula for uniform motion: (x = v_{0x}t), where (t) is the time of flight.## Step 4: Analyze Vertical MotionThe vertical motion of the projectile is under constant acceleration due to gravity ((g)). The vertical component of the velocity at any time (t) is given by (v_y = v_{0y} - gt), using the equation for uniformly accelerated motion. The time it takes for the projectile to reach its maximum height and return to the ground (time of flight) can be found by setting the final vertical velocity to zero at the peak and then to the ground level: (0 = v_{0y} - gt_{text{up}}), which gives (t_{text{up}} = frac{v_{0y}}{g}). The total time of flight is twice this, (t_{text{total}} = 2t_{text{up}} = frac{2v_{0y}}{g}).## Step 5: Calculate Distance TraveledThe horizontal distance traveled (range) can be found by multiplying the horizontal velocity component by the total time of flight: (R = v_{0x} times t_{text{total}} = v_0 cos(theta) times frac{2v_{0y}}{g}). Substituting (v_{0y} = v_0 sin(theta)) gives (R = v_0 cos(theta) times frac{2v_0 sin(theta)}{g}).## Step 6: Simplify the Equation for Distance TraveledSimplifying the equation gives (R = frac{2v_0^2 sin(theta) cos(theta)}{g}). Using the trigonometric identity (sin(2theta) = 2sin(theta)cos(theta)), we can rewrite this as (R = frac{v_0^2 sin(2theta)}{g}).The final answer is: boxed{frac{v_0^2 sin(2theta)}{g}}

🔑:## Step 1: Understand the given electric field intensityThe electric field intensity is given as vec E = hat x 100 x ,,(V/m). This means the electric field varies linearly with the x-coordinate.## Step 2: Recall Gauss's LawGauss's Law states that the total electric flux through a closed surface is proportional to the charge enclosed within that surface. Mathematically, it is expressed as Phi = oint vec E cdot dvec A = frac{Q}{epsilon_0}, where Phi is the electric flux, vec E is the electric field, dvec A is the differential area element, Q is the enclosed charge, and epsilon_0 is the electric constant (permittivity of free space).## Step 3: Determine the surface area of the cubical volumeThe cubical volume has sides of length 100 ,, (mm), which is 0.1 ,, (m). The surface area of one face of the cube is 0.1 times 0.1 = 0.01 ,, (m^2).## Step 4: Calculate the electric flux through the cubeTo find the total electric flux through the cube, we need to integrate the electric field over the surface area of the cube. Since the cube is centered at the origin and is symmetric, we can simplify the calculation by considering the flux through the faces perpendicular to the x-axis. For these faces, the electric field is parallel to the area element dvec A, and its magnitude varies linearly from -0.05 to 0.05 meters from the center.## Step 5: Calculate the electric flux through the faces perpendicular to the x-axisFor the face at x = 0.05, the electric field is vec E = hat x 100 times 0.05 = 5 ,, (V/m). For the face at x = -0.05, the electric field is vec E = hat x 100 times -0.05 = -5 ,, (V/m). The flux through each face is the product of the electric field and the area of the face.## Step 6: Calculate the electric fluxThe electric flux through the face at x = 0.05 is 5 times 0.01 = 0.05 ,, (Vcdot m), and through the face at x = -0.05 is -5 times 0.01 = -0.05 ,, (Vcdot m). However, since we are calculating the total flux through the cube, we must consider that the flux enters through one face and exits through the opposite face, resulting in a net flux of 0.05 - (-0.05) = 0.1 ,, (Vcdot m) through these two faces.## Step 7: Account for all faces of the cubeHowever, since the electric field is only in the x-direction and varies linearly with x, the net flux through the cube is actually determined by the difference in electric field between the two faces perpendicular to the x-axis. The faces perpendicular to the y and z axes do not contribute to the net flux since the electric field is parallel to these faces.## Step 8: Apply Gauss's Law to find the enclosed chargeGiven that the net flux through the cube is 0.1 ,, (Vcdot m) and using Gauss's Law Phi = frac{Q}{epsilon_0}, we can solve for Q. The electric constant epsilon_0 is approximately 8.854 times 10^{-12} ,, (F/m).## Step 9: Calculate the enclosed chargeRearrange Gauss's Law to solve for Q: Q = Phi times epsilon_0. Substituting the values, Q = 0.1 times 8.854 times 10^{-12}.## Step 10: Perform the final calculationQ = 0.1 times 8.854 times 10^{-12} = 8.854 times 10^{-13} ,, (C).The final answer is: boxed{8.854 times 10^{-13}}