🔑:## Step 1: Understanding the Periodic Potential of Metal IonsThe periodic potential of metal ions refers to the regular arrangement of positive ions (metal ions) in a crystal lattice. This arrangement creates a periodic electrostatic potential that influences the behavior of conducting electrons within the metal. The potential is such that it has minima at the positions of the ions, where the potential energy of an electron would be lowest due to the attractive force between the positive ion and the negative electron.## Step 2: Concept of Work FunctionThe work function of a metal is the minimum amount of energy required for an electron to escape from the surface of the metal into a vacuum. It is a measure of how strongly the metal holds onto its electrons. The work function is influenced by the periodic potential of the metal ions and the interactions between electrons themselves, including electron-electron repulsions and the screening effect that reduces the effective charge seen by an electron.## Step 3: Forces Acting on Conducting ElectronsConducting electrons in a metal are subject to several forces:1. Attractive force to metal ions: The positive metal ions in the lattice attract the negatively charged electrons, keeping them within the metal.2. Repulsive forces from other electrons: Due to the Pauli exclusion principle and Coulomb repulsion, electrons also experience repulsive forces from other electrons, which helps to distribute them throughout the metal.3. External electric field: When an external electric field is applied, it exerts a force on the electrons, trying to pull them out of the metal.## Step 4: Why Electrons Do Not Fly Off into SpaceDespite the external electric field, conducting electrons do not fly off into space for several reasons:1. Binding Energy: The electrons are bound to the metal lattice by the attractive forces from the metal ions, which requires a certain amount of energy (the work function) to overcome.2. Screening Effect: The electrons themselves screen the external electric field within the metal, reducing its effect on the electrons deeper in the metal. This screening effect is due to the redistribution of electrons in response to the external field, creating an opposing field that cancels out the external field inside the metal.3. Equilibrium Condition: In the presence of an external field, the electrons at the surface of the metal may be redistributed, but an equilibrium is established where the force due to the external field is balanced by the forces holding the electrons to the metal (attractive forces to ions and repulsive forces from other electrons).## Step 5: ConclusionThe conducting electrons in a metal do not fly off into space under the influence of an external electric field because they are bound by the attractive forces of the metal ions and because of the concept of work function, which defines the minimum energy required for an electron to escape. The screening effect and the establishment of an equilibrium condition also play crucial roles in keeping the electrons within the metal.The final answer is: boxed{0}

🔑:## Step 1: Identify the parametric solution for an open, dust-filled universeThe parametric solution for an open, dust-filled universe is given by the Friedmann equations. For an open universe, the scale factor (a(t)) can be expressed in terms of time (t) using the parametric equations:[a(t) = frac{Omega_m}{2(1 - Omega_m)} (cosh(eta) - 1)][t = frac{Omega_m}{2H_0(1 - Omega_m)^{3/2}} (sinh(eta) - eta)]where (Omega_m) is the matter density parameter, (H_0) is the Hubble constant, and (eta) is a parameter.## Step 2: Calculate the Hubble radius (RH)The Hubble radius, which is the distance light could have traveled since the Big Bang if the expansion had been constant at its current rate, is given by:[R_H = frac{c}{H}]where (c) is the speed of light and (H) is the Hubble parameter. For an open universe, (H) can be expressed as a function of (a(t)) and its derivatives.## Step 3: Express the Hubble parameter (H) in terms of a(t) and its derivativesThe Hubble parameter (H) is defined as:[H = frac{dot{a}}{a}]For an open universe, using the parametric solution, we can express (dot{a}) and (a) in terms of (eta), but since the question focuses on comparing (R_H) and (R_p), we need to relate these quantities directly to the scale factor and its evolution.## Step 4: Calculate the radius of the particle horizon (Rp)The radius of the particle horizon, which is the distance light could have traveled since the Big Bang, is given by:[R_p = a(t) int_{0}^{t} frac{cdt'}{a(t')}]For an open universe, this integral can be evaluated using the parametric solution for (a(t)).## Step 5: Evaluate the condition for RH = RpTo determine if there exists a time when (R_H = R_p), we need to set the two equations equal to each other and solve for (t) or (eta). However, given the complexity of directly solving the parametric equations for (a(t)) and (H(t)) in terms of (eta) and then equating (R_H) and (R_p), we consider the physical implications and the behavior of these quantities over time.## Step 6: Analyze the behavior of RH and Rp over timeIn an expanding universe, (R_H) decreases as (H) increases with the expansion rate, while (R_p) increases as more light reaches us from distant parts of the universe. The condition (R_H = R_p) would imply a specific moment in the universe's evolution where the distance light could travel at the current expansion rate equals the distance light has traveled since the Big Bang.## Step 7: Consider the limitations and simplificationsGiven the complexity of the parametric equations and the integrals involved, an exact analytical solution for when (R_H = R_p) might not be straightforward. The evolution of (R_H) and (R_p) depends on the specific values of (Omega_m) and (H_0), which determine the expansion history of the universe.The final answer is: boxed{RH = Rp at eta = 2 sinh^{-1}(1) = 2 cdot ln(1 + sqrt{2})}

🔑:## Step 1: Understand the problemWe have a uniformly charged rod positioned along the x-axis, and we need to calculate the electric field at a point P on the y-axis. The rod has a length L and a charge Q distributed uniformly along its length.## Step 2: Calculate the electric fieldTo calculate the electric field, we can divide the rod into small segments of length dx and calculate the electric field due to each segment. The electric field due to a small segment of charge is given by Coulomb's law: dE = k * dq / r^2, where k is Coulomb's constant, dq is the charge of the segment, and r is the distance from the segment to point P.## Step 3: Integrate the electric fieldWe can integrate the electric field due to each segment to find the total electric field at point P. Since the rod is symmetric about the origin, the x-component of the electric field will be zero, and the y-component will be non-zero.## Step 4: Calculate the y-component of the electric fieldThe y-component of the electric field is given by E_y = (k * Q / L) * ∫[from -L/2 to L/2] (y / (x^2 + y^2)) dx, where y is the distance from the x-axis to point P.## Step 5: Evaluate the integralEvaluating the integral, we get E_y = (k * Q / L) * [arctan(x/y)] from -L/2 to L/2.## Step 6: Simplify the expressionSimplifying the expression, we get E_y = (2 * k * Q / L) * arctan(L/2y).## Step 7: Consider the x-component of the electric fieldSince the rod is symmetric about the origin, the x-component of the electric field will be zero. However, if we consider a small segment of the rod, the x-component of the electric field due to that segment will be non-zero.## Step 8: Explain the negative sign for the x-componentThe x-component of the electric field due to a small segment of the rod will have a negative sign if the segment is to the left of the origin and a positive sign if the segment is to the right of the origin. This is because the electric field due to a positive charge points away from the charge.## Step 9: ConclusionSince the rod is symmetric about the origin, the x-component of the electric field will be zero when integrated over the entire rod. However, for a small segment of the rod, the x-component of the electric field will have a negative sign if the segment is to the left of the origin.The final answer is: boxed{0}

🔑:The strategy is to use the free-body diagram of the entire assembly, and then use the equations of equilibrium to determine the support reactions.The free-body diagram of the assembly is shown. The angle between the cable and the horizontal is[alpha=tan^{-1}left(frac{12}{16}right)=36.9^{circ}.]The equilibrium equations for the assembly are[sum F_{X} =C_{X}+Tcosalpha=0,] [sum F_{Y} =C_{Y}+Tsinalpha-100=0,] [sum M_{C} =M_{C}+(4)(100)+(8)(Tsinalpha)=0.]Substitute the numerical values and solve:[C_{X}+800cos 36.9^{circ} =0,] [C_{Y}+800sin 36.9^{circ}-100 =0,] [M_{C}+400+(8)(800sin 36.9^{circ}) =0,]from which[C_{X} =-640,mathrm{lb},] [C_{Y} =-481,mathrm{lb},] [M_{C} =-2180,mathrm{ft},mathrm{lb}.]The reactions are (C_{X}=-640,mathrm{lb},C_{Y}=-481,mathrm{lb},M_{C}=-2180,mathrm{ft}, mathrm{lb})