🔑:Given,F = kq1q2/r2New force,F’ = k(2q1)(3q2)/(r/4)2= 96kq1q2/r2= 96FThus, the force between them is increased by a factor of 96.

🔑:## Step 1: Understand the given problem and identify the key components.We are given a bulk metal with anisotropic electrical conductivity, represented by the conductivity tensor σ_ij, and the relationship between the electric field E_j and the electric displacement field D_i is given by D_i = ε_ij E_j, where ε_ij is the permittivity tensor. We need to derive an expression for the electric current density j_i in terms of the electric displacement field D_k.## Step 2: Recall the relationship between electric current density and electric field.The electric current density j_i is related to the electric field E_j by the conductivity tensor σ_ij through the equation j_i = σ_ij E_j.## Step 3: Express the electric field E_j in terms of the electric displacement field D_k.From the given equation D_i = ε_ij E_j, we can solve for E_j by inverting the permittivity tensor: E_j = ε_ij^(-1) D_i, where ε_ij^(-1) represents the inverse of the permittivity tensor.## Step 4: Substitute the expression for E_j into the equation for j_i.Substituting E_j = ε_ij^(-1) D_i into j_i = σ_ij E_j gives j_i = σ_ij ε_jk^(-1) D_k, where we use the Einstein summation convention for repeated indices.## Step 5: Discuss the implications of anisotropy on the boundary conditions for the electric potential.Anisotropy in the conductivity and permittivity tensors means that the electric current density and electric displacement field will depend on the direction. This anisotropy affects the boundary conditions for the electric potential, as the tangential component of the electric field and the normal component of the electric displacement field must be continuous across interfaces between different materials. The anisotropic nature of the tensors complicates these boundary conditions, requiring a more nuanced approach to solving electrostatic problems in such materials.The final answer is: boxed{j_i = σ_ij ε_jk^(-1) D_k}

🔑:## Step 1: Understanding Hubble's LawHubble's Law states that the recessional velocity (v) of a galaxy is directly proportional to its distance (D) from the observer, with the Hubble constant (H_0) being the proportionality constant. The formula is given by (v = H_0D).## Step 2: Calculating the Recessional VelocityGiven (H_0 = 69.32 , text{km/s/Mpc}) and (D = 100 , text{Mpc}), we can substitute these values into Hubble's Law to calculate the recessional velocity (v). Thus, (v = 69.32 , text{km/s/Mpc} times 100 , text{Mpc}).## Step 3: Performing the Calculation(v = 69.32 times 100 = 6932 , text{km/s}).## Step 4: Discussing the ImplicationsThe calculated recessional velocity of approximately (6932 , text{km/s}) is significant in the context of the expansion of the universe. This velocity is a result of the expansion of space itself, as described by the Big Bang theory and observed through the redshift of light from distant galaxies. It does not imply that the galaxy is moving through space at this speed, but rather that the space between us and the galaxy is expanding at a rate that corresponds to this velocity.## Step 5: Considering the Speed of LightThe speed of light ((c)) is approximately (299,792 , text{km/s}). The calculated recessional velocity of (6932 , text{km/s}) is well below the speed of light, which is a critical aspect since, according to the theory of special relativity, no object can reach or exceed the speed of light. The recessional velocity, being a consequence of the expansion of space, does not violate this principle because it is not a velocity through space but rather a velocity of space expansion itself.The final answer is: boxed{6932}

🔑:## Step 1: Identify the given informationThe speed of each beam of protons relative to the earth is given as 0.700c, where c is the speed of light.## Step 2: Determine the relative velocity of the two beamsTo find the magnitude of the velocity of one beam relative to the other, we use the formula for relativistic relative velocity, which is given by:[ v_{rel} = frac{v_1 + v_2}{1 + frac{v_1 v_2}{c^2}} ]Here, (v_1) and (v_2) are the velocities of the two beams relative to the earth, and (c) is the speed of light.## Step 3: Substitute the given values into the formulaGiven that (v_1 = 0.700c) and (v_2 = 0.700c), substituting these values into the formula gives:[ v_{rel} = frac{0.700c + 0.700c}{1 + frac{(0.700c)(0.700c)}{c^2}} ]## Step 4: Simplify the expression[ v_{rel} = frac{1.400c}{1 + frac{0.49c^2}{c^2}} ][ v_{rel} = frac{1.400c}{1 + 0.49} ][ v_{rel} = frac{1.400c}{1.49} ]## Step 5: Calculate the relative velocity[ v_{rel} = frac{1.400c}{1.49} ][ v_{rel} approx 0.939c ]The final answer is: boxed{0.939c}