🔑:To solve this problem, we'll apply the principles of conservation of momentum and energy. Given that the collision is elastic, both momentum and kinetic energy are conserved. Initial Conditions- Puck A's initial velocity, (v_{A_i} = 2.6) m/s in the positive x-direction.- Puck B is initially at rest, so (v_{B_i} = 0) m/s.- After the collision, puck A's velocity, (v_{A_f} = 2.50) m/s at an angle of (+16.3^circ) above the x-axis. Conservation of MomentumMomentum is conserved in both the x and y directions. Since the pucks are identical, we'll denote their mass as (m).# x-DirectionThe initial momentum in the x-direction is (mv_{A_i}) since puck B is at rest. After the collision, the momentum in the x-direction is (mv_{A_{fx}} + mv_{B_{fx}}), where (v_{A_{fx}}) and (v_{B_{fx}}) are the x-components of the final velocities of pucks A and B, respectively.[mv_{A_i} = mv_{A_{fx}} + mv_{B_{fx}}]Given (v_{A_f} = 2.50) m/s at (+16.3^circ), the x-component of (v_{A_f}) is:[v_{A_{fx}} = v_{A_f} cos(16.3^circ) = 2.50 cos(16.3^circ)]Let's calculate (v_{A_{fx}}):[v_{A_{fx}} = 2.50 times cos(16.3^circ) approx 2.50 times 0.9613 approx 2.403, text{m/s}]So, the equation for conservation of momentum in the x-direction becomes:[m times 2.6 = m times 2.403 + m times v_{B_{fx}}][2.6 = 2.403 + v_{B_{fx}}][v_{B_{fx}} = 2.6 - 2.403][v_{B_{fx}} approx 0.197, text{m/s}]# y-DirectionInitially, there's no momentum in the y-direction since both pucks are moving in the x-direction or are at rest. After the collision, the momentum in the y-direction is (mv_{A_{fy}} + mv_{B_{fy}}), where (v_{A_{fy}}) and (v_{B_{fy}}) are the y-components of the final velocities of pucks A and B, respectively.Since there's no initial y-momentum:[0 = mv_{A_{fy}} + mv_{B_{fy}}]Given (v_{A_f} = 2.50) m/s at (+16.3^circ), the y-component of (v_{A_f}) is:[v_{A_{fy}} = v_{A_f} sin(16.3^circ) = 2.50 sin(16.3^circ)]Let's calculate (v_{A_{fy}}):[v_{A_{fy}} = 2.50 times sin(16.3^circ) approx 2.50 times 0.2814 approx 0.7035, text{m/s}]So, the equation for conservation of momentum in the y-direction becomes:[0 = m times 0.7035 + m times v_{B_{fy}}][v_{B_{fy}} = -0.7035, text{m/s}] Speed and Direction of Puck BThe speed of puck B after the collision, (v_B), can be found using the Pythagorean theorem:[v_B = sqrt{v_{B_{fx}}^2 + v_{B_{fy}}^2}][v_B = sqrt{(0.197)^2 + (-0.7035)^2}][v_B = sqrt{0.0388 + 0.4947}][v_B = sqrt{0.5335}][v_B approx 0.731, text{m/s}]The direction of puck B can be found using the arctangent:[theta = arctanleft(frac{v_{B_{fy}}}{v_{B_{fx}}}right)][theta = arctanleft(frac{-0.7035}{0.197}right)][theta = arctan(-3.570)][theta approx -74.7^circ]Since (theta) is negative, it indicates that the direction is below the x-axis, which is consistent with our expectations given the positive angle of puck A's trajectory. ConclusionAfter the collision, puck B moves at a speed of approximately (0.731) m/s in a direction of about (-74.7^circ) relative to the positive x-axis. This solution satisfies both the conservation of momentum and the principles of elastic collisions, where kinetic energy is also conserved.

🔑:To address the question of whether a gas in a mini-universe reaches and remains in a state of maximum entropy, we must consider the principles of thermodynamics, the effects of gravity, and the cosmology of the mini-universe. The concept of entropy and its evolution is central to understanding the direction of spontaneous processes in the universe.## Step 1: Understanding Entropy and the Second Law of ThermodynamicsEntropy (S) is a measure of disorder or randomness in a system. The second law of thermodynamics states that in a closed system, entropy will always increase over time, or at best remain constant in idealized reversible processes. This law is often expressed as (dS geq 0), where (dS) is the change in entropy.## Step 2: Maximum Entropy StateA system reaches its maximum entropy state when it has achieved equilibrium. In the context of a gas, this means that the gas molecules are evenly distributed throughout the available volume, and there are no longer any gradients in temperature, pressure, or concentration. At this point, there are no net macroscopic changes occurring within the system.## Step 3: Effects of Gravity in a Mini-UniverseIn a mini-universe with its own gravity, the effects of gravity on the gas must be considered. Gravity can cause the gas to collapse or become unevenly distributed over time, potentially decreasing entropy locally as it increases globally due to the conversion of potential energy into kinetic energy and eventually into heat.## Step 4: Cosmological ConsiderationsThe cosmology of the mini-universe, including its size, expansion (if any), and the presence of other forms of matter or energy, will also influence the behavior of the gas. For example, in an expanding universe, the gas could continue to expand and cool, potentially affecting its entropy state.## Step 5: Mathematical DerivationTo mathematically derive the change in entropy of a gas in a gravitational field, we consider the Gibbs entropy formula for a system in equilibrium: (S = k ln Omega), where (k) is Boltzmann's constant and (Omega) is the number of microstates. However, in a dynamic, gravitating system, we must also consider the work done by gravity and the conversion of potential energy into other forms.## Step 6: Entropy Change in a Gravitating SystemThe total entropy change ((dS_{total})) in a system includes the entropy change due to internal processes ((dS_{internal})) and the entropy change due to interactions with the surroundings ((dS_{external})). In a closed system like a mini-universe, (dS_{external} = 0), so (dS_{total} = dS_{internal}). The internal entropy change can be influenced by gravitational effects, which can cause local decreases in entropy as matter organizes into structures, but the second law dictates that the total entropy of a closed system will always increase or remain constant.## Step 7: Conclusion on Maximum Entropy StateGiven the principles outlined, once a gas in a mini-universe reaches a state of maximum entropy, it would theoretically remain in that state if the universe is truly closed and no external energy sources or sinks are present. However, the presence of gravity and cosmological evolution could lead to complex behaviors where local regions of lower entropy form (e.g., through gravitational collapse), but the total entropy of the system would still not decrease.The final answer is: boxed{Remains at maximum entropy}

🔑:## Step 1: Understand the given problem and the equations provided.The problem involves a positively charged particle moving between two parallel plate capacitors with different potentials. The particle enters at point A and exits at point B with a change in velocity direction due to the electric field between the capacitors. The equations given relate the force on the particle to the electric field, its acceleration, and its position and velocity as a function of time.## Step 2: Identify the key factors that influence the particle's trajectory.The particle's trajectory is influenced by the electric field (E) between the capacitors, which in turn is determined by the potentials V1 and V2. The particle's mass (m), charge (q), and initial velocity (Vi) also play crucial roles in determining its path.## Step 3: Recognize the limitations of the provided equations for solving the problem directly.The equations provided describe the relationship between the electric field, force, acceleration, and the particle's position and velocity over time. However, without specific values for the particle's charge, mass, initial velocity, or the dimensions of the capacitors, we cannot directly solve for V1 and V2 using these equations alone.## Step 4: Consider the principles of electrostatics and capacitor behavior.In a parallel plate capacitor, the electric field (E) between the plates is given by E = V/d, where V is the potential difference and d is the distance between the plates. For two capacitors with potentials V1 and V2, the electric fields would be E1 = V1/d and E2 = V2/d, assuming the distance between the plates (d) is the same for both capacitors.## Step 5: Analyze the particle's exit condition to infer information about V1 and V2.The particle exits at point B with the same vertical position as point A but with a velocity component Vy, indicating that the electric fields from the two capacitors have caused a deflection. The direction and magnitude of Vy would depend on the difference in electric fields (and thus potentials) between the two regions.## Step 6: Realize the need for additional information or constraints to solve for V1 and V2.Without knowing the specific values of the particle's charge, mass, initial velocity, the dimensions of the capacitors, or the exact trajectory of the particle, we cannot determine the potentials V1 and V2 directly from the given equations.The final answer is: boxed{0}

🔑:## Step 1: Understanding the Wheatstone Bridge Balance ConditionThe Wheatstone Bridge is balanced when the ratio of the resistances in the two arms is equal, given by R1/R2 = R3/R4. At this condition, the voltage across the bridge, between the midpoints of the arms (Vr2 and Vr4), is zero, meaning Vr2 = Vr4.## Step 2: Analyzing the Effect of Changing R4 on BalanceWhen R4 changes, the bridge becomes unbalanced because the ratio R1/R2 is no longer equal to R3/R4. This change disrupts the balance condition, causing a voltage difference between the midpoints of the arms.## Step 3: Understanding the Voltage Distribution in an Unbalanced BridgeIn an unbalanced Wheatstone Bridge, the voltage across each resistor is determined by the voltage divider rule. For resistors R2 and R4, the voltages Vr2 and Vr4 are determined by the ratios of R2 to the total resistance in its arm and R4 to the total resistance in its arm, respectively.## Step 4: Determining the Change in Vr2 and Vr4When R4 changes, the total resistance in its arm changes, affecting the voltage divider ratio and thus the voltage across R4 (Vr4). Since the bridge is now unbalanced, the voltage across R2 (Vr2) will also be affected because the bridge's overall voltage distribution changes.## Step 5: Conclusion on Vr2 and Vr4 ChangeGiven that the bridge becomes unbalanced due to a change in R4, and considering the voltage divider principle, both Vr2 and Vr4 will change. The change in R4 alters the voltage distribution across the bridge, affecting the voltages across both R2 and R4.The final answer is: boxed{Vr2 changes}