🔑:## Step 1: Understanding the ScenarioWe have two magnets, one placed above the other, with both magnets' positive sides facing upwards. The upper magnet is suspended above the lower magnet by a small string, indicating that there is a force acting against gravity to keep it suspended. Since both magnets have their positive sides facing each other, they will repel.## Step 2: Identifying Forces on the Upper MagnetThe upper magnet experiences two primary forces: the downward force of gravity (weight) and the upward force of repulsion from the lower magnet. Since it is suspended in mid-air by a string, the upward force (repulsion from the lower magnet plus the tension in the string) must balance the downward force (weight of the upper magnet).## Step 3: Analyzing Forces on the Lower MagnetThe lower magnet also experiences forces: its weight (downward) and the upward force of repulsion from the upper magnet. However, since the lower magnet is placed on a scale, we are interested in how these forces affect its measured weight.## Step 4: Effect of Repulsion on Measured WeightThe repulsion between the two magnets acts in opposite directions. For the lower magnet, this repulsion is an upward force. According to Newton's third law, the force exerted by the upper magnet on the lower magnet is equal in magnitude and opposite in direction to the force exerted by the lower magnet on the upper magnet.## Step 5: Measured Weight of the Lower MagnetThe measured weight of the lower magnet on the scale is its actual weight minus the upward force (repulsion) exerted by the upper magnet. This is because the scale measures the force exerted by the lower magnet on it, which is the weight of the lower magnet reduced by the upward repulsive force from the upper magnet.## Step 6: Conclusion on Weight MeasurementGiven that the repulsion acts upward on the lower magnet, it reduces the downward force (weight) that the scale measures. Thus, the weight of the lower magnet, as measured by the scale, decreases due to the repulsion between the two magnets.The final answer is: boxed{No}

🔑:## Step 1: Recall Wien's Displacement LawWien's displacement law states that the wavelength at which an object emits most intensely (λ_max) is inversely proportional to its temperature (T) in Kelvin. The formula is λ_max = b / T, where b is Wien's displacement constant, approximately equal to 2.8977685(51) × 10^(-3) m·K.## Step 2: Calculate the Maximum Wavelength Radiated by the SunGiven the surface temperature of the Sun as 5.79 × 10^3 K, we can substitute this value into Wien's displacement law to find the maximum wavelength radiated. λ_max = b / T = (2.8977685 × 10^(-3) m·K) / (5.79 × 10^3 K).## Step 3: Perform the Calculationλ_max = (2.8977685 × 10^(-3) m·K) / (5.79 × 10^3 K) = 5.00 × 10^(-7) m.## Step 4: Convert the Wavelength to Nanometers for Easier Comparison with Visible LightSince 1 m = 10^9 nm, λ_max in nanometers = 5.00 × 10^(-7) m * (10^9 nm / 1 m) = 500 nm.## Step 5: Explain the Discrepancy Between the Calculated Maximum Wavelength and the Apparent Color of the SunThe calculated maximum wavelength is approximately 500 nm, which corresponds to green light. However, the Sun appears yellow to our eyes. This discrepancy can be explained by the effects of scattering in the Earth's atmosphere. Shorter wavelengths (like blue and violet) are scattered more than longer wavelengths (like red and yellow) due to Rayleigh scattering. This scattering effect gives the Sun a yellowish appearance because our eyes perceive the combination of all wavelengths of visible light that reach us, with the scattered shorter wavelengths being distributed across the sky and the direct, longer wavelengths (including the peak emission around 500 nm) dominating our view of the Sun.## Step 6: Consider the Observation of All Wavelengths of Visible LightThe human eye perceives a broad spectrum of light, from approximately 380 nm (violet) to 740 nm (red). The Sun emits a wide range of wavelengths, with the peak emission calculated to be around 500 nm. However, due to the scattering effects and the sensitivity of the human eye, the perceived color of the Sun is yellow, even though the peak emission wavelength would suggest a green color.The final answer is: boxed{500}

🔑:## Step 1: Understand the components of the force on a current-carrying wireThe force on a current-carrying wire can be attributed to both electric and magnetic fields. The electric field exerts a force on the charges on the surface of the wire, while the magnetic field exerts a force on the moving charges within the wire due to the current.## Step 2: Derive the expression for the electric forceThe electric force (F_e) on a charge (q) in an electric field (E) is given by F_e = qE. For a wire with a surface charge density (σ) and length (L), the total charge is q = σL. Thus, the electric force per unit length due to the electric field is F_e/L = σE.## Step 3: Derive the expression for the magnetic forceThe magnetic force (F_m) on a current-carrying wire in a magnetic field (B) is given by F_m = ILB, where I is the current in the wire. The magnetic force per unit length is F_m/L = IB.## Step 4: Combine electric and magnetic forces for the total tensionThe total force (and thus tension, since we're considering forces along the length of the wire) experienced by the wire is the sum of the electric and magnetic forces per unit length. Therefore, the total tension (T) per unit length is T = σE + IB.## Step 5: Consider the relationship between electric field, magnetic field, and the properties of the wireFor a wire with a current I and a surface charge density σ, in the presence of both electric (E) and magnetic (B) fields, the total tension expression derived combines the effects of both fields on the wire. However, without specific values for σ, E, I, or B, the expression remains in terms of these variables.The final answer is: boxed{sigma E + IB}

🔑:## Step 1: Definition of Fissile and Fertile NuclidesFissile nuclides are those that can undergo neutron-induced fission, a process where the nucleus splits into two or more smaller nuclei, releasing a significant amount of energy and more neutrons. Fertile nuclides, on the other hand, cannot undergo fission directly but can be converted into fissile nuclides through neutron capture.## Step 2: Neutron-Induced Fission ProcessIn neutron-induced fission, a neutron collides with the nucleus of a fissile or fertile nuclide. If the neutron's energy is sufficient, it can cause the nucleus to become unstable and split. The energy of the neutron (fast vs. thermal) significantly affects the likelihood of fission occurring.## Step 3: Effect of Neutron Energy on Fission Cross-SectionThe fission cross-section is a measure of the probability of a neutron-induced fission event. For fissile nuclides, the fission cross-section is higher for thermal (slow) neutrons than for fast neutrons. This is because thermal neutrons are more easily captured by the nucleus, increasing the likelihood of fission. Fast neutrons, having higher energies, are less likely to be captured and cause fission.## Step 4: Examples of Fissile RadionuclidesExamples of fissile radionuclides include Uranium-235 (²³⁵U), Plutonium-239 (²³⁹Pu), and Uranium-233 (²³³U). These nuclides can undergo fission when struck by either fast or thermal neutrons.## Step 5: Fission by Fast and Thermal NeutronsDespite the lower fission cross-section for fast neutrons, fissile radionuclides can still undergo fission with fast neutrons because the high energy of these neutrons can overcome the threshold for fission. However, the probability is lower compared to thermal neutrons. The ability of fissile nuclides to undergo fission with both fast and thermal neutrons makes them crucial for nuclear reactors and weapons.## Step 6: Comparison with Fertile NuclidesFertile nuclides, such as Uranium-238 (²³⁸U) and Thorium-232 (²³²Th), have a very low probability of undergoing fission with thermal neutrons and require fast neutrons to induce fission. However, even with fast neutrons, the fission cross-section is significantly lower than that of fissile nuclides. Fertile nuclides are important because they can be converted into fissile materials through neutron capture, followed by beta decay.The final answer is: There is no final numerical answer to this problem as it requires a descriptive response.