🔑:To understand why the Earth doesn't 'fall' towards the Sun, let's break it down using basic principles:1. Newton's Law of Universal Gravitation: Every point mass attracts every other point mass by a force acting along the line intersecting both points. This means the Sun attracts the Earth, and the Earth attracts the Sun, but because the Sun is so much more massive, its pull on the Earth is significantly stronger.2. Newton's First Law (Law of Inertia): An object will remain at rest or in uniform motion in a straight line unless acted upon by an external force. Initially, when the solar system formed, the Earth had a certain velocity. The gravitational pull of the Sun acts as a force that tries to pull the Earth towards it, but because the Earth also has a significant velocity perpendicular to the line connecting it to the Sun, it doesn't fall directly into the Sun.3. Momentum and Energy Conservation: The Earth's momentum (product of its mass and velocity) and the energy (both kinetic due to its motion and potential due to its position in the Sun's gravitational field) are conserved in the absence of external forces. The gravitational force of the Sun continuously changes the direction of the Earth's velocity but doesn't significantly alter its magnitude because the force acts perpendicular to the Earth's velocity vector at any given time. This results in the Earth following a curved path around the Sun, which we call an orbit.Now, let's discuss Kepler's First Law and its implications on orbits:- Kepler's First Law (Law of Ellipses): The orbit of a planet is an ellipse with the Sun at one of the two foci. This law implies that orbits are not perfect circles but can be highly elliptical, depending on the initial conditions (like velocity and distance from the Sun) of the celestial body.The equation `r = k / (1 + e * cos(theta + theta0))` describes the position of an orbiting object in polar coordinates at steady state. Here:- `r` is the distance from the focus (in this case, the Sun) to the object,- `k` is a constant related to the size of the orbit,- `e` is the eccentricity of the orbit, which determines how elliptical the orbit is (0 for a perfect circle, closer to 1 for a highly elliptical orbit),- `theta` is the angle around the orbit from a reference point,- `theta0` is an offset angle that determines the orientation of the orbit in space.This equation shows that the distance of the object from the Sun varies as it orbits, being closest at `theta + theta0 = 0` (periapsis) and farthest at `theta + theta0 = pi` (aphelion). For circular orbits, `e = 0`, and the distance `r` remains constant, simplifying to `r = k`. For highly elliptical orbits, `e` approaches 1, and the variation in distance is more pronounced.In summary, the Earth doesn't fall towards the Sun because its velocity and the gravitational force of the Sun balance out, resulting in a stable orbit. Kepler's First Law and the given equation help us understand the possible shapes and characteristics of orbits, from circular to highly elliptical, depending on the initial conditions and the gravitational interactions between celestial bodies.

🔑:## Step 1: Understand the Problem and Snell's LawSnell's law describes how light bends, or refracts, when traveling from one medium to the next. It is given by the equation n1 sin(θ1) = n2 sin(θ2), where n1 and n2 are the refractive indices of the first and second mediums, respectively, and θ1 and θ2 are the angles of incidence and refraction, respectively.## Step 2: Apply Snell's Law to the ProblemWhen sunlight enters the drop of water at point A, it passes from air into water. Using Snell's law, we have 1 * sin(θ) = n * sin(α), where θ is the incident angle in air, α is the angle of refraction in water, and n is the refractive index of water.## Step 3: Determine the Angle of RefractionFrom Snell's law, we find sin(α) = sin(θ) / n. Thus, α = arcsin(sin(θ) / n).## Step 4: Calculate the Total Angular DeflectionThe total angular deflection Δ of the ray is the sum of the angles of deviation at each interface. After entering the drop, the ray is reflected at point B and exits at point C. The angle of deviation at the air-water interface is θ - α, and at the water-air interface after reflection, it is α - (π/2 - θ) due to the internal reflection, which makes the ray's path symmetric. However, considering the path and applying Snell's law at both interfaces, the total angular deflection Δ can be derived by considering the geometry and the fact that the angle of incidence equals the angle of reflection. Thus, Δ = 2(θ - α) + π - 2θ = π - 2α - 2θ + 2θ = π - 2α. Substituting α from Step 3 gives Δ = π - 2arcsin(sin(θ) / n).## Step 5: Simplify the Expression for ΔGiven that α = arcsin(sin(θ) / n), substituting this into the equation for Δ gives Δ = π - 2arcsin(sin(θ) / n). However, to match the given format of the solution, we recognize that the equation provided in the problem statement, Δ = 2 arcsin(n sin(θ)) - π/2, seems to be a misunderstanding since the correct derivation leads to Δ = π - 2arcsin(sin(θ) / n). The confusion arises from the incorrect representation of the problem's equation. The correct approach to find the stationary point of Δ involves differentiating Δ with respect to θ and setting it equal to zero.## Step 6: Find the Stationary Point of ΔTo find the value of θ for which Δ is stationary, we differentiate Δ with respect to θ. Given the correct expression Δ = π - 2arcsin(sin(θ) / n), the derivative dΔ/dθ involves differentiating the arcsin term, which requires the chain rule. However, recognizing the error in the problem statement's equation, we proceed with the understanding that the actual task is to minimize or find the stationary point of the angular deflection, which typically involves the correct application of calculus to the actual equation describing the physical phenomenon.## Step 7: Correct Approach for Stationary PointThe correct equation to consider for the stationary point should be based on the physical principles of optics and the geometry of the rainbow formation. The equation provided, Δ = 2 arcsin(n sin(θ)) - π/2, suggests a specific form that might be derived from considering the geometry of the rainbow and the refraction/reflection process. To find the θ that makes Δ stationary, we would typically differentiate this expression with respect to θ and set the derivative equal to zero. However, given the confusion in the problem statement, the focus should be on understanding that the formation of a rainbow involves specific angles of incidence and refraction that lead to the phenomenon of total internal reflection and the dispersion of light.## Step 8: Conclusion on Rainbow FormationThe formation of a rainbow is associated with the total internal reflection of light within the water droplet and the subsequent refraction as the light exits the droplet. The angle of incidence that leads to this phenomenon can be found by considering the geometry and the optical properties of water. The specific value of θ that corresponds to the formation of a rainbow can be derived from the principles of optics and the condition for total internal reflection, which typically occurs when the angle of incidence exceeds the critical angle.The final answer is: boxed{theta = arcsinleft(frac{2}{n}sqrt{1-frac{1}{n^2}}right)}

🔑:## Step 1: Identify the forces involved in position AIn position A, the ball is at rest, and the forces acting on it are the tension in the horizontal cord (T_h), the tension in the supporting cord (T_s), and the weight of the ball (mg), where m is the mass of the ball and g is the acceleration due to gravity. Since the ball is at rest, the net force acting on it is zero. The tension in the horizontal cord is balanced by the horizontal component of the tension in the supporting cord.## Step 2: Analyze the forces in position AThe weight of the ball (mg) is acting downward, and the vertical component of the tension in the supporting cord (T_s) is acting upward, balancing the weight. The horizontal component of T_s balances the tension in the horizontal cord (T_h). Since the ball is at rest, T_s cos(θ) = mg and T_s sin(θ) = T_h, where θ is the angle between the supporting cord and the vertical.## Step 3: Identify the forces involved in position BIn position B, the ball is in motion as a pendulum, and the forces acting on it are the tension in the supporting cord (T_s') and the weight of the ball (mg). The horizontal component of the tension in the supporting cord provides the centripetal force needed for the ball to swing in a circular path.## Step 4: Analyze the forces in position BThe weight of the ball (mg) is acting downward, and the vertical component of the tension in the supporting cord (T_s') is acting upward. The horizontal component of T_s' provides the centripetal force (F_c = (m * v^2) / L), where v is the velocity of the ball and L is the length of the pendulum. Since the ball swings in a circular path, T_s' cos(θ) = mg and T_s' sin(θ) = (m * v^2) / L.## Step 5: Determine the relationship between T_s and T_s'To find the ratio of the tensions, we need to relate the forces in position A and position B. At the bottom of the swing (position B), the velocity of the ball is maximum, and the tension in the supporting cord is minimum. Using conservation of energy, the potential energy at position A is converted to kinetic energy at position B. The potential energy at position A is m * g * L, and the kinetic energy at position B is (1/2) * m * v^2.## Step 6: Apply conservation of energySince energy is conserved, m * g * L = (1/2) * m * v^2. Solving for v^2, we get v^2 = 2 * g * L.## Step 7: Calculate the ratio of tensionsThe centripetal force at position B is F_c = (m * v^2) / L = (m * 2 * g * L) / L = 2 * m * g. The tension in the supporting cord at position B is T_s' = mg / cos(θ) + (m * v^2) / (L * cos(θ)) = mg / cos(θ) + 2 * m * g / cos(θ) = 3 * mg / cos(θ). The tension in the supporting cord at position A is T_s = mg / cos(θ). The ratio of the tensions is T_s / T_s' = (mg / cos(θ)) / (3 * mg / cos(θ)) = 1/3.The final answer is: boxed{1/3}

🔑:Yes, sound waves can increase the temperature of the medium they are traveling through, although the effect is typically very small. This phenomenon is known as "acoustic heating" or "sound-induced heating." Here's a breakdown of the effect and the factors that influence it:Mechanism:When a sound wave propagates through a medium, it causes the molecules to oscillate back and forth, creating regions of compression and rarefaction. During the compression phase, the molecules are packed more closely together, increasing the kinetic energy of the molecules and, consequently, the temperature of the medium. This process is known as "viscous heating" or "frictional heating."Factors influencing acoustic heating:1. Intensity of the sound wave: The more intense the sound wave, the greater the temperature increase. Higher intensity sound waves transfer more energy to the medium, leading to greater viscous heating.2. Frequency of the sound wave: Lower frequency sound waves tend to produce more heating than higher frequency sound waves. This is because lower frequency sound waves have longer wavelengths, which means the molecules have more time to interact with each other and transfer energy, leading to greater viscous heating.3. Viscosity of the medium: The viscosity of the medium plays a significant role in acoustic heating. More viscous mediums, such as liquids or gases with high molecular weights, tend to exhibit greater acoustic heating due to the increased friction between molecules.4. Thermal conductivity of the medium: The thermal conductivity of the medium affects how efficiently the heat generated by the sound wave is dissipated. Mediums with high thermal conductivity, such as metals, tend to dissipate heat quickly, reducing the temperature increase.5. Duration of the sound wave: The longer the sound wave is applied, the greater the temperature increase. This is because the viscous heating effect accumulates over time.6. Medium properties: The properties of the medium, such as its specific heat capacity, density, and compressibility, also influence the acoustic heating effect.Examples and applications:1. Medical applications: Acoustic heating is used in medical treatments such as lithotripsy, where high-intensity sound waves are used to break up kidney stones, and in cancer treatment, where sound waves are used to heat and destroy cancer cells.2. Industrial applications: Acoustic heating is used in various industrial processes, such as cleaning, degassing, and mixing of materials.3. Environmental applications: Acoustic heating can be used to enhance the biodegradation of pollutants in soil and groundwater.Quantifying the effect:The temperature increase due to acoustic heating can be estimated using the following equation:ΔT = (α * I * t) / (ρ * c_p)where:ΔT = temperature increase (K)α = absorption coefficient of the medium (m^-1)I = intensity of the sound wave (W/m^2)t = duration of the sound wave (s)ρ = density of the medium (kg/m^3)c_p = specific heat capacity of the medium (J/kg·K)Keep in mind that this equation is a simplification and does not account for all the factors influencing acoustic heating. In practice, the temperature increase due to acoustic heating is often very small, typically on the order of millikelvin or smaller, and may not be significant in many applications.