🔑:## Step 1: Estimate the number density of Helium atoms under standard conditions.The number density of Helium atoms can be calculated using the ideal gas law, (PV = nRT), where (P) is the pressure, (V) is the volume, (n) is the number of moles, (R) is the gas constant, and (T) is the temperature. Under standard conditions, (P = 1) atm (= 101325) Pa, (T = 293.15) K, and (R = 8.3145) J/mol*K. The molar volume of an ideal gas at standard conditions is (V = frac{nRT}{P} = frac{1 times 8.3145 , text{J/mol*K} times 293.15 , text{K}}{101325 , text{Pa}} = 0.02479) m(^3)/mol. The number density (N) (number of atoms per unit volume) can be found by (N = frac{nN_A}{V}), where (N_A) is Avogadro's number ((6.022 times 10^{23}) mol(^{-1})). For one mole of Helium, (n = 1), so (N = frac{1 times 6.022 times 10^{23} , text{mol}^{-1}}{0.02479 , text{m}^3/text{mol}} approx 2.43 times 10^{25}) m(^{-3}).## Step 2: Estimate the polarizability of a Helium atom.The polarizability (alpha) of a Helium atom can be estimated from the Clausius-Mossotti relation or directly from experimental values. For simplicity, we use an approximate value. The polarizability of Helium is approximately (0.205 times 10^{-30}) m(^3) (or (0.205) Å(^3), considering (1) Å (= 10^{-10}) m).## Step 3: Apply the Lorenz-Lorentz model to find the index of refraction.The Lorenz-Lorentz equation is given by (frac{n^2 - 1}{n^2 + 2} = frac{Nalpha}{3epsilon_0}), where (n) is the refractive index, (N) is the number density of atoms, (alpha) is the polarizability of an atom, and (epsilon_0) is the vacuum permittivity ((8.854 times 10^{-12}) F/m). Rearranging for (n) gives (n^2 = frac{2 + frac{Nalpha}{epsilon_0}}{1 - frac{Nalpha}{3epsilon_0}}).## Step 4: Calculate the index of refraction using the given values.Substitute (N = 2.43 times 10^{25}) m(^{-3}), (alpha = 0.205 times 10^{-30}) m(^3), and (epsilon_0 = 8.854 times 10^{-12}) F/m into the equation. First, calculate (frac{Nalpha}{3epsilon_0}): (frac{2.43 times 10^{25} times 0.205 times 10^{-30}}{3 times 8.854 times 10^{-12}} approx frac{4.98 times 10^{-5}}{26.562} approx 1.875 times 10^{-6}). Then, calculate (n^2 = frac{2 + 3 times 1.875 times 10^{-6}}{1 - 1.875 times 10^{-6}} approx frac{2}{1} approx 2), since the terms involving (1.875 times 10^{-6}) are negligible compared to 1 and 2.## Step 5: Solve for (n).Taking the square root of both sides gives (n approx sqrt{2} approx 1.414).## Step 6: Compare with the measured value.The measured index of refraction for Helium at standard conditions is approximately 1.000036, which is very close to that of a vacuum. Our calculation significantly overestimates the index of refraction due to simplifications and the use of approximate values for polarizability and number density.The final answer is: boxed{1.000036}

🔑:## Step 1: Calculate the initial gravitational potential energy of the falling bodyThe gravitational potential energy (U) of an object is given by the formula U = mgh, where m is the mass of the object, g is the acceleration due to gravity (approximately 9.8 m/s^2), and h is the height from which the object falls. Substituting the given values, we get U = 2 kg * 9.8 m/s^2 * 10 m = 196 J.## Step 2: Determine the final kinetic energy of the cart and the falling bodyAs the body falls, its gravitational potential energy is converted into the kinetic energy of both the cart and the falling body, assuming no energy is lost to friction or other forces. The kinetic energy (K) of an object is given by the formula K = 0.5mv^2, where m is the mass of the object and v is its velocity. Since the cart and the body are tied together, they will have the same final velocity.## Step 3: Apply the law of conservation of energyThe total initial energy (gravitational potential energy of the falling body) equals the total final energy (kinetic energy of the cart and the falling body). Therefore, we can set up the equation: m_body * g * h = 0.5 * (m_body + m_cart) * v^2, where m_body is the mass of the falling body (2 kg), m_cart is the mass of the cart (1 kg), g is the acceleration due to gravity (9.8 m/s^2), h is the height from which the body falls (10 m), and v is the final velocity of the cart and the body.## Step 4: Solve for the final velocity of the cartSubstitute the given values into the equation: 2 kg * 9.8 m/s^2 * 10 m = 0.5 * (2 kg + 1 kg) * v^2. This simplifies to 196 J = 0.5 * 3 kg * v^2. Further simplification gives 196 J = 1.5 kg * v^2. Solving for v^2, we get v^2 = 196 J / (1.5 kg) = 130.67 m^2/s^2. Taking the square root of both sides to solve for v, we get v = sqrt(130.67 m^2/s^2).## Step 5: Calculate the square root of 130.67Calculating the square root of 130.67 gives us the final velocity.The final answer is: boxed{11.42}

🔑:No, the black hole will not form. The reason is that the two billiard balls have a large center of mass motion, and this motion cannot be captured by the black hole. The Schwarzschild radius is defined in the rest frame of the object, and if the object has a large motion with respect to the observer, the Schwarzschild radius will not encompass the object.

🔑:A classic problem in physics! Let's break it down step by step.We'll use the principle of conservation of energy, which states that the total energy of a closed system remains constant over time. In this case, the system is the girl and the swing.The girl's energy at the highest point (point A) consists of:1. Potential energy (PE) due to her height above the ground: PE = m × g × h, where m is her mass, g is the acceleration due to gravity (approximately 9.8 m/s²), and h is her height above the ground.2. Kinetic energy (KE) due to her motion: since she's at the highest point, her velocity is momentarily zero, so KE = 0.At the lowest point (point B), her energy consists of:1. Potential energy (PE) due to her height above the ground: again, PE = m × g × h, but with a different height.2. Kinetic energy (KE) due to her motion: since she's moving, KE = (1/2) × m × v², where v is her velocity at the lowest point.The principle of conservation of energy states that the total energy at point A is equal to the total energy at point B:E_A = E_BLet's calculate the energies:At point A (highest point):h_A = 2.0 m (above the ground)PE_A = m × g × h_A = m × 9.8 m/s² × 2.0 m = 19.6 mKE_A = 0 (since she's momentarily at rest)At point B (lowest point):h_B = 0.5 m (above the ground)PE_B = m × g × h_B = m × 9.8 m/s² × 0.5 m = 4.9 mKE_B = (1/2) × m × v² (we want to find v)Now, equate the total energies:E_A = E_B19.6 m + 0 = 4.9 m + (1/2) × m × v²Subtract 4.9 m from both sides:14.7 m = (1/2) × m × v²Divide both sides by m (which cancels out):14.7 = (1/2) × v²Multiply both sides by 2:29.4 = v²Take the square root of both sides:v = √29.4 ≈ 5.43 m/sSo, the girl's speed at the lowest point is approximately 5.43 m/s.Note that we didn't need to know the girl's mass, as it canceled out in the calculation. The length of the swing (4m) wasn't necessary either, as we only considered the energy changes due to the change in height.