🔑:## Step 1: First, we need to understand the given problem and the principles involved.The student falls freely from rest, which means the initial velocity (v_i) is 0 m/s. The student comes to rest in a time of 0.0180 s, so the final velocity (v_f) is also 0 m/s. We are given the average force exerted on the student by the ground as +16000 N. We will use the impulse-momentum theorem, which states that the impulse of a force (J) is equal to the change in momentum (Δp) of an object. The impulse is calculated as the average force (F_avg) multiplied by the time (Δt) over which it acts: J = F_avg * Δt.## Step 2: Calculate the impulse exerted on the student by the ground.Given that the average force (F_avg) is +16000 N and the time (Δt) is 0.0180 s, we can calculate the impulse: J = 16000 N * 0.0180 s = 288 N*s.## Step 3: Apply the impulse-momentum theorem to find the change in momentum.The impulse-momentum theorem states that J = Δp = m * v_f - m * v_i, where m is the mass of the student, v_f is the final velocity, and v_i is the initial velocity. Since v_i = v_f = 0 m/s, the change in momentum Δp = 0. However, this step is to understand the relationship; the actual calculation of momentum change isn't necessary because we are looking for the height from which the student fell, and we can use the impulse to find the velocity just before impact, then use kinematics.## Step 4: Calculate the velocity of the student just before hitting the ground.Since the impulse equals the change in momentum, and knowing that the final momentum is 0 (because the student comes to rest), we can set up the equation as follows: J = m * v - 0, where v is the velocity just before the collision. Rearranging for v gives v = J / m. Substituting the known values: v = 288 N*s / 63 kg = 4.57 m/s.## Step 5: Use kinematic equations to find the height from which the student fell.We can use the equation v^2 = v_i^2 + 2 * a * d, where v is the final velocity (4.57 m/s), v_i is the initial velocity (0 m/s), a is the acceleration due to gravity (-9.8 m/s^2, negative because it's downward), and d is the distance (height) the student fell. Since v_i = 0, the equation simplifies to v^2 = 2 * a * d. Rearranging for d gives d = v^2 / (2 * a). Substituting the known values: d = (4.57 m/s)^2 / (2 * 9.8 m/s^2).## Step 6: Perform the calculation to find the height.d = (4.57 m/s)^2 / (2 * 9.8 m/s^2) = 20.83 / 19.6 = 1.062 m.The final answer is: boxed{1.06}

🔑:## Step 1: Understanding the Coriolis ForceThe Coriolis force is a fictitious force that appears to act on objects moving relative to a rotating frame of reference, such as the Earth. It is given by the equation (F_c = -2m Omega times v), where (m) is the mass of the object, (Omega) is the angular velocity of the Earth, and (v) is the velocity of the object relative to the Earth's surface.## Step 2: Determining the Direction of the Coriolis ForceThe direction of the Coriolis force depends on the hemisphere and the direction of motion. In the Northern Hemisphere, the Coriolis force acts to the right of the direction of motion, and in the Southern Hemisphere, it acts to the left. Since the railway line is being moved from South Africa to Sweden, it spans across both hemispheres.## Step 3: Calculating the Coriolis AccelerationThe Coriolis acceleration (a_c) can be found using the formula (a_c = -2 Omega times v). The angular velocity of the Earth (Omega) is approximately (7.29 times 10^{-5} , text{rad/s}). The velocity (v) of the railway line as it is moved northward would be relatively slow, but for the sake of calculation, let's assume a constant velocity of (1 , text{m/s}) (a rough estimate for a very slow movement over a long period).## Step 4: Considering the Latitude DependenceThe Coriolis force also depends on the latitude (phi) because the effective rotation rate of the Earth varies with latitude. The formula incorporating latitude is (a_c = -2 Omega sin(phi) v), where (phi) is the latitude. For locations near the equator, (sin(phi)) approaches 0, minimizing the Coriolis effect, while at higher latitudes, the effect increases.## Step 5: Applying to the Railway Line MovementGiven the railway line stretches from South Africa to Sweden, it covers a significant range of latitudes. At the equator, the Coriolis acceleration would be zero, but as you move towards either pole, the acceleration increases. For a segment of the track in South Africa (near 30°S latitude) moving north at (1 , text{m/s}), and another segment in Sweden (near 60°N latitude), the Coriolis accelerations would be different due to the latitude dependence.## Step 6: Calculating Coriolis Acceleration for South Africa SegmentFor South Africa (assuming (phi = -30^circ) or (-0.5236 , text{rad})), (a_c = -2 times 7.29 times 10^{-5} times sin(-30^circ) times 1 = 7.29 times 10^{-5} , text{m/s}^2) to the left of the direction of motion.## Step 7: Calculating Coriolis Acceleration for Sweden SegmentFor Sweden (assuming (phi = 60^circ) or (1.0472 , text{rad})), (a_c = -2 times 7.29 times 10^{-5} times sin(60^circ) times 1 = -1.265 times 10^{-4} , text{m/s}^2) to the right of the direction of motion.## Step 8: Considering the Bending of the Railway LineThe difference in Coriolis acceleration between the southern and northern segments of the railway line would indeed cause a bending effect. However, the magnitude of this bending is extremely small due to the low velocities involved and the small accelerations resulting from the Coriolis force.## Step 9: Conclusion on Bending and CalculationsGiven the calculations, the railway line would experience a slight bending due to the Coriolis force, with the southern segment being deflected to the left and the northern segment to the right of their respective directions of motion. However, the effect is negligible for practical purposes due to the small accelerations involved.The final answer is: boxed{0}

🔑:## Step 1: Understanding the ProblemThe problem involves a rigid body spinning on its x-axis with an angular velocity of 2 Hz and tumbling on its z-axis with an angular velocity of 0.1 Hz. We need to determine if it's possible for the body to maintain a clean tumbling motion on the z-axis with no rotation or angular velocity in the y-axis.## Step 2: Angular Momentum ConservationIn the absence of external torques, the total angular momentum of the rigid body is conserved. The angular momentum vector (L) can be expressed as the product of the moment of inertia tensor (I) and the angular velocity vector (ω): L = Iω.## Step 3: Moment of Inertia TensorFor a rigid body, the moment of inertia tensor (I) is a 3x3 matrix that describes the body's mass distribution. The diagonal elements of I represent the moments of inertia about the principal axes (x, y, z), while the off-diagonal elements represent the products of inertia.## Step 4: Angular Velocity VectorThe angular velocity vector (ω) can be expressed as ω = (ωx, ωy, ωz), where ωx, ωy, and ωz are the angular velocities about the x, y, and z axes, respectively.## Step 5: Euler's EquationsEuler's equations describe the rotation of a rigid body in the absence of external torques. They are given by: dL/dt = ω × L, where × denotes the cross product. Substituting L = Iω, we get: I(dω/dt) = ω × (Iω).## Step 6: Simplifying Euler's EquationsAssuming the moment of inertia tensor is diagonal (I = diag(Ix, Iy, Iz)), Euler's equations simplify to: Ix(dωx/dt) = (Iy - Iz)ωyωz, Iy(dωy/dt) = (Iz - Ix)ωzωx, Iz(dωz/dt) = (Ix - Iy)ωxωy.## Step 7: Analyzing the y-ComponentTo maintain a clean tumbling motion on the z-axis with no rotation or angular velocity in the y-axis, we require ωy = 0 and dωy/dt = 0. Substituting these conditions into the middle Euler equation, we get: 0 = (Iz - Ix)ωzωx.## Step 8: ConclusionSince ωz = 0.1 Hz and ωx = 2 Hz are non-zero, the only way to satisfy the equation is if Iz = Ix. However, this is not a general condition for a rigid body, and the moments of inertia about different axes are usually distinct. Therefore, in the absence of external torques, it is not possible for the body to maintain a clean tumbling motion on the z-axis with no rotation or angular velocity in the y-axis, as the angular momentum conservation and Euler's equations would lead to non-zero ωy due to the coupling between the x and z rotations.The final answer is: boxed{No}

🔑:## Step 1: Define the null and alternative hypothesesThe null hypothesis (H0) is that the type of college (public or private) is independent of the dropout rate, meaning there is no significant difference in dropout rates between public and private colleges. The alternative hypothesis (H1) is that the type of college and dropout rate are not independent, indicating a significant difference in dropout rates between public and private colleges.## Step 2: Calculate the expected frequencies for each cell under the assumption of independenceTo perform the chi-square test, we first need to calculate the expected frequencies for each cell in the contingency table under the assumption of independence (H0). The formula for expected frequency is E = (row total * column total) / total sample size.For public college non-dropouts: E = (100 * (100 + 80 - 20 - 10)) / (100 + 80) = (100 * 150) / 180 = 83.33For public college dropouts: E = (100 * 30) / 180 = 16.67For private college non-dropouts: E = (80 * 150) / 180 = 66.67For private college dropouts: E = (80 * 30) / 180 = 13.33## Step 3: Calculate the chi-square statisticThe chi-square statistic (χ²) is calculated using the formula χ² = Σ [(observed frequency - expected frequency)^2 / expected frequency] for all cells in the contingency table.For public college non-dropouts: (80 - 83.33)^2 / 83.33 = (-3.33)^2 / 83.33 = 10.89 / 83.33 = 0.1307For public college dropouts: (20 - 16.67)^2 / 16.67 = (3.33)^2 / 16.67 = 10.89 / 16.67 = 0.6525For private college non-dropouts: (70 - 66.67)^2 / 66.67 = (3.33)^2 / 66.67 = 10.89 / 66.67 = 0.1633For private college dropouts: (10 - 13.33)^2 / 13.33 = (-3.33)^2 / 13.33 = 10.89 / 13.33 = 0.8173χ² = 0.1307 + 0.6525 + 0.1633 + 0.8173 = 1.7638## Step 4: Determine the degrees of freedomThe degrees of freedom (df) for a chi-square test of independence in a 2x2 contingency table is calculated as (number of rows - 1) * (number of columns - 1) = (2 - 1) * (2 - 1) = 1 * 1 = 1.## Step 5: Find the critical value or p-valueUsing a chi-square distribution table with df = 1 and α = 0.05, the critical value is approximately 3.841. Alternatively, we can calculate the p-value associated with the χ² statistic. The p-value is the probability of observing a χ² statistic at least as extreme as the one we calculated, assuming that H0 is true.## Step 6: Compare the calculated χ² statistic to the critical value or interpret the p-valueSince the calculated χ² statistic (1.7638) is less than the critical value (3.841), we fail to reject the null hypothesis. Alternatively, if we were to find the p-value, it would be greater than 0.05, leading to the same conclusion.The final answer is: boxed{1.7638}