🔑:## Step 1: Understand the problem and identify key conceptsThe problem involves a juggler throwing five balls sequentially to a maximum height of 3.0 m. We need to determine the time interval between successive throws, considering the angle of elevation as 90° and neglecting air resistance. Key concepts include the equations of motion, kinetic and potential energy, and the trajectory of projectiles.## Step 2: Determine the time it takes for one ball to reach its maximum heightSince the angle of elevation is 90°, the initial vertical velocity (v0y) is the same as the initial velocity (v0). At the maximum height, the final velocity (vfy) is 0 m/s. We can use the equation vfy = v0y - gt, where g is the acceleration due to gravity (approximately 9.8 m/s^2), to find the time it takes for one ball to reach its maximum height.## Step 3: Calculate the time to reach maximum heightRearranging the equation to solve for time (t), we get t = v0y / g. However, we need to find v0y first. Since the maximum height (h) is 3.0 m, we can use the equation v0y^2 = 2gh to find v0y. Solving for v0y gives v0y = sqrt(2gh).## Step 4: Calculate v0yPlugging in the values, v0y = sqrt(2 * 9.8 m/s^2 * 3.0 m) = sqrt(58.8) m/s ≈ 7.67 m/s.## Step 5: Calculate the time to reach maximum heightNow, we can calculate the time it takes for one ball to reach its maximum height: t = v0y / g = 7.67 m/s / 9.8 m/s^2 ≈ 0.78 s.## Step 6: Calculate the total time for one ball to complete its trajectorySince the ball's trajectory is symmetric, the total time for one ball to complete its trajectory is twice the time it takes to reach its maximum height: total time = 2 * t = 2 * 0.78 s ≈ 1.56 s.## Step 7: Determine the time interval between successive throwsTo keep the balls in the air and maintain a steady juggling pattern, the time interval between successive throws should be equal to the total time it takes for one ball to complete its trajectory divided by the number of balls. Therefore, the time interval = total time / number of balls = 1.56 s / 5.## Step 8: Calculate the time interval between successive throwsPerforming the calculation, the time interval = 1.56 s / 5 ≈ 0.312 s.The final answer is: boxed{0.312}

🔑:## Step 1: Define the Lagrangian of the systemThe Lagrangian of the system is given by L = T - U, where T is the kinetic energy and U is the potential energy. For the rolling cylinder, the kinetic energy is T = frac{1}{2}mv^2 + frac{1}{2}Iomega^2, where I is the moment of inertia and omega is the angular velocity. Since the cylinder rolls without slipping, omega = frac{v}{R}. The potential energy is U = mgh, where h is the height of the cylinder on the inclined plane.## Step 2: Express the kinetic and potential energies in terms of the generalized coordinateThe kinetic energy can be rewritten as T = frac{1}{2}mv^2 + frac{1}{2}Ileft(frac{v}{R}right)^2 = frac{1}{2}mv^2 + frac{1}{2}left(frac{1}{2}mR^2right)left(frac{v}{R}right)^2 = frac{3}{4}mv^2, since I = frac{1}{2}mR^2 for a cylinder. The potential energy is U = mg(L - x)sintheta, where L is the length of the inclined plane and x is the position of the cylinder along the plane.## Step 3: Apply the Lagrangian mechanics to derive the equation of motionThe Lagrangian is L = frac{3}{4}mv^2 - mg(L - x)sintheta. The equation of motion is given by frac{d}{dt}left(frac{partial L}{partial v}right) - frac{partial L}{partial x} = Q, where Q is the generalized force due to the drag. Since v = frac{dx}{dt}, we have frac{partial L}{partial v} = frac{3}{2}mv and frac{partial L}{partial x} = mgsintheta. The generalized force due to the drag is Q = -frac{1}{2}rho v^2 C_d A.## Step 4: Derive the equation of motionSubstituting into the equation of motion, we get frac{d}{dt}left(frac{3}{2}mvright) - mgsintheta = -frac{1}{2}rho v^2 C_d A. Simplifying, we have frac{3}{2}mfrac{dv}{dt} = mgsintheta - frac{1}{2}rho v^2 C_d A.## Step 5: Solve for the velocity of the cylinderRearranging the equation, we get frac{dv}{dt} = frac{2}{3}left(gsintheta - frac{rho v^2 C_d A}{2m}right). This is a nonlinear differential equation.## Step 6: Solve the differential equationTo solve this equation, we can use numerical methods or approximations. However, for the purpose of this problem, we will assume that the drag force is negligible at the beginning of the motion, and the cylinder starts from rest. We can then use the equation v(t) = frac{2}{3}gsintheta t as an approximation for small t.## Step 7: Find the time it takes for the cylinder to roll down the inclined planeThe distance traveled by the cylinder is x = frac{1}{2}at^2, where a = frac{2}{3}gsintheta. Setting x = L, we can solve for t: L = frac{1}{2}left(frac{2}{3}gsinthetaright)t^2. Therefore, t = sqrt{frac{3L}{gsintheta}}.## Step 8: Consider the effect of the drag forceThe drag force becomes significant as the velocity of the cylinder increases. To account for the drag force, we need to solve the nonlinear differential equation numerically.## Step 9: Compare with experimental dataComparing the theoretical result with experimental data would require actual experimental values, which are not provided in the problem statement.The final answer is: boxed{sqrt{frac{3L}{gsintheta}}}

🔑:## Step 1: Understand the circuit configurationThe circuit is an inverting amplifier with resistors R1, R2, and R. R2 and R are in parallel, and the input voltage Vin is applied across R1. The output voltage Vout is across the 50kOhm resistor and is given as -2Vin.## Step 2: Apply Kirchhoff's law for the voltage drop across R1 and the parallel combination of R2 and RGiven that Vout = -2Vin and it's across the 50kOhm resistor, we can use the voltage divider rule and Kirchhoff's law to find the relationship between the resistors. However, the direct relationship between Vout and the resistors needs to consider the inverting amplifier's properties.## Step 3: Recall the properties of an inverting amplifierIn an inverting amplifier, the voltage gain (Av) is given by -R2/R1, where R2 is the feedback resistor and R1 is the input resistor. Given that Vout = -2Vin, this implies the gain Av = 2.## Step 4: Use the gain formula to relate R1 and R2Since Av = -R2/R1 and Av = 2, we have -R2/R1 = 2. This implies R2 = 2*R1.## Step 5: Consider the parallel combination of R2 and RSince R2 and R are in parallel, their combined resistance (let's call it Rp) is given by 1/Rp = 1/R2 + 1/R.## Step 6: Apply the condition for the voltage drop across the 50kOhm resistorHowever, the problem statement directly implies that the gain and thus the relationship between R1 and R2 (or the parallel combination) is key, but it does not directly involve calculating R using the parallel combination formula without more specific information about the circuit's configuration or how R relates to the 50kOhm resistor and the voltage gain.## Step 7: Realize the mistake in the approachGiven the voltage gain and the fact that Vout is -2Vin across a 50kOhm resistor, we need to reconsider how R1, R2, and R are related through the circuit's operation as an inverting amplifier. The problem seems to be asking for the equivalent resistance for R1, but without directly involving R in the calculation of voltage gain or providing its direct relationship to R1 and R2 in terms of voltage drop.## Step 8: Correct the approach based on the inverting amplifier's propertiesFor an inverting amplifier, the input resistance (R1) is directly related to the voltage gain. However, the problem does not provide a direct way to calculate R without knowing how R2 and R are related in terms of resistance value or how they contribute to the overall gain.## Step 9: Consider the role of the 50kOhm resistorIf the 50kOhm resistor is R2, then its relationship with R1 can be directly inferred from the gain. But if it's part of the circuit in a different configuration, more information is needed.## Step 10: Conclude based on given informationGiven the voltage gain Av = 2, and assuming the 50kOhm resistor is part of the feedback (R2), we can infer R1's value based on the gain formula Av = -R2/R1. If R2 = 50kOhm, then 2 = -50kOhm/R1, which implies R1 = -50kOhm/2.The final answer is: boxed{25}

🔑:The contingency table for the given data is:begin{tabular}{l c c c} hline hline & Retired & Not Retired & Total hline Denied promotion & 32 & 83 & 115 Promoted & 9 & 49 & 58 Total & 41 & 132 & 173 hline hline end{tabular} The null and alternative hypotheses are:(H_{0}): There is no significant relationship between the denial of promotions and the increase in retirements.(H_{1}): There is a significant relationship between the denial of promotions and the increase in retirements.The expected frequencies are calculated as follows:begin{tabular}{l c c c} hline hline & Retired & Not Retired & Total hline Denied promotion & (frac{115times 41}{173}=26.47) & (frac{115times 132}{173}=88.53) & 115 Promoted & (frac{58times 41}{173}=14.53) & (frac{58times 132}{173}=43.47) & 58 Total & 41 & 132 & 173 hline hline end{tabular} The chi-square statistic is calculated as follows:[chi^{2} =sumfrac{left(text{observed}-text{expected}right)^{2}}{text {expected}}] [=frac{left(32-26.47right)^{2}}{26.47}+frac{left(83-88.53 right)^{2}}{88.53}+frac{left(9-14.53right)^{2}}{14.53}+frac{left(49-43.4 7right)^{2}}{43.47}] [=0.83+0.37+1.93+0.59] [=3.72]The critical value from the chi-square table with 1 degree of freedom and a 5% level of significance is 3.84. Since the calculated value is less than the critical value, we accept the null hypothesis. Hence, there is no significant relationship between the denial of promotions and the increase in retirements.