🔑:## Step 1: Identify the known variables for both stones.The first stone is dropped from 43.9m, and it takes t seconds to hit the water. The second stone is thrown down 1 second after the first stone, so it travels for t - 1 seconds. Both stones hit the water at the same time.## Step 2: Apply the equation of motion under gravity to the first stone.For the first stone, which is under the sole influence of gravity, we use the equation s = ut + frac{1}{2}at^2, where s is the displacement (43.9m in this case), u is the initial velocity (0, since it's dropped), t is the time it takes to fall, and a is the acceleration due to gravity (approximately 9.81 m/s^2).## Step 3: Calculate the time it takes for the first stone to hit the water.Substitute the known values into the equation: 43.9 = 0 cdot t + frac{1}{2} cdot 9.81 cdot t^2. Simplify to get 43.9 = 4.905 cdot t^2. Solving for t gives t^2 = frac{43.9}{4.905}, and thus t = sqrt{frac{43.9}{4.905}}.## Step 4: Calculate the time it takes for the first stone to hit the water explicitly.t = sqrt{frac{43.9}{4.905}} approx sqrt{8.95} approx 2.99 seconds.## Step 5: Apply the equation of motion to the second stone.For the second stone, which is also under the influence of gravity but has an initial velocity v, the equation becomes s = v(t-1) + frac{1}{2}a(t-1)^2, since it travels for t-1 seconds.## Step 6: Set up the equation for the second stone to hit the water at the same time as the first stone.Given that both stones hit the water at the same time and travel the same distance, we can equate their distances: 43.9 = v(t-1) + frac{1}{2} cdot 9.81 cdot (t-1)^2.## Step 7: Substitute the known time into the equation for the second stone.Given t approx 2.99 seconds, substitute t into the equation: 43.9 = v(2.99-1) + frac{1}{2} cdot 9.81 cdot (2.99-1)^2.## Step 8: Simplify and solve for v.43.9 = v(1.99) + frac{1}{2} cdot 9.81 cdot (1.99)^2. Calculate the terms: 43.9 = 1.99v + frac{1}{2} cdot 9.81 cdot 3.9601. Simplify further: 43.9 = 1.99v + 19.43. Subtract 19.43 from both sides: 24.47 = 1.99v. Solve for v: v = frac{24.47}{1.99}.## Step 9: Calculate the initial speed of the second stone explicitly.v = frac{24.47}{1.99} approx 12.30 m/s.The final answer is: boxed{12.3}

🔑:## Step 1: Understand the Problem and Thevenin's TheoremThevenin's Theorem states that any linear electrical network containing only voltage sources, current sources, and resistors can be replaced by an equivalent circuit consisting of a single voltage source connected in series with a single resistor. To prove this, we'll use Kirchhoff's laws and the superposition principle.## Step 2: Apply Kirchhoff's LawsKirchhoff's laws state that the sum of currents entering a node is equal to the sum of currents leaving the node (KCL), and the sum of voltage changes around a closed loop is zero (KVL). We will apply these laws to the original circuit to find the voltage and current at the output terminals.## Step 3: Use the Superposition PrincipleThe superposition principle allows us to find the contribution of each source to the output voltage and current separately and then sum these contributions. This principle is crucial for simplifying the analysis of circuits with multiple sources.## Step 4: Find the Thevenin Voltage (Vth)To find the Thevenin voltage (Vth), we need to calculate the open-circuit voltage at the output terminals. This involves setting the load resistance to infinity (or removing it) and calculating the voltage across the terminals using the superposition principle and Kirchhoff's laws.## Step 5: Find the Thevenin Resistance (Rth)The Thevenin resistance (Rth) is found by setting all voltage sources to zero (replacing them with short circuits) and all current sources to zero (replacing them with open circuits), then calculating the total resistance seen from the output terminals.## Step 6: Construct the Thevenin Equivalent CircuitWith Vth and Rth determined, we can construct the Thevenin equivalent circuit, which consists of Vth in series with Rth. This circuit is equivalent to the original circuit in terms of voltage and current at the output terminals.## Step 7: Verify ConsistencyTo verify that the Thevenin equivalent circuit is consistent with the original circuit, we compare the voltage and current at the output terminals of both circuits under the same load conditions. The Thevenin equivalent should produce the same voltage and current as the original circuit for any given load.The final answer is: boxed{Vth}

🔑:To find the IRR, we need to find the discount rate that makes the NPV equal to zero. We can use the following equation:NPV = -65,200 + (-31,200) / (1 + IRR) + (-49,100) / (1 + IRR)^2 = 0We can solve for IRR using numerical methods or a financial calculator. The IRR is approximately 12.95%.To find the NPV at a required return of 10%, we can use the following equation:NPV = -65,200 + (-31,200) / (1 + 0.10) + (-49,100) / (1 + 0.10)^2= -65,200 - 28,364 - 40,683= -134,247Since the NPV is negative, the firm should not accept the project.To find the NPV at a required return of 0%, we can use the following equation:NPV = -65,200 - 31,200 - 49,100= -145,500To find the NPV at a required return of 20%, we can use the following equation:NPV = -65,200 + (-31,200) / (1 + 0.20) + (-49,100) / (1 + 0.20)^2= -65,200 - 26,000 - 34,222= -125,422Note that the NPV is negative at all required returns, indicating that the project is not profitable.

🔑:A great question about the celestial mechanics of Venus!The reason behind Venus being approximately 72 degrees away from its previous position in the sky after each synodic period of 584 days is due to the difference between the Venusian synodic period and the Earth's orbital period.To understand this, let's define some terms:* Synodic period: The time it takes for a planet to return to the same position in the sky relative to the Sun, as seen from Earth. For Venus, this period is approximately 584 days.* Orbital period: The time it takes for a planet to complete one orbit around the Sun. For Venus, this period is approximately 225 days, and for Earth, it's approximately 365.25 days.Now, let's calculate the angular displacement of Venus in the sky after each synodic period:1. Calculate the number of Earth orbits that occur during one Venusian synodic period:584 days (Venusian synodic period) ÷ 365.25 days (Earth's orbital period) ≈ 1.6 Earth orbitsThis means that during one Venusian synodic period, Earth completes approximately 1.6 orbits around the Sun.2. Calculate the angular displacement of Venus in the sky due to its own orbital motion:Venus completes one orbit around the Sun in approximately 225 days. During this time, it moves through an angle of 360 degrees (one full orbit).Now, let's calculate the angular displacement of Venus during one synodic period:(584 days ÷ 225 days) × 360 degrees ≈ 936 degreesHowever, this is not the angular displacement we observe from Earth, since Earth is also moving around the Sun. To account for this, we need to subtract the angular displacement of Earth during the same period:(584 days ÷ 365.25 days) × 360 degrees ≈ 576 degrees (Earth's angular displacement)Now, subtract the Earth's angular displacement from the Venusian angular displacement:936 degrees - 576 degrees ≈ 360 degreesThis result might seem surprising, but it's not the final answer. We need to consider that Venus is moving in the opposite direction of Earth during part of its synodic period (when it's on the other side of the Sun). To account for this, we need to subtract the angular displacement of Venus from the angular displacement of Earth during the time when Venus is on the other side of the Sun.This occurs for approximately 290 days (half of the synodic period). During this time, Earth moves through an angle of:(290 days ÷ 365.25 days) × 360 degrees ≈ 288 degreesNow, subtract this angle from the previous result:360 degrees - 288 degrees ≈ 72 degreesTherefore, after each synodic period of 584 days, Venus is approximately 72 degrees away from its previous position in the sky. This calculation assumes a simplified model of the orbits and neglects other effects, such as the elliptical shape of the orbits and the gravitational interactions between the planets.Keep in mind that this is an approximate calculation, and the actual angular displacement of Venus might vary slightly due to various astronomical factors. Nonetheless, this calculation provides a good understanding of the underlying celestial mechanics that govern the motion of Venus in the sky.