🔑:## Step 1: Understanding the Mission RequirementsThe New Horizons spacecraft is on a mission to Pluto, requiring significant speed to cover the vast distance between Earth and Pluto efficiently. The spacecraft's initial speed was 23 km/s, which was increased to 27 km/s after a gravity-assisted flyby of Jupiter.## Step 2: Ion Drive BasicsAn ion drive, or ion thruster, is a type of propulsion system that uses electrical energy to accelerate ions, typically xenon gas, to high speeds, generating a low but continuous thrust. This type of propulsion is more fuel-efficient than traditional chemical rockets but provides much less thrust.## Step 3: Benefits of Ion Drive for New Horizons- Fuel Efficiency: Ion drives are highly fuel-efficient, which could extend the mission duration by reducing the amount of fuel needed for a given delta-v (change in velocity).- Continuous Thrust: Although the thrust is low, an ion drive can operate for thousands of hours, providing a continuous acceleration that, over time, can achieve significant speeds.- Low Mass: The propulsion system itself and the fuel required for a mission can be significantly less massive than traditional propulsion systems, which is beneficial for deep space missions where launch costs are directly related to the mass of the spacecraft.## Step 4: Drawbacks of Ion Drive for New Horizons- Low Thrust: The acceleration provided by an ion drive is very low, meaning it takes a long time to achieve significant speed increases. This could be a drawback for missions where time is critical.- Power Consumption: Ion drives require a significant amount of electrical power to operate, which can be a challenge for spacecraft that rely on solar panels or nuclear reactors for power.- Risk of Engine Failure: The operation of an ion engine is complex and involves risks such as grid erosion, cathode failure, and thrust vectoring issues. A failure could leave the spacecraft without propulsion.## Step 5: Trade-offs for New Horizons Mission- Fuel Mass vs. Mission Duration: Using an ion drive could reduce the fuel mass significantly but would increase the mission duration due to the low thrust.- Mission Duration vs. Risks: A longer mission duration increases the risks associated with space travel, such as cosmic radiation exposure, electronic component failure, and the challenges of maintaining communication with Earth.- Risks vs. Benefits: The benefits of increased fuel efficiency and potentially lower launch costs must be weighed against the risks of engine failure and the challenges of operating an ion drive over extended periods.## Step 6: ConclusionFor a mission like New Horizons, which requires a significant speed increase to reach Pluto in a reasonable time frame, the use of an ion drive after the initial gravity-assisted speed boost could offer benefits in terms of fuel efficiency and potentially lower mass. However, the low thrust and power requirements of an ion drive, along with the increased mission duration and associated risks, are significant drawbacks. The decision to use an ion drive would depend on the specific mission requirements, the trade-offs between fuel mass, mission duration, and the acceptable level of risk.The final answer is: boxed{Ion drive benefits: fuel efficiency, continuous thrust, and low mass. Drawbacks: low thrust, power consumption, and risk of engine failure. Trade-offs involve balancing fuel mass, mission duration, and risks.}

🔑:To calculate the percentage of the original energy that exists after the light travels 30 meters through the material, we first need to understand the given loss in decibels (dB) and how it relates to the distance the light travels.Given:- Losses = 10^3 dB/km- Distance = 30 meters = 0.03 km (since 1 km = 1000 meters)The formula to calculate the loss in decibels over a certain distance is:[ text{Loss (dB)} = alpha times d ]where (alpha) is the attenuation coefficient in dB/km, and (d) is the distance in kilometers.However, to find the percentage of the original energy that remains, we need to convert the decibel loss into a ratio of the remaining energy to the original energy. The formula to relate decibel loss to the ratio of powers (or energies, since power is energy per unit time and we're considering the energy here) is:[ text{Loss (dB)} = 10 log_{10} left( frac{P_{text{out}}}{P_{text{in}}} right) ]where (P_{text{out}}) is the output power (or energy in this context) and (P_{text{in}}) is the input power (or energy).Rearranging this formula to solve for the ratio ( frac{P_{text{out}}}{P_{text{in}}} ) gives:[ frac{P_{text{out}}}{P_{text{in}}} = 10^{frac{-text{Loss (dB)}}{10}} ]First, calculate the loss over 30 meters (0.03 km):[ text{Loss (dB)} = 10^3 , text{dB/km} times 0.03 , text{km} = 10^3 times 0.03 = 30 , text{dB} ]Now, calculate the ratio of the output energy to the input energy:[ frac{P_{text{out}}}{P_{text{in}}} = 10^{frac{-30}{10}} = 10^{-3} ]This means that ( frac{1}{1000} ) of the original energy remains after the light travels 30 meters through the material.To find the percentage, multiply by 100:[ frac{1}{1000} times 100 = 0.1% ]Therefore, 0.1% of the original energy exists after the light travels 30 meters through the material.

🔑:## Step 1: Calculate the mass per unit length of the stringTo find the mass per unit length (μ) of the string, we use the formula μ = m / L, where m is the mass of the string and L is its length. Given that a 3-meter length of the string has a mass of 18g, we can calculate μ as follows: μ = 18g / 3m = 6g/m = 0.006 kg/m.## Step 2: Calculate the speed of the wave in the stringThe speed (v) of a wave in a string is given by the formula v = √(T / μ), where T is the tension in the string and μ is the mass per unit length. Given that the tension T is 9N and μ is 0.006 kg/m, we can calculate v as follows: v = √(9N / 0.006 kg/m) = √(1500) = 38.73 m/s.## Step 3: Determine the wavelength for the first two standing wave modesFor a string fixed at both ends, the wavelength (λ) of the nth standing wave mode is given by λ = 2L / n, where L is the length of the string and n is the mode number (n = 1, 2, 3, ...). Given that L = 0.7m, we can calculate λ for the first two modes: For n = 1, λ1 = 2*0.7m / 1 = 1.4m; for n = 2, λ2 = 2*0.7m / 2 = 0.7m.## Step 4: Calculate the frequency of the first two standing wave modesThe frequency (f) of a wave is given by the formula f = v / λ, where v is the speed of the wave and λ is its wavelength. Using the speed calculated in step 2 and the wavelengths for the first two modes from step 3, we can calculate the frequencies: For the first mode (n = 1), f1 = 38.73 m/s / 1.4m = 27.66 Hz; for the second mode (n = 2), f2 = 38.73 m/s / 0.7m = 55.33 Hz.The final answer is: boxed{27.66, 55.33}

🔑:In a tug-of-war match, placing the strongest members at the back of the rope can provide a significant advantage. From a physics perspective, this strategic placement can be explained by considering the application of horizontal force, the role of the anchor, and the concept of impulse.Application of Horizontal Force:When a team member pulls on the rope, they apply a horizontal force to the rope. According to Newton's second law of motion (F = ma), the force applied to the rope is proportional to the mass (m) of the team member and their acceleration (a). The strongest members, typically with larger masses, can generate a greater force on the rope. By placing them at the back of the rope, they can apply this force over a longer distance, resulting in a greater torque (rotational force) on the rope. This increased torque allows the team to generate more rotational motion, which can help to pull the opposing team towards them.Role of the Anchor:The anchor, typically the team member at the back of the rope, plays a crucial role in maintaining the team's stability and generating force. By placing the strongest members at the back, they can serve as a more effective anchor, providing a stable base for the rest of the team to pull from. The anchor's primary function is to resist the opposing team's force, allowing the rest of the team to generate a net force in the desired direction. A stronger anchor can better resist the opposing team's force, enabling the team to maintain a consistent pulling motion and generate more force overall.Concept of Impulse:Impulse (J) is the product of force (F) and time (t), and it's a measure of the change in momentum (Δp) of an object. In a tug-of-war match, the team that applies the greatest impulse to the rope will win. By placing the strongest members at the back of the rope, they can generate a greater force over a longer period, resulting in a larger impulse. This increased impulse allows the team to transfer more momentum to the rope, which can help to overcome the opposing team's resistance and ultimately pull them towards the winning team.Additional Advantages:Placing the strongest members at the back of the rope also provides other advantages, including:1. Leverage: The back of the rope is typically farther from the center of rotation, providing a longer lever arm. This allows the strongest members to generate more torque and rotational motion, making it easier to pull the opposing team.2. Stability: A stronger anchor at the back of the rope helps to maintain the team's stability, reducing the likelihood of the team being pulled off balance.3. Coordination: With the strongest members at the back, the rest of the team can focus on generating a consistent pulling motion, while the anchor provides the necessary stability and force to overcome the opposing team's resistance.In conclusion, placing the strongest members at the back of the rope in a tug-of-war match provides a strategic advantage by allowing them to apply a greater horizontal force, serve as a more effective anchor, and generate a larger impulse. This placement enables the team to generate more torque, maintain stability, and ultimately transfer more momentum to the rope, increasing their chances of winning the match.