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The luxury liner Queen Elizabeth 2 has a diesel-electric power plant with a maximum power of \(92 \mathrm{MW}\) at a cruising speed of 32.5 knots. What forward force is exerted on the ship at this speed? \((1 \mathrm{knot}=1.852 \mathrm{~km} / \mathrm{h} .)\)

Short Answer

Expert verified
The forward force is approximately 5,501,437 N.

Step by step solution

01

Convert Speed to Meters per Second

First, we need to convert the ship's speed from knots to meters per second. We know that 1 knot equals 1.852 km/h. So, the speed in km/h is \(32.5 \times 1.852 = 60.17 \) km/h. Next, convert this speed to meters per second by dividing by 3.6 (since 1 km/h is approximately 0.27778 m/s). Thus, the speed is \( \frac{60.17}{3.6} \approx 16.714 \, \text{m/s} \).
02

Recall Power Formula

We use the formula relating power, force, and velocity: \[ \text{Power} = \text{Force} \times \text{Velocity} \]With power \( P = 92 \, \text{MW} = 92,000,000 \, \text{W} \) and velocity \( v = 16.714 \, \text{m/s} \).
03

Solve for Force

Rearrange the power formula to solve for force: \[ \text{Force} = \frac{\text{Power}}{\text{Velocity}} \]Substituting the given values, we have:\[ \text{Force} = \frac{92,000,000 \, \text{W}}{16.714 \, \text{m/s}} \approx 5,501,437.45 \, \text{N} \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Diesel-Electric Power Plant
A diesel-electric power plant combines a diesel engine with an electric generator to provide power. This setup is often favored in maritime vessels like large ships due to its efficiency and flexibility. The diesel engine converts fuel into mechanical energy, which is then transformed into electrical energy by the generator. This electrical energy powers electric motors that drive the ship's propellers.

The Queen Elizabeth 2, a renowned luxury liner, uses such a system and can produce a massive power output of 92 megawatts. This level of power is crucial for propelling the ship at high speeds, especially over long distances across the sea. Diesel-electric systems are advantageous for such applications because they offer consistent power and can easily adjust to the varying energy demands of a ship.

Moreover, the modular nature of these systems allows for better maintenance and more efficient operation, making them an ideal choice for the demanding conditions at sea.
Unit Conversion
Unit conversion is a fundamental process in solving physics problems, especially when dealing with various measurement systems. In this exercise, we need to convert speed from knots to meters per second. It's important because consistency in units is crucial for accurate calculations.

Start by understanding that 1 knot corresponds to 1.852 kilometers per hour. Multiply this value by the ship's cruising speed, in knots, to get the speed in kilometers per hour:
  • Speed in km/h: \(32.5 \times 1.852 = 60.17 \text{ km/h}\)
Next, convert kilometers per hour to meters per second. Since there are 1000 meters in a kilometer and 3600 seconds in an hour, divide by 3.6 (i.e., 1000/3600):
  • Speed in m/s: \(\frac{60.17}{3.6} \approx 16.714 \text{ m/s} \)
Converting between units ensures calculations are in the correct form to apply physics equations properly, like those involving power or force.
Velocity and Force Relationship
The relationship between velocity and force is an essential part of understanding motion dynamics in physics. Power, velocity, and force are interrelated through the formula:
  • \( \text{Power} = \text{Force} \times \text{Velocity} \)
For a moving object, like the Queen Elizabeth 2, this means the power it generates not only depends on the force exerted but also on its speed or velocity.

Given the power output of 92 megawatts and a calculated velocity of approximately 16.714 meters per second, we can solve for the force exerted by rearranging the power formula:
  • \( \text{Force} = \frac{\text{Power}}{\text{Velocity}} \)
  • \( \text{Force} = \frac{92,000,000 \text{ W}}{16.714 \text{ m/s}} \approx 5,501,437.45 \text{ N} \)
Understanding this relationship helps in determining how much force is required to maintain the linear motion of the ship at a constant speed.

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Most popular questions from this chapter

Figure 8 - 34 shows a thin rod, of length \(L=2.00 \mathrm{~m}\) and negligible mass, that can pivot about one end to rotate in a vertical circle. A ball of mass \(m=5.00 \mathrm{~kg}\) is attached to the other end. The rod is pulled aside to angle \(\theta_{0}=30.0^{\circ}\) and released with initial velocity \(\vec{v}_{0}=0 .\) As the ball descends to its lowest point, (a) how much work does the gravitational force do on it and (b) what is the change in the gravitational potential energy of the ball-Earth system? (c) If the gravitational potential energy is taken to be zero at the lowest point, what is its value just as the ball is released? (d) Do the magnitudes of the answers to (a) through (c) increase, decrease, or remain the same if angle \(\theta_{0}\) is increased?

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