/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 52 The aircraft carrier John \(F .\... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 6: Problem 52

The aircraft carrier John \(F .\) Kennedy has mass \(7.4 \times 10^{7} \mathrm{kg}\). When its engines are developing their full power of \(280,000 \mathrm{hp}\), the John \(F\) . Kennedy travels at its top speed of 35 knots \((65 \mathrm{km} / \mathrm{h})\). If 70\(\%\) of the power output of the engines is applied to pushing the ship through the water, what is the magnitude of the force of water resistance that opposes the carrier's motion at this speed?

Short Answer

Expert verified
The magnitude of the force of water resistance is approximately 8,098,884 N.

Step by step solution

01

Convert Horsepower to Watts

First, we need to convert the engine's power from horsepower to watts. We know that 1 horsepower equals approximately 746 watts. Therefore, the total engine power in watts is given by \(280,000 \times 746 = 208,880,000 \text{ watts}\).
02

Calculate Useful Power Applied to Moving the Ship

Determine the portion of power used for pushing the ship through the water. Since 70% of the engine's power is used for this purpose, we calculate \(0.7 \times 208,880,000 = 146,216,000 \text{ watts}\).
03

Convert Speed from Kilometers per Hour to Meters per Second

The speed of the ship is given as 65 km/h. Convert this speed to meters per second using the conversion: \(1\, \text{km/h} = \frac{1,000}{3,600} \, \text{m/s}\). So, \(65 \times \frac{1,000}{3,600} = 18.0556\, \text{m/s}\).
04

Calculate the Force of Water Resistance

Using the formula for power, \( P = F \times v \), where \(P\) is the power, \(F\) is the force, and \(v\) is the velocity, we can solve for the force \(F\). Substitute the known values to get \( 146,216,000 = F \times 18.0556\). Solving for \(F\) gives \( F = \frac{146,216,000}{18.0556} \approx 8,098,884\, \text{newtons}\).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

Power Conversion
Power conversion is essential when working with systems that use different units to describe power. In many parts of the world, horsepower is a common unit for indicating the power of engines, particularly in the automotive and marine industries. However, scientific calculations often require power to be expressed in watts for consistency with other SI units.
  • To convert from horsepower to watts, we utilize the known conversion rate where 1 horsepower equals approximately 746 watts.
  • In our exercise, the total power of the aircraft carrier's engines in watts is calculated by multiplying 280,000 hp by 746, which results in 208,880,000 watts.
This conversion ensures precise and consistent calculations in physics problems involving power and energy.
Force Calculation
Force calculation involves determining the force exerted by or on an object, which can be solved through the relation between power, force, and velocity. The formula used is: \[ P = F \times v \]where \(P\) is power, \(F\) is the force, and \(v\) is the velocity. In scenarios where power and velocity are known, the force can be easily calculated.
  • In the exercise, we're interested in the water resistance force, an opposing force acting on the moving ship.
  • Given the useful power used to push the ship through water and the ship's speed, the force can be calculated by rearranging the formula: \(F = \frac{P}{v}\).
  • The calculation involves substituting the useful power (146,216,000 watts) and the converted speed (18.0556 m/s) to find the force.
  • The estimated force opposing the ship's motion is approximately 8,098,884 newtons.
This connection between these three variables provides a way to determine unknowns when dealing with moving objects.
Unit Conversion
Unit conversion is critical for ensuring that all parts of a calculation use consistent units, simplifying problem-solving. Often, measurements need to be converted to their SI equivalents before they鈥檙e used in formulas.
  • Speed is typically given in kilometers per hour, but for calculations in physics, it's converted to meters per second.
  • The conversion factor from kilometers per hour to meters per second is \( \frac{1,000}{3,600} \) (or roughly 0.2778).
  • For the aircraft carrier moving at 65 km/h, we multiply by this factor to find the speed in SI units: \( 18.0556\, \text{m/s}\).
Before performing any calculation in physics, always verify units are compatible. This careful initial conversion prevents errors and maintains the clarity and accuracy of results.
Water Resistance Force
The water resistance force is the counteracting force that water exerts against the ship as it moves forward. This force is a type of drag that needs to be overcome by the power produced by the ship's engines to maintain motion.
  • It's influenced by factors such as the ship's shape, speed, and the viscosity of the water.
  • In our exercise, it's calculated based on the useful power applied to overcoming this resistance and the ship's velocity.
  • The calculated force opposing the aircraft carrier's motion is critical for determining engine efficiency and operational power requirements.
Understanding water resistance is vital for maritime engineers and helps in designing ships that are both efficient and capable of cutting through water with minimal energy loss.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

An object is attracted toward the origin with a force given by \(F_{x}=-k / x^{2} .\) (Gravitational and electrical forces have this distance dependence.) (a) Calculate the work done by the force \(F_{x}\) when the object moves in the \(x\) -direction from \(x_{1}\) to \(x_{2}\) . If \(x_{2}>x_{1}\) , is the work done by \(F_{x}\) positive or negative? (b) The only other force acting on the object is a force that you exert with your hand to move the object slowly from \(x_{1}\) to \(x_{2}\) . How much work do you do? If \(x_{2}>x_{1},\) is the work you do positive or negative? (c) Explain the similarities and differences between your answers to parts (a) and (b).

Use the work-energy theorem to solve each of these problems. You can use Newton's laws to check your answers. Neglect air resistance in all cases. (a) A branch falls from the top of a 95.0-m-tall redwood tree, starting from rest. How fast is it moving when it reaches the ground? (b) A volcano ejects a boulder directly upward 525 \(\mathrm{m}\) into the air. How fast was the boulder moving just as it left the volcano? (c) A skier moving at 5.00 \(\mathrm{m} / \mathrm{s}\) encounters a long, rough horizontal patch of snow having coefficient of kinetic friction 0.220 with her skis. How far does she travel on this patch before stopping? (d) Suppose the rough patch in part (c) was only 2.90 \(\mathrm{m}\) long? How fast would the skier be moving when she reached the end of the patch? (e) At the base of a frictionless icy hill that rises at \(25.0^{\circ}\) above the horizontal, a toboggan has a speed of 12.0 \(\mathrm{m} / \mathrm{s}\) toward the hill. How high vertically above the base will it go before stopping?

Half of a Spring. (a) Suppose you cut a massless ideal spring in half. If the full spring had a force constant \(k\) , what is the force constant of each half, in terms of \(k ?\) (Hint: Think of the original spring as two equal halves, each producing the same force as the entire spring. Do you see why the forces must be equal? (b) If you cut the spring into three equal segments instead, what is the force constant of each one, in terms of \(k ?\)

The gravitational pull of the earth on an object is inversely proportional to the square of the distance of the object from the center of the earth. At the earth's surface this force is equal to the object's normal weight \(m g,\) where \(g=9.8 \mathrm{m} / \mathrm{s}^{2},\) and at large distances, the force is zero. If a \(20,000-k g\) asteroid falls to earth from a very great distance away, what will be its minimum speed as it strikes the earth's surface, and how much kinetic energy will it impart to our planet? You can ignore the effects of the earth's atmosphere.

The Grand Coulee Dam is 1270 \(\mathrm{m}\) long and 170 \(\mathrm{m}\) high. The electrical power output from generators at its base is approximately 2000 MW. How many cubic meters of water must flow from the top of the dam per second to produce this amount of power if 92\(\%\) of the work done on the water by gravity is converted to electrical energy? (Each cubic meter of water has a mass of 1000 \(\mathrm{kg}\) .)
See all solutions

Recommended explanations on Physics Textbooks

Scientific Method Physics

Read Explanation

Nuclear Physics

Read Explanation

Measurements

Read Explanation

Radiation

Read Explanation

Force

Read Explanation

Classical Mechanics

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.