/*! 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 80 Calculate the magnitude of the d... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 6: Problem 80

Calculate the magnitude of the drag force on a missile \(53 \mathrm{~cm}\) in diameter cruising at \(250 \mathrm{~m} / \mathrm{s}\) at low altitude, where the density of air is \(1.2 \mathrm{~kg} / \mathrm{m}^{3} .\) Assume \(C=0.75\)

Short Answer

Expert verified
The drag force is approximately 6206.25 N.

Step by step solution

01

Calculate the Area

First, calculate the cross-sectional area of the missile. The missile is a cylinder, so we use the formula for the area of a circle, which is given by \[ A = \pi \left( \frac{d}{2} \right)^2 \]where \( d \) is the diameter. Replace \( d = 53 \ cm = 0.53 \ m \) into the equation, we get:\[ A = \pi \left( \frac{0.53}{2} \right)^2 = \pi \times (0.265)^2 \approx 0.2208 \ m^2 \]
02

Use Drag Force Formula

Now, use the drag force formula \[ F_d = \frac{1}{2} C \rho v^2 A \]where \( C \) is the drag coefficient, \( \rho \) is the air density, \( v \) is the velocity, and \( A \) is the area.Substitute in the known values \( C = 0.75 \), \( \rho = 1.2 \ kg/m^3 \), \( v = 250 \ m/s \), and \( A = 0.2208 \ m^2 \):\[ F_d = \frac{1}{2} \times 0.75 \times 1.2 \times (250)^2 \times 0.2208 \]
03

Calculate the Drag Force

Perform the calculations:First, calculate \( v^2 \):\[ v^2 = (250)^2 = 62500 \]Then, calculate the drag force:\[ F_d = 0.375 \times 1.2 \times 62500 \times 0.2208 = 0.45 \times 62500 \times 0.2208 \]Simplify:\[ F_d = 28125 \times 0.2208 \approx 6206.25 \ N \]
04

Result and Conclusion

The magnitude of the drag force on the missile is approximately calculated as\[ F_d \approx 6206.25 \ N \]. This result indicates the resistance experienced by the missile due to air drag.

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.

Understanding the Drag Coefficient
The drag coefficient, often denoted as \( C \), is a dimensionless number that quantifies the resistance an object faces as it moves through a fluid, such as air. It's crucial in calculating the drag force, as seen in the formula \( F_d = \frac{1}{2} C \rho v^2 A \).

The drag coefficient is influenced by several factors:
  • Shape of the object: Sleeker shapes usually have lower drag coefficients.
  • Surface roughness: A smoother surface results in lower drag.
  • Reynolds Number: A measure of the flow conditions around the object.
In this exercise, a value of 0.75 is assumed for the missile. This indicates a moderate level of drag, typical for streamlined shapes. A perfect sphere would have a lower coefficient, while a more boxy shape would have a higher one.
Cross-Sectional Area
The cross-sectional area is the surface area of the slice of the object perpendicular to the direction of the flow of fluid. It is a fundamental factor in determining drag force.

For a cylindrical object like a missile, the cross-sectional area is calculated using the area formula for a circle: \[ A = \pi \left( \frac{d}{2} \right)^2 \] where \( d \) is the diameter.
In our example, with a diameter of 53 cm, converting to meters (\( 0.53 \ m \)), the area is approximately \( 0.2208 \ m^2 \).
The larger the cross-sectional area, the more air the object has to push against, leading to greater drag force. Thus, reducing the diameter, while maintaining aerodynamic efficiency, can result in significantly lower drag forces.
The Role of Air Density
Air density, symbolized as \( \rho \), is a measure of how many air molecules are present in a given volume. It is expressed in \( kg/m^3 \) and is essential in the drag force calculation.

Denser air contains more molecules, thus increasing the likelihood of these molecules colliding with the moving body, which increases the drag force.
  • At higher altitudes, air density decreases, reducing drag.
  • Bigger and heavier gas molecules make air denser.
In our problem, the air density is given as \( 1.2 \ kg/m^3 \). This is typical for low altitudes where the composition of the atmosphere remains relatively thick. As a result, drag forces are more substantial at low altitudes compared to higher ones. Understanding air density helps in predicting and controlling the drag an object may encounter during its movement.

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

A ski that is placed on snow will stick to the snow. However, when the ski is moved along the snow, the rubbing warms and partially melts the snow, reducing the coefficient of kinetic friction and promoting sliding. Waxing the ski makes it water repellent and reduces friction with the resulting layer of water. A magazine reports that a new type of plastic ski is especially water repellent and that, on a gentle \(200 \mathrm{~m}\) slope in the Alps, a skier reduced his top-to-bottom time from 61 s with standard skis to 42 s with the new skis. Determine the magnitude of his average acceleration with (a) the standard skis and (b) the new skis. Assuming a \(3.0^{\circ}\) slope, compute the coefficient of kinetic friction for (c) the standard skis and (d) the new skis.

A filing cabinet weighing \(556 \mathrm{~N}\) rests on the floor. The coefficient of static friction between it and the floor is \(0.68,\) and the coefficient of kinetic friction is \(0.56 .\) In four different attempts to move it, it is pushed with horizontal forces of magnitudes (a) \(222 \mathrm{~N}\), (b) \(334 \mathrm{~N},\) (c) \(445 \mathrm{~N},\) and (d) \(556 \mathrm{~N}\). For each attempt, calculate the magnitude of the frictional force on it from the floor. (The cabinet is initially at rest.) (e) In which of the attempts does the cabinet move?

As a \(40 \mathrm{~N}\) block slides down a plane that is inclined at \(25^{\circ}\) to the horizontal, its acceleration is \(0.80 \mathrm{~m} / \mathrm{s}^{2},\) directed up the plane. What is the coefficient of kinetic friction between the block and the plane?

You testify as an expert witness in a case involving an accident in which car \(A\) slid into the rear of car \(B,\) which was stopped at a red light along a road headed down a hill (Fig. 6 -25). You find that the slope of the hill is \(\theta=12.0^{\circ},\) that the cars were separated by distance \(d=24.0 \mathrm{~m}\) when the driver of car \(A\) put the car into a slide (it lacked any automatic anti-brake-lock system), and that the speed of car \(A\) at the onset of braking was \(v_{0}=18.0 \mathrm{~m} / \mathrm{s} .\) With what speed did car \(A\) hit car \(B\) if the coefficient of kinetic friction was (a) 0.60 (dry road surface) and (b) 0.10 (road surface covered with wet leaves)?

The coefficient of static friction between Teflon and scrambled eggs is about \(0.04 .\) What is the smallest angle from the horizontal that will cause the eggs to slide across the bottom of a Teflon-coated skillet?
See all solutions

Recommended explanations on Physics Textbooks

Electricity and Magnetism

Read Explanation

Work Energy and Power

Read Explanation

Physical Quantities And Units

Read Explanation

Rotational Dynamics

Read Explanation

Conservation of Energy and Momentum

Read Explanation

Particle Model of Matter

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.