/*! 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 8 Conceptual Example 2 provides so... [FREE SOLUTION] | 91Ó°ÊÓ

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Conceptual Example 2 provides some relevant background for this problem. A jet is circling an airport control tower at a distance of \(18.0 \mathrm{km}\) An observer in the tower watches the jet cross in front of the moon. As seen from the tower, the moon subtends an angle of \(9.04 \times 10^{-3}\) radians. Find the distance traveled (in meters) by the jet as the observer watches the nose of the jet cross from one side of the moon to the other.

Short Answer

Expert verified
The jet travels 162.72 meters across the moon.

Step by step solution

01

Understand the Problem

The observer sees the jet cross the moon, which subtends an angle of \(9.04 \times 10^{-3}\) radians. We need to find the arc length (distance traveled by the jet as it crosses the moon) at a distance of \(18.0\) km from the observer.
02

Use Arc Length Formula

Use the formula for arc length, \(s = r\theta\), where \(s\) is the arc length, \(r\) is the radius (distance from the control tower to the jet), and \(\theta\) is the subtended angle in radians. Here, \(r = 18.0\, \text{km} = 18,000\, \text{m}\) and \(\theta = 9.04 \times 10^{-3}\) radians.
03

Calculate the Distance

Substitute the given values into the arc length formula: \[s = 18,000 \times 9.04 \times 10^{-3} = 162.72 \text{ meters}.\]This calculation gives the distance the jet travels as it crosses from one side of the moon to the other as seen from the tower.

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Key Concepts

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

Radians
Radians are a way of measuring angles based on the radius of a circle. Unlike degrees, which divide a circle into 360 parts, radians use the radius as a unit of measurement. This means that when you measure an angle in radians, you're measuring the length of the arc that the angle creates when drawn on a circle with a radius of 1 unit.
To convert an angle from degrees to radians, use the formula \[ ext{Radians} = rac{ ext{Degrees} imes oldsymbol{ rac{oldsymbol{ ext{Ï€}}}}{oldsymbol{180}}}\]In this exercise, the subtended angle of the moon is given as a very small value of \(9.04 \times 10^{-3}\) radians. This small radian measure is typical for astronomical observations, where angles appear tiny due to the vast distances involved. Radians are particularly useful in circle geometry because they make calculations involving circles simpler and more intuitive, especially in formulas such as those for arc length and sector area.
Circle Geometry
Circle geometry is the study of the properties and measurements of circles and the shapes that form within them. When dealing with problems involving circles, there are a few key terms and formulas to understand:
  • Radius (\[r\]): The distance from the center of the circle to any point on its circumference.

  • Diameter: Twice the radius, or the longest possible line segment that can fit within the circle.

  • Circumference: The total distance around the circle, calculated using \[ C = 2\pi r \].
  • Arc Length: The distance between two points along the circumference. This is found using \[s = r\theta\], where \[r\] is the radius and \[\theta\] is the angle in radians.
Understanding circle geometry helps solve problems involving arcs, such as determining how far the jet travels while crossing in front of the moon. This application of circle geometry is seen in the exercise by using the arc length formula to find the path the jet takes across the moon from the perspective of the observer.
Distance Calculation
Distance calculation often applies circle geometry principles to solve practical problems. In this context, distance refers to the length of the path traveled by an object, such as a jet, as it moves along a circular path, in this case, across the view of the moon.
The key formula for calculating this distance, or arc length, is \[s = r\theta\]. Here, \[s\] is the arc length (or the distance the jet travels), \[r\] is the radius (the distance from the observer to the jet), and \[\theta\] is the angle in radians subtended by the section of the circle in question.
In the provided exercise, using the values:
  • \[r = 18,000 \text{ m}\] (18 km converted into meters), and

  • \[\theta = 9.04 \times 10^{-3} \text{ radians}\], the calculation yields an arc length:

  • \[s = 18,000 \times 9.04 \times 10^{-3} = 162.72 \text{ meters}\].
This formula and computation allow observers to determine precisely how much of a given circular path has been traveled, providing insights into the movement of objects along circular paths efficiently and accurately.

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Most popular questions from this chapter

A Ferris wheel rotates at an angular velocity of \(0.24 \mathrm{rad} / \mathrm{s}\). Starting from rest, it reaches its operating speed with an average angular acceleration of \(0.030 \mathrm{rad} / \mathrm{s}^{2} .\) How long does it take the wheel to come up to operating speed?

The sun has a mass of \(1.99 \times 10^{30} \mathrm{kg}\) and is moving in a circular orbit about the center of our galaxy, the Milky Way. The radius of the orbit is \(2.3 \times 10^{4}\) light-years (1 light-year \(\left.=9.5 \times 10^{15} \mathrm{m}\right)\), and the angular speed of the sun is \(1.1 \times 10^{-15} \mathrm{rad} / \mathrm{s}\) (a) Determine the tangential speed of the sun. (b) What is the magnitude of the net force that acts on the sun to keep it moving around the center of the Milky Way?

A rider on a mountain bike is traveling to the left in the figure. Each wheel has an angular velocity of \(+21.7 \mathrm{rad} / \mathrm{s},\) where, as usual, the plus sign indicates that the wheel is rotating in the counterclockwise direction. (a) To pass another cyclist, the rider pumps harder, and the angular velocity of the wheels increases from +21.7 to \(+28.5 \mathrm{rad} / \mathrm{s}\) in a time of 3.50 s. (b) After passing the cyclist, the rider begins to coast, and the angular velocity of the wheels decreases from +28.5 to +15.3 rad/s in a time of 10.7 s. Concepts: (i) Is the angular acceleration positive or negative when the rider is passing the cyclist and the angular speed of the wheels is increasing? (ii) Is the angular acceleration positive or negative when the rider is coasting and the angular speed of the wheels is decreasing? Calculations: In both instances, (a) and (b), determine the magnitude and direction of the angular acceleration (assumed constant) of the wheels.

At the local swimming hole, a favorite trick is to run horizontally off a cliff that is \(8.3 \mathrm{m}\) above the water. One diver runs off the edge of the cliff, tucks into a "ball," and rotates on the way down with an average angular speed of 1.6 rev/s. Ignore air resistance and determine the number of revolutions she makes while on the way down.

Our sun rotates in a circular orbit about the center of the Milky Way galaxy. The radius of the orbit is \(2.2 \times 10^{20} \mathrm{m},\) and the angular speed of the sun is \(1.1 \times 10^{-15} \mathrm{rad} / \mathrm{s}\). How long (in years) does it take for the sun to make one revolution around the center?

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