/*! 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 In T.V. transmission tower at a ... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

In T.V. transmission tower at a perpendicular station has a height of \(160 \mathrm{~m}\) What is the population covered by the transmission, if the average population density around the tower is 1200 per square \(\mathrm{km}\). (A) \(5.22 \times 10^{6}\) (B) \(7.72 \times 10^{6}\) (C) \(3.52 \times 10^{6}\) (D) \(9.22 \times 10^{6}\)

Short Answer

Expert verified
The population covered by the transmission is approximately \(7.72 \times 10^{6}\) people.

Step by step solution

01

Determine Line of Sight Distance

To calculate the line of sight distance (d) of a signal from a tower of height (h), we can use the formula: \(d = \sqrt{2 \times h \times R}\) where R is the radius of the Earth, which is approximately 6371 km. In our case, the height of the transmission tower is 160 m, so we have: \(d = \sqrt{2 \times (160) \times (6371)}\)
02

Calculate the Area Covered

Now that we have the line of sight distance, we can calculate the area covered by the transmission. Assuming it broadcasts the signal in a circular area, we can use the formula: Area (A) = π × \(d^2\) First, we calculate the square of the line of sight distance (d): \(d^2 = (160 \times 6371)^2\) Now, we find the area: \(A = π \times (160 \times 6371)^2\)
03

Calculate the Population Covered

Now that we have the area covered by the transmission tower, we can calculate the population covered by multiplying the area by the population density (given as 1200 per square kilometer). We have: Population Covered = Area × Population Density Population Covered = \(π \times (160 \times 6371)^2 \times 1200\) Now, we evaluate our result and compare it to the provided options. (A) \(5.22 \times 10^{6}\) (B) \(7.72 \times 10^{6}\) (C) \(3.52 \times 10^{6}\) (D) \(9.22 \times 10^{6}\) If our result matches one of these options, that is the correct answer for the population covered by the transmission from the 160 m tower.

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 91Ó°ÊÓ!

Key Concepts

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

Line of Sight Distance
When it comes to TV transmission towers, understanding the concept of "Line of Sight Distance" is crucial. This is the maximum distance over which two points can directly see each other without any obstruction considering the curvature of the Earth. In simpler words, it is how far the transmission from the tower can reach.

The formula to calculate this distance is \(d = \sqrt{2 \times h \times R}\), where \(h\) stands for the height of the tower, and \(R\) is the Earth's radius. For instance, if a tower is 160 meters high, and the Earth's radius is 6371 km, you would plug these values into the formula to estimate how far the broadcast can potentially reach.
  • Height of tower \(h\) = 160 m
  • Earth’s radius \(R\) = 6371 km
  • Line of Sight Distance \(d\) = \(\sqrt{2 \times 160 \times 6371}\)
This neat calculation allows network planners to determine how much area can successfully receive a signal from the TV tower without interruptions.
Population Density
Population Density is another significant concept in understanding how many people a TV transmission can potentially reach. It refers to the average number of people living per square kilometer in a specific area surrounding the tower.

For example, if the population density around a TV tower is 1200 people per square kilometer, and you already know the area the tower covers, you use the density to approximate the total population that could receive the transmission. Calculating with density provides a realistic estimate of how many individuals will benefit from the broadcast.

Here's how it works:
  • Calculate the area covered by the transmission using the Line of Sight distance.
  • Multiply the area by the population density.
  • Get the total population potentially covered by the broadcast.
This approach ensures that the tower's reach is not only a function of distance but also accounts for how densely people are distributed around that area.
Earth's Radius
The Earth's Radius is a constant value used in various geographical and signal coverage calculations, including determining the Line of Sight Distance for a TV transmission tower. The average radius of Earth is approximately 6371 km, which is a standardized measure used to simplify various complex calculations.

This radius plays a pivotal role in calculating the potential reach of signals over the curvature of the Earth. When you know the height of the tower and the radius of the earth, applying these in the line of sight formula helps to determine the effective range of the broadcast signal.

The Earth's radius ensures the calculations take into account the planet's curvature, providing a more accurate estimation of how far the TV tower's signal can spread. It essentially corrects the misconception that signals travel in straight lines when, in reality, they bend along with the curvature of the Earth.

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 silicon diode requires a minimum current of \(2 \mathrm{~mA}\) to be above the knee point \((0.7 \mathrm{~V})\) of its \(\mathrm{V}-1\) characteristics. Assume that the voltage across the diode is independent of current above the knee point. Find the maximum value of \(R\) so that voltage across the diode is above the knee point. (A) \(2.25 \Omega\) (B) \(4.65 \Omega\) (C) \(6.25 \Omega\) (D) \(2.65 \Omega\)

Read the following questions and choose (A) If both Assertion and Reason are true and Reason is the correct explanation of the Assertion. (B) If both Assertion and reason are true but Reason is not correct explanation of the Assertion. (C) If Assertion is true but Reason is false. (D) If Assertion is false but Reason is true. Assertion: An ideal junction diode offers zero resistance, when forward biased and infinite resistance, when reverse biased. Reason: Resistance is low when forward biased and high when reverse biased. Ideal of low is zero and ideal of high is infinity. (A) \(\mathrm{A}\) (B) \(\mathrm{B}\) (C) \(\mathrm{C}\) (D) D

The maximum efficiency of full wave rectifier is (A) \(100 \%\) (B) \(25.20 \%\) (C) \(40.2 \%\) (D) \(81.2 \%\)

Read the following questions and choose (A) If both Assertion and Reason are true and Reason is the correct explanation of the Assertion. (B) If both Assertion and reason are true but Reason is not correct explanation of the Assertion. (C) If Assertion is true but Reason is false. (D) If Assertion is false but Reason is true. Assertion: The temperature coefficient of resistance is positive for metals and negative for p-type semiconductor. Reason: The effective charge carriers in metals are negatively charged whereas in p-type semiconductors they are positively charged. (A) \(\mathrm{A}\) (B) \(\mathrm{B}\) (C) \(\mathrm{C}\) (D) D

A solid which is transparent to visible light and whose conductivity increases with temperature is formed by (A) Metallic binding (B) Ionic binding (C) Covalent binding (D) Vander Waals binding

See all solutions

Recommended explanations on Physics Textbooks

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.