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erik [133]
3 years ago
14

A mobile base station in an urban environment has a power measurement of 25 µW at 225 m. If the propagation follows an inverse 4

th-power law, assuming a distance of 0.9 km from the base station, what would be a reasonable power value, in µW?
Give your answer in scientific notation to 2 decimal places
Computers and Technology
1 answer:
Dmitry_Shevchenko [17]3 years ago
6 0

Answer:

The answer is below

Explanation:

The inverse fourth power law states that the intensity varies inversely to the fourth power of the distance from the source. That is as the distance increases, the intensity decreases.

Let I represent the intensity and let d represent the distance from the source, hence:

I ∝ 1 / d⁴

I = k / d⁴

Where k is the constant of proportionality.

Given that at a power of 25W = 25*10⁻⁶ W, the distance (d) = 225 m. We can find the value of k.

25*10⁻⁶  = k / 225⁴

k = 225⁴ * 25*10⁻⁶ = 64072.27

Hence:

I = 64072.27 / d⁴

At a distance (d) = 0.9km = 900 m, we can find the corresponding power:

I = 64072.27 / 900⁴

I = 9.77 * 10⁻⁸ W

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Answer:

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Explanation:

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g write a recursive function that prints out all the even numbers between x and y inclusive if x is odd then the print out will
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Answer:

The function written in python is as follows:

def EvenNumbers(x, y):

    if (y < x):

         return

    if (y % 2 == 0):

         EvenNumbers(x, y - 2)

    else:

         EvenNumbers(x, y - 1)

    if (y % 2 == 0):

         print(y)

Explanation:

x represents the lower limit while y represents the upper limit.

So, the recursion will be repeated as long as x <= y

This defines the function

def EvenNumbers(x, y):

This checks if y < x. If yes, the recursion stops execution and return to the main

<em>     if (y < x): </em>

<em>          return </em>

The following if and else condition checks determines the appropriate range of the recursion.

<em>     if (y % 2 == 0): </em>

<em>          EvenNumbers(x, y - 2) </em>

<em>     else: </em>

<em>          EvenNumbers(x, y - 1) </em>

The above takes care of printing starting at x + 1 if x is odd

This checks and prints the even numbers in the range

<em>     if (y % 2 == 0): </em>

<em>          print(y)</em>

<em />

To call the function from main, make use of

EvenNumbers(x, y)

Where x and y are integers

e.g EvenNumbers(2,10), EvenNumbers(1,17), EvenNumbers(3,21)

<em>See attachment</em>

3 0
3 years ago
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