Please look at the attachments and help me with these questions

1. Consider a city of 10 square kilometers. A macro cellular system design divides the city up into square cells of 1 square kil

kakasveta [241]

Answer:

a) $n = 1000\,users$ , b) $\Delta t_{min} = \frac{1}{30}\,h$ , $\Delta t_{max} = \frac{\sqrt{2} }{30}\,h$ , $\Delta t_{mean} = \frac{1 + \sqrt{2} }{60}\,h$ , c) $n = 10000000\,users$ , $\Delta t_{min} = \frac{1}{3000}\,h$ , $\Delta t_{max} = \frac{\sqrt{2} }{3000}\,h$ , $\Delta t_{mean} = \frac{1 + \sqrt{2} }{6000}\,h$

Explanation:

a) The total number of users that can be accomodated in the system is:

$n = \frac{10\,km^{2}}{1\,\frac{km^{2}}{cell} }\cdot (100\,\frac{users}{cell} )$

$n = 1000\,users$

b) The length of the side of each cell is:

$l = \sqrt{1\,km^{2}}$

$l = 1\,km$

Minimum time for traversing a cell is:

$\Delta t_{min} = \frac{l}{v}$

$\Delta t_{min} = \frac{1\,km}{30\,\frac{km}{h} }$

$\Delta t_{min} = \frac{1}{30}\,h$

The maximum time for traversing a cell is:

$\Delta t_{max} = \frac{\sqrt{2}\cdot l }{v}$

$\Delta t_{max} = \frac{\sqrt{2} }{30}\,h$

The approximate time is giving by the average of minimum and maximum times:

$\Delta t_{mean} = \frac{1+\sqrt{2} }{2}\cdot\frac{l}{v}$

$\Delta t_{mean} = \frac{1 + \sqrt{2} }{60}\,h$

c) The total number of users that can be accomodated in the system is:

$n = \frac{10\times 10^{6}\,m^{2}}{100\,m^{2}}\cdot (100\,\frac{users}{cell} )$

$n = 10000000\,users$

The length of each side of the cell is:

$l = \sqrt{100\,m^{2}}$

$l = 10\,m$

Minimum time for traversing a cell is:

$\Delta t_{min} = \frac{l}{v}$

$\Delta t_{min} = \frac{0.01\,km}{30\,\frac{km}{h} }$

$\Delta t_{min} = \frac{1}{3000}\,h$

The maximum time for traversing a cell is:

$\Delta t_{max} = \frac{\sqrt{2}\cdot l }{v}$

$\Delta t_{max} = \frac{\sqrt{2} }{3000}\,h$

The approximate time is giving by the average of minimum and maximum times:

$\Delta t_{mean} = \frac{1+\sqrt{2} }{2}\cdot\frac{l}{v}$

$\Delta t_{mean} = \frac{1 + \sqrt{2} }{6000}\,h$

8 0

3 years ago

A flow rate sensing device used on a liquid transport pipeline functions as follows. The device provides a 5-bit output where al

marysya [2.9K]

Answer:

Explanation:

The step by step analysis is as shown in the attached files.

8 0

3 years ago

This code initially states that n=10 and f=n. Once the code enters the while loop, it stays in the while loop until the conditio

zavuch27 [327]

Answer:

n = 10;
f = n;
while(n > 1)
n = n - 1;
f = f * n;
end
disp(f);

Explanation:

The solution code is written in Matlab.

Firstly, we initialize n to 10 and set n to f variable (Line 1-2).

Next, create a while loop that will continue to loop until n is equal or smaller than one. In the loop, decrement n by one and multiply n with current f_variable.

At last display value of f. The output is 3628800.

6 0

3 years ago

PLSSSSS Help !!!!!!!!!!!. It's due today. <br> I will give brainliest to correct answer

4vir4ik [10]

Answer:

b

Explanation:

3 0

3 years ago

Read 2 more answers

Why exam is important in our life

qwelly [4]

Answer:

I am also finding this ans but it is difficult

Explanation:

well to live in this world exam r important

we have to come with every difficulty

so that we can learn

6 0

2 years ago

Read 2 more answers