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4 years ago

15

Consider the medium of air as defined by the use of radio frequency. What made some of the early standards so slow compared to t

oday's 802.11n 300Mbps and higher bandwidth

1 answer:

Tanzania [10]4 years ago

3 0

Answer:

Explanation:

Earliest standards were dependent on a single frequency/channel to both send and receive. This shared medium creates the same problem as half-duplex coax cable. Because receivers had to wait for the signal before sending a response, this reduced the overall bandwidth.

Other factors affect wireless signal propagation, too, including RF interference, antenna choice, and obstacles such as walls, trees, and even weather (precipitation, for example).

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A stereo speaker is rated at P1000 = 52 W of output at 1000 Hz. At 20 Hz, the sound intensity level LaTeX: \betaβ decreases by 1

baherus [9]

Answer:

The value of the power is $P_c = 38.55 \ W$

Explanation:

From the question we are told that

The power rating $P_{1000} =P_b= 52 \ W$

The frequency is $f = 1000 \ Hz$

The frequency at which the sound intensity decreases $f_k = 20 \ Hz$

The decrease in intensity is by $\beta = 1.3 dB$

Generally the initial intensity of the speaker is mathematically represented as

$\beta_1 = 10 log_{10} [\frac{P_b}{P_a} ]$

Generally the intensity of the speaker after it has been decreased is

$\beta_2 = 10 log_{10} [\frac{P_c}{P_a} ]$

So

$\beta_1-\beta_2 = 10 log_{10} [\frac{P_c}{P_a} ]- 10 log_{10} [\frac{P_b}{P_a} ]$

=> $\beta = 10 log_{10} [\frac{P_c}{P_a} ]- 10 log_{10} [\frac{P_b}{P_a} ]= 1.3$

=> $\beta =10log_{10} [\frac{\frac{P_b}{P_a}}{\frac{P_c}{P_a}} ] = 1.3$

=> $\beta =10log_{10} [\frac{P_b}{P_c} ] = 1.3$

=> $10log_{10} [\frac{P_b}{P_c} ] = 1.3$

=> $log_{10} [\frac{P_b}{P_c} ] = 0.13$

taking atilog of both sides

$[\frac{P_b}{P_c} ] = 10^{0.13}$

=> $[\frac{52}{P_c} ] = 10^{0.13}$

=> $P_c = \frac{52}{1.34896}$

=> $P_c = 38.55 \ W$

3 0

3 years ago

You are presented with several wires made of the same conducting material. The radius and drift speed are given for each wire in

jekas [21]

$\begin{array}{ccc}&\text{Radius} & \text{Drift Speed}\\d) & 2 \; r &2.5 \; v\\a) & 3 \; r &1 \; v\\b) & 4 \; r &0.5 \; v\\c) & 1 \; r &5 \; v\\\end{array}$

<h3>Explanation</h3>

$I = v \cdot A \cdot n \cdot q$ ,

where

$I$ is the current;
$v$ is the drift speed;
$A$ is the cross-section area of the wire,
$n$ is the number of charge carrier per unit volume, and
$q$ is the charge on each charge carrier.

Area of a circular cross-section:

$A = \pi \cdot r^{2}$ ,

where

$r$ is the radius of the wire.

$n$ and $q$ are the same for all four samples, for they are made out of the same material.

As a result, $I$ of each wire is directly proportional to $v \cdot r^{2}$ where the value of $\pi \cdot n \cdot q$ is constant.

For each of the four wires:

$\begin{array}{ccc|c}\\& r & v &I \propto v\cdot r^{2}\\a) & 3 & 1 & 9\\b) & 4 & 0.5 & 8\\c) & 1 & 5 & 5\\d) & 2 & 2.5 & 10\\\end{array}$ .

How do the four wires rank by their current?

d > a > b > c.

3 0

4 years ago

What is represented by the slope of this graph

Paha777 [63]

Answer:

D

Explanation:

on the left u see it says velocity

6 0

3 years ago

What type of tissue in the heart pumps blood throughout the body?

slavikrds [6]

Answer:

Myocardium. That is the type. (srry i was in a rush hope this helps)

7 0

3 years ago

A box with the mass of 20 kg at 5 m is lifted to 20 m. How much work was 7 points<br> done?

kirza4 [7]

Answer:

$2,900\: \mathrm{J}$

Explanation:

Work is given by the equation $W=F\Delta x$ where $F$ is force and $\Delta x$ is displacement.

Displacement is defined as change in position. In this case, the box moves from 5m to 20m. Its displacement is $20-5=15\:\mathrm{m}$ .

The force acting on the box is the force of gravity, given as $F_g=mg$ .

Therefore, the total work done is:

$W=F\Delta x=mg\Delta x = 20\cdot 9.81\cdot 15=2,943=\fbox{$2,900\:\mathrm{J}$}$ (two significant figures).

3 0

3 years ago