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2 years ago

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How does error detection take place

1 answer:

sattari [20]2 years ago

6 0

Answer: To detect and correct errors, additional bits are added to the data bits at the time of transmission. The additional bits are called parity bits. They allow detection or correction of errors. The data bits along with the parity bits form a code word.

Explanation:

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Alicia had to get over her fear of heights in order to become comfortable maintaining the generators in wind turbines. professio

kow [346]

Answer:a

Explanation:

5 0

3 years ago

Using the Tsai-Hill failure criterion, determine the strength of a lamina under equal biaxial tension and shear at 45o to the fi

Alinara [238K]

Answer:

Explanation:

solution to the question

5 0

3 years ago

Four eight-ohm speakers are connected in parallel to an audio power amplifier. The amplifier can supply a maximum driver output

Vera_Pavlovna [14]

Answer:

112.5 watts

Explanation:

The output voltage (V) = 15 volts

The equivalent resistance for the speakers connected in parallel ( $R_T$ ) is gotten by using the formula:

$\frac{1}{R_T}=\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}+\frac{1}{R_4}\\\\But\ R_1=R_2=R_3=R_4=8\ ohm\\\\\frac{1}{R_T} =\frac{1}{8} +\frac{1}{8} +\frac{1}{8} +\frac{1}{8} \\\\\frac{1}{R_T} =\frac{1+1+1+1}{8} \\\\\frac{1}{R_T} =\frac{4}{8} \\\\R_T=\frac{8}{4} \\\\R_T=2\ ohm$

The current flowing through the amplifier (I) is:

I = V / $R_T$ = 15 / 2 = 7.5 A

The audio power (P) when outputting this maximum voltage is given by:

P = V² / $R_T$ = I² $R_T$ . Therefore:

P = V² / $R_T$ = 15² / $R_T$ = 112.5 watts

6 0

3 years ago

Water is flowing steadily through a 3-m long, 20 mm innerdiameter cast iron pipe. The water enters at a uniform velocity U and e

devlian [24]

Answer:

Hello your question is incomplete attached below is the complete question

answer :

U/r = R = $U_{max}$ [ 1 - $(\frac{R}{R} )^2$ ] = 0

∴ No slip condition is satisfied

Explanation:

Given that

Outlet velocity U(r) = $U_{max}$ [ 1 - $(\frac{r}{R} )^2$ ]

<u>prove that the output velocity profile satisfies the no slip condition</u>

at no slip u = 0 ( i.e. at the pipe's inner surface )

at , r = R ( inference is at center )

hence U/r = R = $U_{max}$ [ 1 - $(\frac{R}{R} )^2$ ] = 0

∴ No slip condition is satisfied

8 0

3 years ago

Purely resistive loads of 24kW, 18kW and 12kW are connected between the neutral and the red, yellow and the blue respectively of

seraphim [82]

Answer:

The phase current in each line conductor are;

$I_{R} = 100.17 < 0A$

$I_{Y} = 75.13< - 120A$

$I_{B} = 50.08$

Explanation:

Given the following data;

Red phase = 24kW,

Yellow phase = 18kW

Blue phase = 12kW

Line voltage = 415V

For a star connected system, we have;

$Phase voltage (V_{p} ) = \frac{Line voltage}{\sqrt{3}}$

$Phase voltage (V_{p} ) = \frac{415}{\sqrt{3}}$

$Phase voltage (V_{p} ) = 239.6V$

The phase sequence for RYB is given by;

$V_{R} = 239.6$

$Phase current (I) = \frac{Phase power}{Phase voltage}$

$Hence, I = \frac{P}{V}$

<em>For the Red phase;</em>

$I_{R} = \frac{24000}{239.6$

$I_{R} = 100.17 < 0A$

<em>For the Yellow phase;</em>

$I_{Y} = \frac{18000}{239.6$

$I_{Y} = 75.13< - 120A$

<em>For the Blue phase;</em>

$I_{B} = \frac{12000}{239.6$

$I_{B} = 50.08$

For the line neutral;

$I_{N} =\sqrt{ (I_{R}^{2} +I_{Y}^{2}+I_{B}^{2}-I_{R}I_{Y}-I_{Y}I_{B}-I_{R}I_{B}$

Substituting we have, $I_{N} = 43.29A$

6 0

3 years ago