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expeople1 [14]
3 years ago
9

Consider a 6-bit cyclic redundancy check (CRC) generator, G = 100101, and suppose that D = 1000100100. 1. What is the value of R

? (10 pts)
Engineering
1 answer:
dmitriy555 [2]3 years ago
3 0

Answer:

The value of R is 10101

Explanation:

As per the given data

D = 1000100100

G = 100101

Redundant bit = 6-bits - 1-bit = 5-bits

No add fice zero to D

D = 100010010000000

Now calculate R as follow

R = D / G

R = 100010010000000 / 100101

R = 10101

Workings are attached with this question

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Answer:

The frequency that the sampling system will generate in its output is 70 Hz

Explanation:

Given;

F = 190 Hz

Fs = 120 Hz

Output Frequency = F - nFs

When n = 1

Output Frequency = 190 - 120 = 70 Hz

Therefore, if a system samples a sinusoid of frequency 190 Hz at a rate of 120 Hz and writes the sampled signal to its output without further modification, the frequency that the sampling system will generate in its output is 70 Hz

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I need help on the Coderz Challenge missions 3 part 3. PLEASE HELP!
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Answer:

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A hot air balloon is used as an air-vehicle to carry passengers. It is assumed that this balloon is sealed and has a spherical s
monitta

Answer:

a. \dfrac{D_{1}}{ D_{2}}  =  \left (\dfrac{   \left{D_1}  }{ {D_2}}   \right )^{-3\times n} which is constant therefore, n = constant

b. The temperature at the end of the process is 109.6°C

c. The work done by the balloon boundaries = 10.81 MJ

The work done on the surrounding atmospheric air = 10.6 MJ

Explanation:

p₁ = 100 kPa

T₁ = 27°C

D₁ = 10 m

v₂ = 1.2 × v₁

p ∝ α·D

α = Constant

v_1 = \dfrac{4}{3} \times  \pi \times r^3

\therefore v_1 = \dfrac{4}{3} \times  \pi \times  \left (\dfrac{10}{2}  \right )^3 = 523.6 \ m^3

v₂ = 1.2 × v₁ = 1.2 × 523.6 = 628.32 m³

Therefore, D₂ = 10.63 m

We check the following relation for a polytropic process;

\dfrac{p_{1}}{p_{2}} = \left (\dfrac{V_{2}}{V_{1}}   \right )^{n} = \left (\dfrac{T_{1}}{T_{2}}   \right )^{\dfrac{n}{n-1}}

We have;

\dfrac{\alpha \times D_{1}}{\alpha \times D_{2}} = \left (\dfrac{ \dfrac{4}{3} \times  \pi \times  \left (\dfrac{D_2}{2}  \right )^3}{\dfrac{4}{3} \times  \pi \times  \left (\dfrac{D_1}{2}  \right )^3}   \right )^{n} = \left (\dfrac{   \left{D_2}  ^3}{ {D_1}^3}   \right )^{n}

\dfrac{D_{1}}{ D_{2}} = \left (\dfrac{   \left{D_2}  }{ {D_1}}   \right )^{3\times n} =  \left (\dfrac{   \left{D_1}  }{ {D_2}}   \right )^{-3\times n}

\dfrac{ D_{1}}{ D_{2}} = \left ( 1.2  \right )^{n} = \left (\dfrac{   \left{D_2}  ^3}{ {D_1}^3}   \right )^{n}

log  \left (\dfrac{D_{1}}{ D_{2}}\right )  =  -3\times n \times log\left (\dfrac{   \left{D_1}  }{ {D_2}}   \right )

n = -1/3

Therefore, the relation, pVⁿ = Constant

b. The temperature T₂ is found as follows;

\left (\dfrac{628.32 }{523.6}   \right )^{-\dfrac{1}{3} } = \left (\dfrac{300.15}{T_{2}}   \right )^{\dfrac{-\dfrac{1}{3}}{-\dfrac{1}{3}-1}} = \left (\dfrac{300.15}{T_{2}}   \right )^{\dfrac{1}{4}}

T₂ = 300.15/0.784 = 382.75 K = 109.6°C

c. W_{pdv} = \dfrac{p_1 \times v_1 -p_2 \times v_2 }{n-1}

p_2 = \dfrac{p_{1}}{ \left (\dfrac{V_{2}}{V_{1}}   \right )^{n} } =  \dfrac{100\times 10^3}{ \left (1.2) \right  ^{-\dfrac{1}{3} } }

p₂ =  100000/0.941 = 106.265 kPa

W_{pdv} = \dfrac{100 \times 10^3 \times 523.6 -106.265 \times 10^3  \times 628.32 }{-\dfrac{1}{3} -1} = 10806697.1433 \ J

The work done by the balloon boundaries = 10.81 MJ

Work done against atmospheric pressure, Pₐ, is given by the relation;

Pₐ × (V₂ - V₁) = 1.01×10⁵×(628.32 - 523.6) = 10576695.3 J

The work done on the surrounding atmospheric air = 10.6 MJ

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3 years ago
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Is this a question or a statement?

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III. During January, at a location in Alaska winds at −27°C can be observed. However, several meters below ground the temperatur
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Answer:

Not possible.

Explanation:

According to second law of thermodynamics, the maximum efficiency any heat engine could achieve is Carnot Efficiency η defined by:

\eta=1-\frac{T_{cold}}{T_{hot}}

Where

T_{hot} and T_{cold} are temperature (in Kelvin) of heat source and heatsink respectively

In our case (I will be using K = 273+°C) :

\eta=1-\frac{-27+273}{14+273}\\=0.1428

In percentage, this is 14.28% efficiency, which is the <em>maximum</em> theoretical efficiency <em>any</em> heat engine could have while working between -27 and 14 °C temperature. Any claim of more efficient heat engine between these 2 temperature are violates the second law of thermodynamics. Therefore, the claim must be false.

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3 years ago
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