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Alenkinab [10]
3 years ago
12

Select a research proposal topic that relates to electrical and electronics engineering and write a proposal report taking into

consideration the following;
Introduction,problem statement,objection,justification,literature review,materials and method,References.​
Engineering
1 answer:
fiasKO [112]3 years ago
8 0

Expand your technical knowledge, form global networks and balance life & work commitments. Our advanced diplomas remain current with technological and industry developments.

You might be interested in
Help please if you don't know don't give wrong answer please​
mrs_skeptik [129]

4.75cm * 5.22cm is 24.795cm²

but let's do this with m² instead:

0.0475m * 0.0522m = 0.0024795

now we can compare it with the 49780 much easier.

devide. that's the <u>roughly</u> 20,000,000fold of the plot, but it's still squared, so let's take the root

so the representative fraction is 1:4,480.

the exact value is in the screens

have a nice day:)

6 0
3 years ago
1 UNREAD MESSAGE
myrzilka [38]

Answer:

triangular trade

Explanation:

triangular trade

7 0
2 years ago
RC4 has a secret internal state which is a permutation of all the possible values of the vector S and the two indices i and j.
Stolb23 [73]

Answer:

a. Using a straight forward scheme, the RC4 algorithm stores 2064 bits in the interval state.

b. 1700 bits

Explanation:

See RC4 Algorithm attached with details

3 0
3 years ago
A wide-sense stationary process X(t) is the input to a linear system whose impulse response is h(t)=2e^(-7t), t≥0.
zimovet [89]

Answer:

The power spectral density of Y(t) is Syy(w) = \frac{32}{(4+w^2)(16+w^2)}

Explanation:

4 0
3 years ago
: During a heavy rainstorm, water from a parking lot completely fills an 18-in.- diameter, smooth, concrete storm sewer. If the
Montano1993 [528]

Answer

diameter of parking lot = 18 in

flowrate = 10 ft³/s

pressure drop = 100 ft

using general equation

\dfrac{P_1}{\gamma}+\dfrac{v_1^2}{2g}+Z_1 = \dfrac{P_2{\gamma} + \dfrac{v_2^2}{2g} + Z_2 +\dfrac{fLV^2}{2\rho D}

V = \dfrac{Q}{A} = \dfrac{10}{\dfrac{\pi}{4}(\dfrac{18}{12})^2} = 5.66\ ft/s

\Delta P = \gamma (Z_2-Z_1) +\dfrac{fLV^2}{2\rho D}

taking f = 0.0185

at Z₁ = Z₂

\Delta P = \dfrac{0.0185 \times 100\times 1.94\times 5.66^2}{2\dfrac{18}{12} (2)}

ΔP = 0.266 psi

b) when flow is uphill z₂-z₁ = 2

\Delta P =62.4\times 2 \times \dfrac{1}{144} +0.266

\Delta P= 1.13\ psi

c) When flow is downhill  z₂-z₁ = -2

\Delta P =62.4\times 2 \times \dfrac{1}{144} +0.266

\Delta P=-0.601\ psi

7 0
3 years ago
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