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ikadub [295]
3 years ago
6

Just to let you know Christmas is in 10 days<3 lol

Engineering
2 answers:
Harrizon [31]3 years ago
7 0

Answer:

yay yay

Explanation:

im so excited i cant wait

Annette [7]3 years ago
7 0

Answer:

Thanks

Explanation:

for reminding us

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Use a passband of 0 to 5 kHz with 5 kHz cutoff frequency and filter to attenuate all frequencies at and above 10 kHz by at least
Alinara [238K]

Answer:

See attached picture for answer.

Explanation:

See attached picture for explanation.

5 0
3 years ago
A thin aluminum sheet is placed between two very large parallel plates that are maintained at uniform temperatures T1 = 900 K, T
Maru [420]

The net radiation heat transfer between the two plates per unit surface area of the plates with shield and without shied are respectively; 2282.76 W/m² and 9766.75 W/m²

<h3>How to find the net radiation heat transfer?</h3>

We are given;

Temperature 1; T₁

Temperature 2; T₂

Temperature 3; T₃

Emissivity 1; ε₁ = 0.3

Emissivity 2; ε₂ = 0.7

Emissivity 3; ε₃ = 0.2

The net rate of radiation heat transfer with a thin aluminum shield per unit area of the plates with shield is;

Q'₁₂ = σ(T₁⁴ - T₂⁴)]/[((1/ε₁) + (1/ε₂) - 1) + ((1/ε₃,₁) + (1/ε₃,₂) - 1)]

Q'₁₂ = 5.67 * 10⁻⁸(900⁴ - 300⁴)/[((1/0.3) + (1/0.7) - 1) + ((1/0.15) + (1/0.15) - 1)]

Q'₁₂,shield = 2282.76 W/m²

The net rate of radiation heat transfer with a thin aluminum shield per unit area of the plates with no shield is;

Q'₁₂,no shield = σ(T₁⁴ - T₂⁴)]/((1/ε₁) + (1/ε₂) - 1))

Q'₁₂,no shield = 5.67 * 10⁻⁸(900⁴ - 300⁴)/[(1/0.3) + (1/0.7) - 1)]

Q'₁₂,no shield = 9766.75 W/m²

Then the ratio of radiation heat transfer for the two cases becomes;

Q'₁₂,shield/Q'₁₂,no shield = 2282.76/9766.75 = 0.2337 or 4/17

Read more about Net Radiation Heat Transfer at; brainly.com/question/14148915

#SPJ1

8 0
2 years ago
Two vertical parallel plates are spaced 0.01 ft apart. If the pressure decreases at a rate of 60 psf/ft in the vertical z-direct
Whitepunk [10]

Answer:

umax = 0.1259ft/s

Explanation:

Given:

•Distance between plates, B = 0.01ft

•Pressure difference decrease, \frac{dp}{dz}=60ps/ft

•Fluid viscosity, u = 10^-³lbf-s/ft²

Specific gravity, S = 0.80

Max velocity in the z-direction will be:

u_max= [\frac{B^2y}{8u}]\frac{dh}{ds}

But h = \frac{P}{y}+z

Substituting for h in the first equation, we have:

\frac{d}{dz}[\frac{p}{y}+z]

\frac{dh}{dz}=\frac{1}{y}\frac{dp}{ds}+\frac{dz}{dz}

= \frac{1}{0.8*62.4}(-60)+1

= -0.20192

Substituting dh/dz value in the first equation (umax), we have:

umax = \frac{0.01^2(0.8*62.4)}{8*10^-^3}(-0.20192)

umax = 0.1259ft/s

4 0
3 years ago
A ladder logic program and the associated physical input/output components are given below. Lighting changes from full darkness
Katena32 [7]

Answer:

O2 is true.

Explanation:

8 0
3 years ago
I don't know help me​
sergey [27]

Answer:

for...?

thenks for the points :))

4 0
3 years ago
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