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3 years ago

If you touch a downed power line, covered or bare, what's the likely outcome?

Engineering

1 answer:

muminat3 years ago

5 0

You are going to die most likely from being electrocuted

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Consider the velocity boundary layer profile for flow over u flat plate to be of the form u = C_1 + C_2 y. Applying appropriate

ra1l [238]

Answer:

The result in terms of the local Reynolds number ⇒ Re = [μ_∞ · x] / v

Explanation:

See below my full workings so you can compare the results with those obtained from the exact solution.

4 0

4 years ago

Compute the theoretical density of ZnS given that the Zn-S distance and bond angle are 0.234 nm and 109.5o, respectively. The at

andriy [413]

Answer: the theoretical density is 4.1109 g/cm³

Explanation:

first the image of one set of ZnS bonding in the crystal structure, we calculate the value of angle θ

θ + ∅ + 90° = 180°

θ = 90° - ∅

θ = 90° - ( 109.5° / 2 )

θ = 35.25°

next we calculate the value of x from the geometry

given that; distance angle d = 0.234

x = dsinθ

= 0.234 × sin35.25°)

= 0.135 nm = 0.135 × 10⁻⁷ cm

next we calculate the length of the unit cell

a = 4x

a = 4(0.135)

a = 0.54 nm = 0.54 × 10⁻⁷ cm

next we calculate number of formula units

n' = (no of corner atoms in unit ell × contribution of each corner atom in unit cell) + ( no of face center atom in a unit cell × contribution of each face center atom in a unit cell)

n' = 8 × 1/8) + ( 6 × 1/2)

= 1 + 3

= 4

next we calculate the theoretical density using this equation

P = [n'∑(Ac + AA)] / [Vc.NA]

= [n'∑(Ac + AA)] / [(a)³NA]

where the ∑Ac is sum of atomic weights of all cations in the formula unit( 65.41 g/mol)

∑AA is the sum of weights of all anions in the formula unit( 32.06 g/mol)

Na is the Avogadro’s number( 6.023 × 10²³ units/mole)

so we substitute

P = [4( 65.41 + 32.06)] / [ ( 0.54 × 10⁻⁷ )³ × (6.023 × 10²³)]

= 389.88 / 94.84

= 4.1109 g/cm³

therefore the theoretical density is 4.1109 g/cm³

5 0

3 years ago

In order to defend against side channel power analysis, we should: ______________

wariber [46]

Answer:

Some examples of predator and prey are lion and zebra, bear and fish, and fox and rabbit. ... The words "predator" and "prey" are almost always used to mean only animals that eat animals, but the same concept also applies to plants: Bear and berry, rabbit and lettuce, grasshopper and leaf

Explanation:

8 0

3 years ago

Problem 9.11 A structural component in the form of a wide plate is to be fabricated from a steel alloy that has a plane strain f

BabaBlast [244]

Answer:

the critical flaw length is 10.06 mm

Explanation:

Given the data in the question;

plane strain fracture toughness $K_{tc$ = 92 Mpa√m

yield strength σ $_y$ = 900 Mpa

design stress is one-half of the yield strength ( 900 Mpa / 2 ) 450 Mpa

Y = 1.15

we know that;

Critical crack length $a_c$ = 1/π( $K_{tc$ / Yσ )²

we substitute

$a_c$ = 1/π( 92 Mpa√m / (1.15 × 450 Mpa )²

$a_c$ = 1/π( 92 Mpa√m / (517.5 Mpa )²

$a_c$ = 1/π( 0.177777 )²

$a_c$ = 1/π( 0.03160466 )

$a_c$ = 0.01006 m = 10.06 mm

Therefore, the critical flaw length is 10.06 mm

{ $a_c$ = ( 10.06 mm ) > 3 mm

The critical flow is subject to detection

5 0

3 years ago

A square aluminum plate 5 mm thick and 150 mm on a side is heated while vertically suspended in quiescent air at 75°c. determine

Doss [256]

By using the boundary layer equation, the average heat transfer coefficient for the plate is equal to 4.87 W/m²k.

<u>Given the following data:</u>

Surface temperature = 15°C

Bulk temperature = 75°C

Side length of plate = 150 mm to m = 0.15 meter.

<h3>How to calculate the average heat transfer coefficient.</h3>

Since we have a quiescent room air and a uniform pole surface temperature, the film temperature is given by:

$T_f=\frac{T_{s} + T_{\infty} }{2} \\\\T_f=\frac{15 + 75 }{2} \\\\T_f = 45$

Film temperature = 45°C to K = 273 + 45 = 318 K.

For the coefficient of thermal expansion, we have:

$\beta =\frac{1}{T_f} \\\\\beta =\frac{1}{318}$

From table A-9, the properties of air at a pressure of 1 atm and temperature of 45°C are:

Kinematic viscosity, v = $1.750 \times 10^{-5}$ m²/s.

Thermal conductivity, k = 0.02699 W/mk.

Thermal diffusivity, α = $2.416 \times 10^{-5}$ m²/s.

Prandtl number, Pr = 0.7241.

Next, we would solve for the Rayleigh number to enable us determine the heat transfer coefficient by using the boundary layer equations:

$R_{aL}=\frac{g\beta \Delta T l^3}{v\alpha } \\\\R_{aL}=\frac{9.8 \;\times \;\frac{1}{318} \;\times \;(75-15) \;\times \;0.15^3 }{1.750 \times 10^{-5}\; \times \;2.416 \times 10^{-5} } \\\\R_{aL}=\frac{9.8\; \times 0.00315 \;\times \;60\; \times\; 0.003375 }{4.228 \times 10^{-10} }\\\\R_{aL}=1.48 \times 10^{7}$

Also take note, g(Pr) is given by this equation:

$g(P_r)=\frac{0.75P_r}{[0.609 \;+\;1.221\sqrt{P_r}\; +\;1.238P_r]^\frac{1}{4} } \\\\g(P_r)=\frac{0.75(0.7241)}{[0.609 \;+\;1.221\sqrt{0.7241}\; +\;1.238(0.7241)]^\frac{1}{4} }\\\\g(P_r)=\frac{0.543075}{[0.609 \;+\;1.221\sqrt{0.7241}\; +\;1.238(0.7241)]^\frac{1}{4} }\\\\g(P_r)=\frac{0.543075}{[2.5444]^\frac{1}{4} }\\\\g(P_r)=\frac{0.543075}{1.2630 }$

g(Pr) = 0.430

For GrL, we have:

$G_{rL}=\frac{R_{aL}}{P_r} \\\\G_{rL}=\frac{1.48 \times 10^7}{0.7241} \\\\G_{rL}=1.99 \times 10^7$

Since the Rayleigh number is less than 10⁹, the flow is laminar and the condition is given by:

$N_{uL}=\frac{h_{L}L}{k} = \frac{4}{3} (\frac{G_{rL}}{4} )^\frac{1}{4} g(P_r)\\\\h_{L}=\frac{0.02699}{0.15} \times [\frac{4}{3} \times (\frac{1.99 \times 10^7}{4} )^\frac{1}{4} ]\times 0.430\\\\h_{L}= 0.1799 \times 62.9705 \times 0.430\\\\h_{L}=4.87\;W/m^2k$

Based on empirical correlation method, the average heat transfer coefficient for the plate is given by this equation:

$N_{uL}=\frac{h_{L}L}{k} =0.68 + \frac{0.670 R_{aL}^\frac{1}{4}}{[1+(\frac{0.492}{P_r})^\frac{9}{16}]^\frac{4}{19} } \\\\h_{L}=\frac{0.02699}{0.15} \times ( 0.68 + \frac{0.670 (1.48 \times 10^7)^\frac{1}{4}}{[1+(\frac{0.492}{0.7241})^\frac{9}{16}]^\frac{4}{19} })\\\\h_{L}=4.87\;W/m^2k$

Read more on heat transfer here: brainly.com/question/10119413

3 0

2 years ago