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3 years ago

What is Class Night in Puerto Rico?

Engineering

1 answer:

valentina_108 [34]3 years ago

8 0

Answer:

Explanation:

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Need help right now pls

LenaWriter [7]

Put them in 2,4,1,3 and got in 100 in the test

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3 years ago

True False. First angle projection type used in United states.

Sloan [31]

Answer:

FALSE.

Explanation:

the correct answer is FALSE.

Projection is the process of representing the 3 D object on the flat surface.

there are four ways of representing the projection

1) First angle projection

2) second angle projection

3) third angle projection

4) fourth angle projection.

Generally, people prefer First and third angle projection because there is no overlapping of the projection take place.

In USA people uses the third angle of projection.

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3 years ago

If you are exposed to potentially infectious material via a sharps injury, what should you do immediately?

Sergio039 [100]

The correct answer is "immediately flood the area with water and clean the wound."

Further Explanation:

If someone is exposed to a potentially infectious material with a sharps item such a needle, they need to wash the area with clean water immediately. It must be a steady flow of water to get out any potential infectious material.

After flushing the injury with water, you will then wash the area or exposed skin with soap, if at all possible use a disinfectant on the area.

The person who was injured should seek medical assistance for tests to be sure that they have not been affected.

Learn more about infectious material at brainly.com/question/10928952

#LearnwithBrainly

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3 years ago

When using a Hammer I should always keep an eye on may fingers so as not to hit them.?

nlexa [21]

False you should watch where you’re hitting the hammer

6 0

3 years ago

Read 2 more answers

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