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2 years ago

The team needs to choose a primary view for the part drawing. Three team members make suggestions:

Engineering

1 answer:

dexar [7]2 years ago

6 0

Answer:

<u>Option 1</u>

Explanation:

As the team has already submitted the plans for the part drawing, the best way to proceed would be how it was given in the plans. Hence, the option to be selected :

<u>Team member 1 suggests an orthographic top view because that is how the plans for the part were submitted.</u>

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Technician A says that gear lube and bearing grease cannot be substituted due to the different housing and seal designs each lub

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Whether a lube and grease are interchangeable depends on the characteristics of each because they come in many different grades.

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3 years ago

Ayden read 84 pages in 2 hours. At that rate, how many pages can he read in 5 hours

Papessa [141]

Answer:210

Explanation:

Divide 84 by 2 and then multiply by 5 by that number

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4 years ago

Read 2 more answers

A device that helps increase field worker productivity by providing reliable location and time

frutty [35]

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<u>Explanation:</u>

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3 years ago

For laminar flow over a flat plate, the local heat transfer coefficient hx is known to vary as x−1/2, where x is the distance fr

Reika [66]

Answer:

Explanation:

So for solving this problem we need the local heat transfer coefficient at distance x,

$h_x=cx^{-1/2}$

We integrate between 0 to x for obtain the value of the coefficient, so $\bar{h}_x =\frac{1}{x} \int\limit^x_0 h_x dx\\\bar{h}_x = \frac{c}{x} \int\limit^x_0 \frac{1}{\sqrt{x}}dx\\\bar{h}_x = \frac{c}{c} (2x^{1/2})\\\bar{h}_x = 2cx^{-1/2}$

Substituing

$\bar{h}_x=2h_x\\\frac{\bar{h}_x}{h_X}=2$

The ratio of the average convection heat transfer coefficient over the entire length is 2

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4 years ago

PLEASE HELP QUICK!!

ivolga24 [154]

R01= 14.1 Ω

R02= 0.03525Ω

<h3>Calculations and Parameters</h3>

Given:

K= E2/E1 = 120/2400

= 0.5

R1= 0.1 Ω, X1= 0.22Ω

R2= 0.035Ω, X2= 0.012Ω

The equivalence resistance as referred to both primary and secondary,

R01= R1 + R2

= R1 + R2/K2

= 0.1 + (0.035/9(0.05)^2)

= 14.1 Ω

R02= R2 + R1

=R2 + K^2.R1

= 0.035 + (0.05)^2 * 0.1

= 0.03525Ω