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3 years ago

5

Viscous effects are negligible outside of the hydrodynamic boundary layer. (3 points) a. True b. False

1 answer:

Valentin [98]3 years ago

8 0

Answer:

I would say false but I am not for sure

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Who can use NIST resources?

sukhopar [10]

Answer:

Federal agencies

Explanation:

NIST (National Institute of Standards and Technology) also called between 1901 and 1988 National Bureau of Standards (NBS), it is an agency of the Technology Administration of the United States Department of Commerce. The mission of this institute is to promote innovation and industrial competition in the United States through advances in metrology, standards and technology in ways that improve economic stability and quality of life.

As part of this mission, NIST scientists and engineers continually refine the science of measurement (metrology) by creating precise engineering and manufacturing required for most current technological advances. They are also directly involved in the development and testing of standards made by the private sector and government agencies. The NIST was originally called the National Bureau of Standards (NBS), a name it had from 1901 to 1988. The progress and technological innovation of the United States depends on the abilities of the NIST, especially if we talk about four areas: biotechnology , nanotechnology, information technologies and advanced manufacturing.

7 0

3 years ago

A 3-phase , 1MVA, 13.8kV/4160V, 60 Hz, transformer with Y-Delta winding connection is supplying a3-phase, 0.75 p.u. load on the

Tanya [424]

Answer:

a) 23.89 < -25.84 Ω

b) 31.38 < 25.84 A

c) 0.9323 leading

Explanation:

A) Calculate the load Impedance

current on load side = 0.75 p.u

power factor angle = 25.84

$I_{load}$ = 0.75 < 25.84°

attached below is the remaining part of the solution

<u>B) Find the input current on the primary side in real units </u>

load current in primary = 31.38 < 25.84 A

<u>C) find the input power factor </u>

power factor = 0.9323 leading

<em></em>

<em>attached below is the detailed solution </em>

8 0

3 years ago

A NC drill press is to perform a series of through-hole drilling operations on a 1.75 in thick aluminum plate that is a componen

jekas [21]

Answer:

26.7 min

Explanation:

First, we will find the <u>time required to drill each hole</u>:

N = 300 x 12/0.75 $\pi$ = 1527.7 rev/min

fr = 1527.7 (0.015) = 22.916 in/min

Formula for <u>distance per hole</u>: 0.5 + A + 1.75

A = 0.5 (0.75) tan (90-100 / 2) = 0.315 in
Tm = (0.5 + 0.315 + 1.75) / 22.916 = 0.112 min

Now, we will calculate the <u>time required to draw back the drill form hole</u>:

= 0.112 / 2 = 0.056 min

Time to move between holes = 1.5 / 15 = 0.1 min

For 100 holes, the number of moves between holes = 99

Total time required to drill 100 holes (t):

t = 100 (0.112 + 0.056) + 99 (0.1) = 26.7 min

7 0

3 years ago

Read 2 more answers

The time to failure for a gasket follows the Weibull distribution with ß = 2.0 and a characteristic life of 300 days. What is th

Aleks04 [339]

Answer:

64.11% for 200 days.

t=67.74 days for R=95%.

t=97.2 days for R=90%.

Explanation:

Given that

β=2

Characteristics life(scale parameter α)=300 days

We know that Reliability function for Weibull distribution is given as follows

$R(t)=e^{-\left(\dfrac{t}{\alpha}\right)^\beta}$

Given that t= 200 days

$R(200)=e^{-\left(\dfrac{200}{300}\right)^2}$

R(200)=0.6411

So the reliability at 200 days 64.11%.

When R=95 %

$0.95=e^{-\left(\dfrac{t}{300}\right)^2}$

by solving above equation t=67.74 days

When R=90 %

$0.90=e^{-\left(\dfrac{t}{300}\right)^2}$

by solving above equation t=97.2 days

7 0

3 years ago

A westbound section of freeway currently has three 12-ft wide lanes, a 6-ft right shoulder, and no ramps within 3 miles upstream

iogann1982 [59]

Answer:

The level of the service is loss and the density is 34.2248 pc/mi/ln

Explanation:

the solution is attached in the Word file

6 0

3 years ago