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Setler79 [48]
2 years ago
10

A master precision square is used to validate the

Engineering
1 answer:
kumpel [21]2 years ago
4 0

Answer:

me you same I didn't understand good luck

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A simply supported wood roof beam is loaded with single point dead and roof live loads applied at midspan (PD = 400 lb, PLr = 16
Lynna [10]

Answer:mold i belive

Explanation:

3 0
3 years ago
The B-pillar may also be called the:
slega [8]

Answer:

if you're talking about the car b-post, the answer is "posts"

Explanation:

looked it up

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2 years ago
A fill covering a wide area is to be placed at the surface of this profile. The fill has a total unit weight of 20 kN/m^3 and is
goblinko [34]

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hello your question lacks some information attached below is the complete question with the required information

answer : 81.63 mm

Explanation:

settlement of the surface due to compression of the clay ( new consolidated )

= 81.63 mm

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3 years ago
Which of the following is NOT true about hydraulic valves? A. Directional control valves determine the path of a fluid in a give
Lelechka [254]

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5 0
3 years ago
Calculate the reluctance of a 4-meter long toroidal coil made of low-carbon steel with an inner radius of 1.75 cm and an outer r
My name is Ann [436]

Answer:

R = 31.9 x 10^(6) At/Wb

So option A is correct

Explanation:

Reluctance is obtained by dividing the length of the magnetic path L by the permeability times the cross-sectional area A

Thus; R = L/μA,

Now from the question,

L = 4m

r_1 = 1.75cm = 0.0175m

r_2 = 2.2cm = 0.022m

So Area will be A_2 - A_1

Thus = π(r_2)² - π(r_1)²

A = π(0.0225)² - π(0.0175)²

A = π[0.0002]

A = 6.28 x 10^(-4) m²

We are given that;

L = 4m

μ_steel = 2 x 10^(-4) Wb/At - m

Thus, reluctance is calculated as;

R = 4/(2 x 10^(-4) x 6.28x 10^(-4))

R = 0.319 x 10^(8) At/Wb

R = 31.9 x 10^(6) At/Wb

8 0
3 years ago
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