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3 years ago

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1. In an assembly two flanges are held together by a 1/2"" bolt that is threaded into one of the flanges. (a) If the clearance h

ole in one of the flanges used a 33/64"" drill what positional tolerance should be used for the flanges? (b) If a nut and a bolt are used and both flanges have a 33/64"" clearance hole what positional tolerance should be used?

1 answer:

creativ13 [48]3 years ago

5 0

Answer:

The solution and complete explanation for the above question and mentioned conditions is given below in the attached document.i hope my explanation will help you in understanding this particular question.

Explanation:

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Document the XSS stored exploit script: Use the View Source feature of the web page and create a screenshot of the few lines cod

Natali [406]

Answer:

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Explanation:

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2 years ago

When a circuit has more resistance, what happens to current flow ?

zloy xaker [14]

The relationship between resistance and the area of the cross section of a wire is inversely proportional . When resistance is increased in a circuit , for example by adding more electrical components , the current decreases as a result.

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2 years ago

An aggregate blend is composed of 55% aggregate A (Sp. Gr. 2.631), 25% aggregate B (Sp. Gr. 2.331) and 20% sand (Sp. Gr. 2.609).

Phoenix [80]

Answer:

The right choice would be Option b (2.545).

Explanation:

The given values are:

The aggregate blend will be:

$W_A$ = 55%

$G_ A$ = 2.631

$W_ B$ = 25%

$G_B$ = 2.331

$W_C$ = 20%

$G_C$ = 2.609

Now,

On applying the formula, we get

⇒ $G_{BA}=\frac{W_A+W_B+W_C}{\frac{W_A}{G_A} +\frac{W_B}{G_B} +\frac{W_C}{G_C}}$

On substituting the values, we get

⇒ $=\frac{55+25+20}{\frac{55}{2.631} +\frac{25}{2.331} +\frac{20}{2.609}}$

⇒ $= \frac{100}{\frac{55}{2.631} +\frac{25}{2.331} +\frac{20}{2.609}}$

⇒ $=2.545$

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2 years ago

A large plate is fabricated from a steel alloy that has a plane strain fracture toughness of 55 MPa √m (50 ksi √in.). If, during

astra-53 [7]

Answer:

0.024 m = 24.07 mm

Explanation:

1) Notation

$\sigma_c$ = tensile stress = 200 Mpa

$K$ = plane strain fracture toughness= 55 Mpa $\sqrt{m}$

$\lambda$ = length of a surface crack (Variable of interest)

2) Definition and Formulas

The Tensile strength is the ability of a material to withstand a pulling force. It is customarily measured in units (F/A), like the pressure. Is an important concept in engineering, especially in the fields of materials and structural engineering.

By definition we have the following formula for the tensile stress:

$\sigma_c=\frac{K}{Y\sqrt{\pi\lambda}}$ (1)

We are interested on the minimum length of a surface that will lead to a fracture, so we need to solve for $\lambda$

Multiplying both sides of equation (1) by $Y\sqrt{\pi\lambda}$

$\sigma_c Y\sqrt{\pi\lambda}=K$ (2)

Sequaring both sides of equation (2):

$(\sigma_c Y\sqrt{\pi\lambda})^2=(K)^2$

$\sigma^2_c Y^2 \pi\lambda=K^2$ (3)

Dividing both sides by $\sigma^2_c Y^2 \pi$ we got:

$\lambda=\frac{1}{\pi}[\frac{K}{Y\sigma_c}]^2$ (4)

Replacing the values into equation (4) we got:

$\lambda=\frac{1}{\pi}[\frac{55 Mpa\sqrt{m}}{1.0(200Mpa)}]^2 =0.02407m$

3) Final solution

So the minimum length of a surface crack that will lead to fracture, would be 24.07 mm or more.

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3 years ago

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Explanation:

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