Ask question

Ask question

All categories

2 years ago

5

determine the moments of inertia of the quartercircular area about the x- and y- axes, and find the polar radius of gyration abo

ut point o.

1 answer:

ASHA 777 [7]2 years ago

7 0

I can’t take a full photo of step one so sorry but this is the step 2 and step 3 with the final answer

You might be interested in

Which Finance jobs can someone pursue with only a high school diploma? Check all that apply.

Zolol [24]

Answer:

Teller, Loan Officer, and Tax Preparer

Explanation:

3 0

2 years ago

Read 2 more answers

Physical properties of minerals

lord [1]

Most minerals can be characterized and classified by their unique physical properties: hardness, luster, color, streak, specific gravity, cleavage, fracture, and tenacity.

<em>-</em><em> </em><em>BRAINLIEST</em><em> answerer</em><em> ❤️</em><em>✌</em>

6 0

2 years ago

Read 2 more answers

Tech A says you can find the typical angle of a V-block engine by dividing the number of cylinders by 720

Lady_Fox [76]

Answer:

Tech A is correct

Explanation:

Tech A is right as its V- angle is identified by splitting the No by 720 °. Of the piston at the edge of the piston.

Tech B is incorrect, as the V-Angle will be 720/10 = 72 for the V-10 motor, and he says 60 °.

8 0

2 years ago

A 132mm diameter solid circular section

Ganezh [65]

Answer:

not sure if this helps but

5 0

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.

7 0

3 years ago