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2 years ago

Question 2

Engineering

1 answer:

ValentinkaMS [17]2 years ago

3 0

Answer:

r h ch y TC GG ch hh h h

No se

Explanation:

No seibibk h j h uh t t t

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A manufacturer makes two types of drinking straws: one with a square cross-sectional shape, and the other type the typical round

Harlamova29_29 [7]

Answer:

$\frac{Q_{square}}{Q_{circle}} = 0.785$

Explanation:

given data

types of drinking straws

square cross-sectional shape
round shape

solution

we know that both perimeter of the cross section are equal

so we can say that

perimeter of square = perimeter of circle

4 × S = π × D

here S is length and D is diameter

S = $\frac{\pi D}{4}$ ....................1

and

ratio of flow rate through the square and circle is here

$\frac{Q_{square}}{Q_{circle}} = \frac{AV^2}{AV^2}$

$\frac{Q_{square}}{Q_{circle}} = \frac{S^2}{\frac{\pi D^2}{4}}$

$\frac{Q_{square}}{Q_{circle}} = \frac{(\frac{\pi D}{4})^2}{\frac{\pi D^2}{4}}$

$\frac{Q_{square}}{Q_{circle}} = \frac{\pi }{4}$

$\frac{Q_{square}}{Q_{circle}} = 0.785$

4 0

4 years ago

A part made from annealed AISI 1018 steel undergoes a 20 percent cold-work operation. Obtain the yield strength and ultimate str

Charra [1.4K]

Answer:

yield strength before cold work = 370 MPa

yield strength after cold work = 437.87 MPa

ultimate strength before cold work = 440 MPa

ultimate strength after cold work = 550 MPa

Explanation:

given data

AISI 1018 steel

cold work factor W = 20% = 0.20

to find out

yield strength and ultimate strength before and after the cold-work operation

solution

we know the properties of AISI 1018 steel is

yield strength σy = 370 MPa

ultimate tensile strength σu = 440 MPa

strength coefficient K = 600 MPa

strain hardness n = 0.21

so true strain is here ∈ = $ln\frac{1}{1-0.2}$ = 0.223

yield strength after cold is

yield strength = $K \varepsilon ^n$

yield strength = $600*0.223^{0.21)$

yield strength after cold work = 437.87 MPa

and

ultimate strength after cold work is

ultimate strength = $\frac{\sigma u}{1-W}$

ultimate strength = $\frac{440}{1-0.2}$

ultimate strength after cold work = 550 MPa

8 0

3 years ago

Manufacturing employees who perform assembly line work are referred to as

mamaluj [8]

Answer:

C. assembly line workers.

Explanation:

8 0

3 years ago

Read 2 more answers

A rectangular block of material with shear modulus G= 620 MPa is fixed to rigid plates at its top and bottom surfaces. Thelower

PIT_PIT [208]

Answer:

γ $_{xy}$ =0.01, P=248 kN

Explanation:

Given Data:

displacement = 2mm ;

height = 200mm ;

l = 400mm ;

w = 100 ;

G = 620 MPa = 620 N//mm²; 1MPa = 1N//mm²

a. Average Shear Strain:

The average shear strain can be determined by dividing the total displacement of plate by height

γ $_{xy}$ = displacement / total height

= 2/200 = 0.01

b. Force P on upper plate:

Now, as we know that force per unit area equals to stress

τ = P/A

Also, τ = Gγ $_{xy}$

By comapring both equations, we get

P/A = Gγ $_{xy}$ ------------ eq(1)

First we need to calculate total area,

A = l*w = 400 * 100= 4*10^4mm²

By putting the values in equation 1, we get

P/40000 = 620 * 0.01

P = 248000 N or 2.48 *10^5 N or 248 kN

6 0

3 years ago

What happens when the arms of the milky move away from the center of the galaxy

Alina [70]

Well this question is though because we have never seen such a thing ! and to be quite frank when that happens , nothing good comes from it. Black Holes

6 0

3 years ago