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LuckyWell [14K]
2 years ago
6

A book sitting on a shelf is an example of ____________.

Engineering
2 answers:
lutik1710 [3]2 years ago
8 0

Answer:

A) inertia

Explanation:

inertia: "A object at rest stay at rest A object in motion will stay in motion unless a outside force meet." - Sir Issac Newton

lyudmila [28]2 years ago
5 0
Answer is A.) Inertia
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I need help!!! Because this is due
xenn [34]
I don’t understand either
7 0
3 years ago
Read 2 more answers
g A thin-walled pressure vessel 6-cm thick originally contained a small semicircular flaw (radius 0.50-cm) located at the inner
galben [10]

This question is not complete, the complete question is;

A thin-walled pressure vessel 6-cm thick originally contained a small semicircular flaw (radius 0.50-cm) located at the inner surface and oriented normal to the hoop stress direction. Repeated pressure cycling enabled the crack to grow larger. If the fracture toughness of the material is 88 Mpam^\frac{1}{2} , the yield strength equal to 1250 MPa, and the hoop stress equal to 300 MPa, would the vessel leak before it ruptured

Answer:

length of crack is 5.585 cm

we will observe that, the length of crack (5.585 cm) is less than the vessel thickness (6 cm) Hence, vessel will not leak before it ruptures

Explanation:

Given the data in the question;

vessel thickness = 6 cm

fracture toughness k = 88 Mpam^\frac{1}{2}

yield strength = 1250 MPa

hoop stress equal = 300 MPa

we know that, the relation between fracture toughness and crack length is expressed as;

k = (1.1)(2/π)(r√(πa))  

where k is the fracture toughness, r is hoop stress and a is length of crack

so we rearrange to find  length of crack

a = 1/π[( k / 1.1(r)(2/π)]²

a = 1/π[( kπ / 1.1(r)(2)]²

so we substitute  

a = 1/π [( 88π / 1.1(300)(2/π)]²    

a = 1/π[ 0.1754596 ]

a = 0.05585 m

a = 0.05585 × 100 cm

a = 5.585 cm  

so, length of crack is 5.585 cm

we will observe that, the length of crack (5.585 cm) is less than the vessel thickness (6 cm) Hence, vessel will not leak before it ruptures

8 0
3 years ago
Suppose you are asked to design an office building. Explain what type of drawing you would use and why.
ohaa [14]

Answer:

birds-eye view perspective

Explanation:

If someone asked me to design an office building, I would draw it from a birds-eye view perspective. I would draw it this way so I could map out where everything in the office would go and make sure I have enough space for everything. I would also draw it this way in order to clearly see where everything would go in the office. For instance, cubicles/desks could go in the bottom left corner, while the boss's office could go in the top right. It would be easier to organize and it would be easier for me to look back on when I need to actually design the office later.

(i'm not sure if this is what your question is asking for so i just made my best guess)

7 0
3 years ago
Read 2 more answers
A wastewater treatment plant has two primary clarifiers, each 20m in diameter with a 2-m side-water depth. the effluent weirs ar
jasenka [17]

Answer:

overflow rate 20.53 m^3/d/m^2

Detention time 2.34 hr

weir loading  114.06 m^3/d/m

Explanation:

calculation for single clarifier

sewag\  flow Q = \frac{12900}{2} = 6450 m^2/d

surface\  area =\frac{pi}{4}\times diameter ^2 = \frac{pi}{4}\times 20^2

surface area = 314.16 m^2

volume of tankV  = A\times side\ water\ depth

                             =314.16\times 2 = 628.32m^3

Length\ of\  weir = \pi \times diameter of weir

                       = \pi \times 18 = 56.549 m

overflow rate =v_o = \frac{flow}{surface\ area} = \frac{6450}{314.16} = 20.53 m^3/d/m^2

Detention timet_d = \frac{volume}{flow} = \frac{628.32}{6450} \times 24 = 2.34 hr

weir loading= \frac{flow}{weir\ length} = \frac{6450}{56.549} = 114.06 m^3/d/m

6 0
3 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

so

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
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