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tensa zangetsu [6.8K]
3 years ago
5

In the engineering design process, testing is the least important step. True False

Engineering
2 answers:
Wittaler [7]3 years ago
7 0

Answer:

false

Explanation:

Romashka-Z-Leto [24]3 years ago
4 0

i think the answer is false

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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
Solve the inequality. Then graph your solution.<br> -9v – 10 &lt; 7y +6
Cerrena [4.2K]
16

if you add 9+10 you get 18 - 7+6
5 0
3 years ago
What is the angular velocity (in rad/s) of a body rotating at N r.p.m.?
Darina [25.2K]

Answer:

0.1047N

Explanation:

To solve this problem we must remember the conversion factors, remembering that 1 revolution equals 2π radians and 1min equals 60s

N\frac{rev}{min} \frac{2\pi }{1rev} \frac{1min}{60} =N\frac{2\pi }{60} =0.1047N

in conclusion, to know how many rad / s an element rotates which is expressed in Rev / min we must only multiply by 0.1047

3 0
3 years ago
What does polarity give you information about?
Brrunno [24]

Well, I do know that polarity affects the voltage.

6 0
3 years ago
For installations where the nonlinear load is huge, most consulting engineers will specify ____-rated transformers.
Scorpion4ik [409]

Answer:

K

Explanation:

For installations where the nonlinear load is huge, most consulting engineers will specify K-rated transformers.

8 0
2 years ago
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