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3 years ago

Is a unit of measurement for angles

Engineering

1 answer:

dimulka [17.4K]3 years ago

3 0

Answer:

Degrees or Radians

Explanation:

Degrees are used for normal angles while radians are used in a unit circle.

Hope this helped!

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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

Prove the following languages are nonregular, once using the pumping lemma and once using the Myhill-Nerode theorem. When using

VashaNatasha [74]

Answer:

For any string, we use $s = xyz$

Explanation:

The pumping lemma says that for any string s in the language, with length greater than the pumping length p, we can write s = xyz with |xy| ≤ p, such that xyi z is also in the language for every i ≥ 0. For the given language, we can take p = 2.

Here are the cases:

Consider any string a i b j c k in the language. If i = 1 or i > 2, we take $x = \epsilon$ and y = a. If i = 1, we must have j = k and adding any number of a’s still preserves the membership in the language. For i > 2, all strings obtained by pumping y as defined above, have two or more a’s and hence are always in the language.

For i = 2, we can take and y = aa. Since the strings obtained by pumping in this case always have an even number of a’s, they are all in the language.
Finally, for the case i = 0, we take $x = \epsilon$ , and y = b if j > 0 and y = c otherwise. Since strings of the form b j c k are always in the language, we satisfy the conditions of the pumping lemma in this case as well.

8 0

3 years ago

A saturated 1.5 ft3 clay sample has a natural water content of 25%, shrinkage limit (SL) of 12% and a specific gravity (GS) of 2

Svetllana [295]

$79 f t^{3}$ is the volume of the sample when the water content is 10%.