Answer:
the critical flaw length is 10.06 mm
Explanation:
Given the data in the question;
plane strain fracture toughness 
= 92 Mpa√m
yield strength σ
= 900 Mpa
design stress is one-half of the yield strength ( 900 Mpa / 2 ) 450 Mpa
Y = 1.15
we know that;
Critical crack length 
= 1/π( / Yσ )²
we substitute
= 1/π( 92 Mpa√m / (1.15 × 450 Mpa )²
= 1/π( 92 Mpa√m / (517.5 Mpa )²
= 1/π( 0.177777 )²
= 1/π( 0.03160466 )
= 0.01006 m = 10.06 mm
Therefore, the critical flaw length is 10.06 mm
{ = ( 10.06 mm ) > 3 mm
The critical flow is subject to detection
Answer:
The speed is the same at 1.5 m/s while
The work done by the force F is 0.4335 J
Explanation:
Here we have angular acceleration α = v²/r
Force = ma = 2.8 × 1.5²/r₁
and ω₁ = v₁/r₁ = ω₂ = v₁/r₂
The distance moved by the force = 600 - 300 = 300 mm = 0.3 m
If the velocity is constant
The speed is 1.5 m/s while the work done is
2.8 × 1.5²1/(effective radius) ×0.3
r₁ = effective radius
2.8*9.81 = 2.8 × 1.5²/r₁
r₁ = 0.229
The work done by the force = 2.8 × 1.5²*1/r₁ *0.3 = 0.4335 J
Answer:
<em>d. </em><em>reducing </em><em>the </em><em>amount </em><em>of </em><em>recourses </em><em>spent </em><em>on </em><em>one </em><em>want </em><em>to </em><em>spend </em><em>more </em><em>on </em><em>another </em><em>want </em>
<em>brainliest</em><em>? </em><em>plz! </em>
Answer: They showed that many people worked on the project and demonstrate how hard they worked.
Explanation:
Question
The mean weight of a breed of yearling cattle is 1187 pounds. Suppose that weights of all such animals can be described by the Normal model N(1187,78).
a) How many standard deviations from the mean would a steer weighing 1000 pounds be?
b) Which would be more unusual, a steer weighing 1000 pounds, or one weighing 1250 pounds?
Answer:
a. z = -2.40
A sleet weighing 1,000 pounds is 2.40 standard deviations below the mean.
b. z = 0.81
1000 is more unusual because its contained on the extreme end from the mean
Explanation:
a.
Let weight (in pounds) of the cattle be denoted by letter x:
z = (x - u)/ σ
Where u = mean and σ = standard deviation
u = 1187
σ = 78
x = 1000
Use z score formula to standardize the value of x:
z = (1000 - 1187)/78
z = -187/78
z = -2.397436
z = -2.40 ------_ Approximated
A sleet weighing 1,000 pounds is 2.40 standard deviations below the mean.
b.
x= 1250
z= (1250 - 1187)/78
z = 63/78
z = 0.807692
z = 0.81 --------- Approximated
1000 is more unusual because its contained on the extreme end from the mean
