Answer:
1790 μrad.
Explanation:
Young's modulus, E is given as 10000 ksi,
μ is given as 0.33,
Inside diameter, d = 54 in,
Thickness, t = 1 in,
Pressure, p = 794 psi = 0.794 ksi
To determine shear strain, longitudinal strain and circumferential strain will be evaluated,
Longitudinal strain, eL = (pd/4tE)(1 - 2μ)
eL = (0.794 x 54)(1 - 0.66)/(4 x 1 x 10000)
eL = 3.64 x 10-⁴ radians
Circumferential strain , eH = (pd/4tE)(2-μ)
eH = (0.794 x 54)(2 - 0.33)/(4 x 1 x 10000)
eH = 1.79 x 10-³ radians
The maximum shear strain is 1790 μrad.
Explanation:
A.
H = Aeσ^4
Using the stefan Boltzmann law
When we differentiate
dH/dT = 4AeσT³
dH/dT = 4(0.15)(0.9)(5.67)(10^-8)(650)³
= 8.4085
Exact error = 8.4085x20
= 168.17
H(650) = 0.15(0.9)(5.67)(10^-8)(650)⁴
= 1366.376watts
B.
Verifying values
H(T+ΔT) = 0.15(0.9)(5.67)(10)^-8(670)⁴
= 1542.468
H(T+ΔT) = 0.15(0.9)(5.67)(10^-8)(630)⁴
= 1205.8104
Error = 1542.468-1205.8104/2
= 168.329
ΔT = 40
H(T+ΔT) = 0.15(0.9)(5.67)(10)^-8(690)⁴
= 1735.05
H(T-ΔT) = 0.15(0.9)(5.67)(10^-8)(610)⁴
= 1735.05-1059.83/2
= 675.22/2
= 337.61
Answer:
The mass flow rate of the mixture in the manifold is 6.654 kg/min
Explanation;
In this question, we are asked to calculate mass flow rate of the mixture in the manifold
Please check attachment for complete solution and step by step explanation.
If particleboard is used as wall sheathing, the grade mark with type M1 or M2 should be stamped on it.
<h3>What is particle board?</h3>
Particle board is notably used as floors underlayment or as a base for parquet floors, timber floors, or for carpets. For this purpose, the particle forums are dealt with with unique chemical compounds and resins to cause them to water-resistant or termite proof.
Waferboard, OSB, and composite plywood, while carried out as wall sheathing, offer a nail base for software of shingle siding.
Read more about the sheathing:
brainly.com/question/5029827
#SPJ1
Answer:
P ( 2.5 < X < 7.5 ) = 0.7251
Explanation:
Given:
- The pmf for normal distribution for random variable x is given:
f(x)=0.178 exp(-0.100(x-4.51)^2)
Find:
the fraction of individuals demonstrating a response in the range of 2.5 to 7.5.
Solution:
- The random variable X follows a normal distribution with mean u = 4.51, and standard deviation s.d as follows:
s.d = sqrt ( 1 / 0.1*2)
s.d = sqrt(5) =2.236067
- Hence, the normal distribution is as follows:
X ~ N(4.51 , 2.236)
- Compute the Z-score values of the end points 2.5 and 7.5:
P ( (2.5 - 4.51) / 2.236 < Z < (7.5 - 4.51 ) / 2.236 )
P ( -0.898899327 < Z < 1.337168651 )
- Use the Z-Table for the probability required:
P ( 2.5 < X < 7.5 ) = P ( -0.898899327 < Z < 1.337168651 ) = 0.7251