2.1 What is the minimum number of pins required for a so-called dual-op-amp IC package, one containing two op amps? What is the
1 answer:
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Answer:
See attachments for step by step explanation towards getting answer.
Explanation:
Given that;
College Engineering 10+5 pts
When -iron is subjected to an atmosphere of hydrogen gas, the concentration of hydrogen in the iron, CH (in weight percent), is a function of hydrogen pressure, (in MPa), and absolute pH2temperature (T) according to(5.14)Furthermore, the values of D0 and Qd for this diffusion system are 1.4 10-7 m2/s and 13,400 J/mol, respectively. Consider a thin iron membrane 1 mm thick that is at 250C. Compute the diffusion flux through this membrane if the hydrogen pressure on one side of the membrane is 0.15 MPa (1.48 atm), and on the other side 7.5 MPa (74 atm).
See attachlent for complete solving.
Answer:
For block A, a = 9.66 ft/s²
For block B, a = 15 ft/s²
Explanation:
A free body diagram for this force system is attached to this solution
Mass of block A = m₁ = 8 lb
Mass of block B = m₂ = 6 lb
Coefficient of kinetic friction = μ
Normal reaction on the blocks = N
Spring stiffness of the spring btw block A and B = k = 20 lb/ft
Compression of the spring = 0.2 ft
Analysing Block A first
The forces on block A include, the weight, normal reaction, frictional force and the elastic force due to the spring
Sum of forces in the y-direction = 0
So, the weight of the block = Normal reaction of the surface on the block
N = W = 8 lb
Sum of forces in the x-direction = maₓ
(k × x) - (μ × N) = maₓ
m = W/g = 8/32.2 = 0.248 lbm
(20×0.2) - (0.2 × 8) = (0.248) aₓ
aₓ = 9.66 ft/s²
The forces on block B include, the weight, normal reaction, frictional force and the elastic force due to the spring
Sum of forces in the y-direction = 0
So, the weight of the block = Normal reaction of the surface on the block
N = W = 6 lb
Sum of forces in the x-direction = maₓ
(k × x) - (μ × N) = maₓ
m = W/g = 6/32.2 = 0.186 lbm
(20×0.2) - (0.2 × 6) = (0.186) aₓ
aₓ = 15 ft/s²
Answer:
A) ΔS_refrigerant = 0.70754 Kj/K
B) ΔS_space = -0.68441 Kj/K
C) ΔS_total = 0.02313 Kj/K
Explanation:
A) From he table attached, at Pressure of 140 KPa, and by interpolation, we get, Temperature of T = -18.77°C
Converting to degree kelvin yields;
T = -18.77 + 273 = 255.23 K
Formula for entropy change of refrigerant is given as;
ΔS_refrigerant = Q_in/T_refrigerant
We are given Q = 180 KJ
Thus, ΔS_refrigerant = 180/255.23 = 0.70754 Kj/K
B) Formula for entropy change of cooled space is given as;
ΔS_space = Q_out/T_s pace
T_space = -10°C = 273 - 10 = 263K
Thus, ΔS_space = -180/263 = -0.68441 Kj/K
C) the total entropy change would be;
ΔS_total = ΔS_refrigerant + ΔS_space
Thus,
ΔS_total = 0.70754 - 0.68441 = 0.02313 Kj/K
