Answer:
In the steel: 815 kPa
In the aluminum: 270 kPa
Explanation:
The steel pipe will have a section of:
A1 = π/4 * (D^2 - d^2)
A1 = π/4 * (0.8^2 - 0.7^2) = 0.1178 m^2
The aluminum core:
A2 = π/4 * d^2
A2 = π/4 * 0.7^2 = 0.3848 m^2
The parts will have a certain stiffness:
k = E * A/l
We don't know their length, so we can consider this as stiffness per unit of length
k = E * A
For the steel pipe:
E = 210 GPa (for steel)
k1 = 210*10^9 * 0.1178 = 2.47*10^10 N
For the aluminum:
E = 70 GPa
k2 = 70*10^9 * 0.3848 = 2.69*10^10 N
Hooke's law:
Δd = f / k
Since we are using stiffness per unit of length we use stretching per unit of length:
ε = f / k
When the force is distributed between both materials will stretch the same length:
f = f1 + f2
f1 / k1 = f2/ k2
Replacing:
f1 = f - f2
(f - f2) / k1 = f2 / k2
f/k1 - f2/k1 = f2/k2
f/k1 = f2 * (1/k2 + 1/k1)
f2 = (f/k1) / (1/k2 + 1/k1)
f2 = (200000/2.47*10^10) / (1/2.69*10^10 + 1/2.47*10^10) = 104000 N = 104 KN
f1 = 200 - 104 = 96 kN
Then we calculate the stresses:
σ1 = f1/A1 = 96000 / 0.1178 = 815000 Pa = 815 kPa
σ2 = f2/A2 = 104000 / 0.3848 = 270000 Pa = 270 kPa
Answer: C
Explanation:
Uploading a huge file on the internet will slow it down alot
Answer:
You are a historian who is compiling information on the changing materials used in a product in one of the categories listed below. Choose a product that uses a variety of materials that have been discussed in this lesson (for example, a car or a tennis racket). Using online or library resources, research past materials used to create the product. Be sure to discuss the materials currently being used for the design and include information on the sustainability of these resources. Your research must also explain why the materials changes were made (for example, safety, cost, or weight). Keep track of your references so that you can document all of the resources used for your report.
Answer:
for...?
thenks for the points :))
Answer: a 8143.71 kJ/kg
b 393.15 K
Explanation:
This system is an isobaric process in which there is no change in pressure a quasistatic process where a pressure distribution exists
a since no change in pressure =0 the system does work thus
FOR HELIUM properties in standard thermodynamic chart
cv = 3.1 kJ/kgK
M = Molar mass = 4 kg/kmol
R = Universal gas constant = 8.314 kJ/kg K
cp ≈ cv +R /M = 3.1 + 8.314 /4 = 5.1785 kJ/kgK
Cp = cp * M = 5.1785 kJ/kgK * 4 kg/kmol = 20.714 kJ/kgkmol
T = 120 °C to Kelvin = 120 + 273.15k = 393.15 K
W =n Cp ΔT = 1 kmol * 20.714 kJ/kg kmol* 393.15 K = 8143.71 kJ/kg
b convert T °C = T K thus 120 + 273.15 K = 393.15 K
P₁/T₁ = P₂/T₂
200 kPa/ 393.15 K = 200 kPa/T₂
T₂ = 200 kPa * 393.15 K/ 200 kPa = 393.15 K or 120 k
