A particle, originally at rest and located at point (3 ft, 2 ft, 5 ft), is subjected to an acceleration a = {6ti + 12t^2k} ft/s^
1 answer:
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Rubber block is not shown. I have attached an image of it.
Answer:
A) ε_x = 0.0075
B) ε_y = 0.00375
C) γ_xy = 0.0122 rad
Explanation:
We are given;
δ = 0.03 in
L = 4 in
ν_r = 0.5
θ = 89.3° = 89.3π/180 rad
Let's calculate ε_x in the direction of axis x
Thus, ε_x = δ/L = 0.03/4 = 0.0075
Let's calculate ε_y in the direction of axis y;
ε_y = v•ε_x = 0.5 x 0.0075 = 0.00375
Now, shear strain is angle between π/2 rad surfaces at that point.
Thus,
γ_xy = π/2 - θ = π/2 - 89.3π/180
γ_xy = π(0.003889) = 0.0122 rad
Answer:
Thermal resistance for a wall depends on the material, the thickness of the wall and the cross-section area.
Explanation:
Current flow and heat flow are very similar when we are talking about 1-dimensional energy transfer. Attached you can see a picture we can use to describe the heat flow between the ends of the wall. First of all, a temperature difference is required to flow heat from one side to the other, just like voltage is required for current flow. You can also see that represents the thermal resistance. The next image explains more about the parameters which define the value of the thermal resistances which are the following:
- Wall Thickness. More thickness, more thermal resistance.
- Material thermal conductivity (unique value for each material). More conductivity, less thermal resistance.
- Cross-section Area. More cross-section area, less thermal resistance.
A expression to define the thermal resistance for the wall is as follows: , where l is the distance between the tow sides of the wall, that is to say the wall thickness; A is the cross-section area and k is the material conducitivity.
