Answer:
a) 0.3
b) 3.6 mm
Explanation:
Given
Length of the pads, l = 200 mm = 0.2 m
Width of the pads, b = 150 mm = 0.15 m
Thickness of the pads, t = 12 mm = 0.012 m
Force on the rubber, P = 15 kN
Shear modulus on the rubber, G = 830 GPa
The average shear strain can be gotten by
τ(average) = (P/2) / bl
τ(average) = (15/2) / (0.15 * 0.2)
τ(average) = 7.5 / 0.03
τ(average) = 250 kPa
γ(average) = τ(average) / G
γ(average) = 250 kPa / 830 kPa
γ(average) = 0.3
horizontal displacement,
δ = γ(average) * t
δ = 0.3 * 12
δ = 3.6 mm
Examples
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Problem Solving-Solve situation
Critical Thinking-Sturdy fast pace thinking
Leadership-The ability to guide others
Communication-The ability to talk to others
Collaboration-Working with some one
Answer:
a diameter of D₂ = 0.183 inches would be required
Explanation:
appyling pascal's law
P applied to the hydraulic jack = P required to lift the rock
F₁*A₁ = F₂*A₂
since A₁= π*D₁²/4 , A₂= π*D₂²/4
F₁*π*D₁²/4 = F₂* π*D₂²/4
F₁*D₁²=F₂*D₂²
D₂ = D₁ *√(F₁/F₂)
replacing values
D₂ = D₁ *√(F₁/F₂) = 6 in * √(120 lbf/(4000 lbm * 32.174 (lbf/lbm)) = 0.183 inches
Answer:
The required mechanical work is required to reduce each day by 1.05×10^8 Joules.
Explanation:
Coefficient of Performance (COP) = Q/W
Q is thermal energy absorbed by the air conditioner
W is mechanical work done
Q = 3.9×10^8 J
COP of old air conditioner = 2.3
W = Q/COP = 3.9×10^8/2.3 = 1.70×10^8 J
COP of new air conditioner = 6
W = Q/COP = 3.9×10^8/6 = 6.5×10^7 J
Reduction in mechanical work = (1.7×10^8) - (6.5×10^7) = 1.05×10^8 J
Answer:
The entity relationship (ER) data model has existed for over 35 years. It is well suited to data modelling for use with databases because it is fairly abstract and is easy to discuss and explain. ER models are readily translated to relations. ER models, also called an ER schema, are represented by ER diagrams.
