Except the Table of Contents
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
Answer:
The right solution is "2625 kN".
Explanation:
According to the question,
The average pressure will be:
= 
By putting values, we get
= 
= 
= 
hence,
The average force will be:
= 
= 
= 
Or,
= 
Answer:
A,C, and D
Explanation:
Potible ladders have to configure with many designs in mind but the most evedent is that they are usally unstable
BRAINLIEST PLS
Answer: The answer is four; four
Explanation: This is because of the mixture of material used and the number of directions it causes strain I directly proportional to the number of times it causes stress.
