Alexis is studying how lenses work. She looks through a
1 answer:
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Answer:
Rectangular path
Solution:
As per the question:
Length, a = 4 km
Height, h = 2 km
In order to minimize the cost let us denote the side of the square bottom be 'a'
Thus the area of the bottom of the square, A =
Let the height of the bin be 'h'
Therefore the total area,
The cost is:
C = 2sh
Volume of the box, V = (1)
Total cost, (2)
From eqn (1):
Using the above value in eqn (1):
Differentiating the above eqn w.r.t 'a':
For the required solution equating the above eqn to zero:
a = 4
Also
The path in order to minimize the cost must be a rectangle.
Answer:
(a) Elongation of the rod==5.61×10⁻⁹m
(b) Change in diameter=1.640×10⁻⁸m
Explanation:
Given data
Diameter d=78 in=1.9812 m
Cross Area is:
Applied Load P=17 KN=17×10³N
E=29 × 106 psi=1.99947961×10¹¹Pa
Stress and Strain in x direction
Stress in x direction
σ=P/A
σ=5517.25 Pa
Strain in x direction
ε=σ/E
ε=2.76×10⁻⁸
Part (a)
Elongation of the rod=Lε
=(0.2032)(2.76×10⁻⁸)
Elongation of the rod==5.61×10⁻⁹m
Part(b) Change in diameter
Strain in y direction
ε₁= -vε
ε₁= -(0.30)(2.76×10⁻⁸)
ε₁=-8.28×10⁻⁹
Change in diameter=d×ε₁
Change in diameter=(1.9812m)×(-8.28×10⁻⁹)
Change in diameter=1.640×10⁻⁸m