Answer:
The horizontal conductivity is 41.9 m/d.
The vertical conductivity is 37.2 m/d.
Explanation:
Given that,
Thickness of A = 8.0 m
Conductivity = 25.0 m/d
Thickness of B = 2.0 m
Conductivity = 142 m/d
Thickness of C = 34 m
Conductivity = 40 m/d
We need to calculate the horizontal conductivity
Using formula of horizontal conductivity
Put the value into the formula
We need to calculate the vertical conductivity
Using formula of vertical conductivity
Put the value into the formula
Hence, The horizontal conductivity is 41.9 m/d.
The vertical conductivity is 37.2 m/d.
Answer:
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Explanation: if you don't know what it is look it up on .
Restraining devices and barriers shall be visually inspected on the rim wheel components or sudden release of contained air.
Restraining device means an apparatus such as a <em>cage, rack, assemblage of bars and other components</em> that will constrain all rim wheel components.
Restraining devices and barriers shall be visually inspected on the rim wheel components or sudden release of contained air. Any restraining device or barrier exhibiting damage such as the following defects shall be immediately removed from service.
Find out more on Restraining devices at: brainly.com/question/24647450
Answer:
R = 31.9 x 10^(6) At/Wb
So option A is correct
Explanation:
Reluctance is obtained by dividing the length of the magnetic path L by the permeability times the cross-sectional area A
Thus; R = L/μA,
Now from the question,
L = 4m
r_1 = 1.75cm = 0.0175m
r_2 = 2.2cm = 0.022m
So Area will be A_2 - A_1
Thus = π(r_2)² - π(r_1)²
A = π(0.0225)² - π(0.0175)²
A = π[0.0002]
A = 6.28 x 10^(-4) m²
We are given that;
L = 4m
μ_steel = 2 x 10^(-4) Wb/At - m
Thus, reluctance is calculated as;
R = 4/(2 x 10^(-4) x 6.28x 10^(-4))
R = 0.319 x 10^(8) At/Wb
R = 31.9 x 10^(6) At/Wb
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