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:
No, they need to be somewhat flexible so that forces such as turbulance don't shear the wing off.
Answer:
The current through the coil is 2.05 A
Explanation:
Given;
number of turns of the coil, N = 1
radius of the coil, r = 9.8 cm = 0.098 m
magnetic moment of the coil, P = 6.2 x 10⁻² A m²
The magnetic moment is given by;
P = IA
Where;
I is the current through the coil
A is area of the coil = πr² = π(0.098)² = 0.03018 m²
The current through the coil is given by;
I = P / A
I = (6.2 x 10⁻² ) / (0.03018)
I = 2.05 A
Therefore, the current through the coil is 2.05 A
Answer:
$7,778.35
Explanation:
At year 3, the final payment of the remaining balance is equal to the present worth P of the last three payments.
First, calculate the uniform payments A:
A = 12000(A/P, 4%, 5)
= 12000(0.2246) = 2695.2 (from the calculator)
Then take the last three payments as its own cash flow.
To calculate the new P:
P = 2695.2 + 2695.2(P/A, 4%, 2) = 2695.2 + 2695.2(1.886) = 7778.35
Therefore, the final payment is $7,778.35
Answer:
Part a: The yield moment is 400 k.in.
Part b: The strain is 
Part c: The plastic moment is 600 ksi.
Explanation:
Part a:
As per bending equation
Here
- M is the moment which is to be calculated
- I is the moment of inertia given as
Here
- b is the breath given as 0.75"
- d is the depth which is given as 8"
The yield moment is 400 k.in.
Part b:
The strain is given as
The stress at the station 2" down from the top is estimated by ratio of triangles as
Now the steel has the elastic modulus of E=29000 ksi
So the strain is
Part c:
For a rectangular shape the shape factor is given as 1.5.
Now the plastic moment is given as
The plastic moment is 600 ksi.
