Answer:
Control mechanisms
Explanation:
Organizational chart of any company will give details of different aspects of the company such as the major sub-units of the organization with the names of team leaders for different sub-units, it can also give you the span of control and the division of work within the company. However, the chart can't show you control mechanisms of different departments.
Answer:
a)
b)Mass in lb = 376.8 lb
Mass in slug = 11.71 slug
c)
d)
Explanation:
Given that
d= 2 ft
r= 1 ft
h= 3 ft
Density

a)
We know that volume V given as



b)
Mass = Density x volume

mass= 376.8 lb
We know that
1 lb = 0.031 slug
So 376.8 lb= 11.71 slug
Mass in lb = 376.8 lb
Mass in slug = 11.71 slug
c)
we know that specific volume(v) is the inverse of density.



d)
Specific weight(w) is the product of density and the gravity(g).
w= ρ X g
w = 40 x 31.9

Answer:
a.) -147V
b.) -120V
c.) 51V
Explanation:
a.) Equation for potential difference is the integral of the electrical field from a to b for the voltage V_ba = V(b)-V(a).
b.) The problem becomes easier to solve if you draw out the circuit. Since potential at Q is 0, then Q is at ground. So voltage across V_MQ is the same as potential at V_M.
c.) Same process as part b. Draw out the circuit and you'll see that the potential a point V_N is the same as the voltage across V_NP added with the 2V from the other box.
Honestly, these things take practice to get used to. It's really hard to explain this.
Answer: 0.93 mA
Explanation:
In order to calculate the current passing through the water layer, as we have the potential difference between the ends of the string as a given, assuming that we can apply Ohm’s law, we need to calculate the resistance of the water layer.
We can express the resistance as follows:
R = ρ.L/A
In order to calculate the area A, we can assume that the string is a cylinder with a circular cross-section, so the Area of the water layer can be written as follows:
A= π(r22 – r12) = π( (0.0025)2-(0.002)2 ) m2 = 7.07 . 10-6 m2
Replacing by the values, we get R as follows:
R = 1.4 1010 Ω
Applying Ohm’s Law, and solving for the current I:
I = V/R = 130 106 V / 1.4 1010 Ω = 0.93 mA
The magnitude of applied stress in the direction of 101 is 12.25 MPA and in the direction of 011, it is not defined.
<u>Explanation</u>:
<u>Given</u>:
tensile stress is applied parallel to the [100] direction
Shear stress is 0.5 MPA.
<u>To calculate</u>:
The magnitude of applied stress in the direction of [101] and [011].
<u>Formula</u>:
zcr=σ cosФ cosλ
<u>Solution</u>:
For in the direction of 101
cosλ = (1)(1)+(0)(0)+(0)(1)/√(1)(2)
cos λ = 1/√2
The magnitude of stress in the direction of 101 is 12.25 MPA
In the direction of 011
We have an angle between 100 and 011
cosλ = (1)(0)+(0)(-1)+(0)(1)/√(1)(2)
cosλ = 0
Therefore the magnitude of stress to cause a slip in the direction of 011 is not defined.