Answer:
A continuity test
Explanation:
A continuity test is used to verified that current will flow in an electrical circuit, it performed by placing a small voltage across the chosen path. continuity test ensure that the equipment grounding conductor is electrically continuous and this test is perform on all the cord sets, receptacles that aren't part of a building or structure's permanent wiring, and cord-and-plug connected equipment required to be grounded. example of equipment used in testing current flow in continuity test are Analog multi-meter, voltage/continuity tester etc.
Continuity test and terminal connection test are the two test required by OSHA on all electrical equipment
Answer:
Disaggregation
Explanation:
In a company it is a way to create operational plans that are focused, either by time or by section.
Answer:
option B.
Explanation:
The correct answer is option B.
Principal stress is the maximum normal stress a body can have. In principal stress, there is purely normal stress. On principal plane shear stress is zero.
In-plane shear stress are the shear stress which is acting on the plane.
The statement which is correct regarding principle plane and shear stress is that The shear stress over principal stress planes is always zero.
Answer:
affects the flow of electricity
Explanation:
A loose battery terminal affects the flow of electricity. There is less power going to the electrical systems and the vehicle will not start or start sluggishly. Also, a loose battery terminal causes the car's electrical components like navigation, car lights, and audio among others to dim or fail completely.
Answer:
(b) Given the Weibull parameters of example 11-3, the factor by which the catalog rating must be increased if the reliability is to be increased from 0.9 to 0.99.
Equation 11-1: F*L^(1/3) = Constant
Weibull parameters of example 11-3: xo = 0.02 (theta-xo) = 4.439 b = 1.483
Explanation:
(a)The Catalog rating(C)
Bearing life:
Catalog rating: 
From given equation bearing life equation,

we Dividing eqn (2) with (1)

The Catalog rating increased by factor of 1.26
(b) Reliability Increase from 0.9 to 0.99

Now calculating life adjustment factor for both value of reliability from Weibull parametres


Similarly

Now calculating bearing life for each value

Now using given ball bearing life equation and dividing each other similar to previous problem

Catalog rating increased by factor of 0.61