Answer:
Electroosmotic velocity will be equal to 
Explanation:
We have given applied voltage v = 100 volt
Length of capillary L = 5 mm = 0.005 m
Zeta potential of the capillary surface 
Dielectric constant of glass is between 5 to 10 here we are taking dielectric constant as 
Viscosity of glass is 
Electroosmotic velocity is given as 
So Electroosmotic velocity will be equal to
Answer:
The value of heat transferred watt per foot length Q = 54.78 Watt per foot length.
Explanation:
Diameter of pipe = 2 in = 0.0508 m
Steam temperature 
= 300 F = 422.04 K
Duct temperature 
= 70 F = 294.26 K
Emmisivity of surface 1 = 0.79
Emmisivity of surface 2 = 0.276
Net emmisivity of both surfaces ∈ = 0.25
Stefan volazman constant 
= 5.67 × 
Heat transfer per foot length is given by
Q = ∈ A ( 
) ------ (1)
Put all the values in equation (1) , we get
Q = 0.25 × 5.67 × × 3.14 × 0.0508 × 1 × ( 
)
Q = 54.78 Watt per foot.
This is the value of heat transferred watt per foot length.
According to O*NET, the common work contexts for Licensing Examiners and Inspectors include:
- Telephone
- Face-to-face discussions
- Contact with others
- Importance of being exact or accurate.
O*NET is an acronym for occupational information network and it refers to a free resource center or online database that is updated from time to time with several occupational definitions, so as to help the following categories of people understand the current work situation in the United States of America:
- Workforce development professionals
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On O*NET, work contexts are typically used to describe the physical and social elements that are common to a particular profession or occupational work. Also, the less common work contexts are listed toward the bottom while common work contexts are listed toward the top.
According to O*NET, the common work contexts for Licensing Examiners and Inspectors include:
1. Telephone
2. Face-to-face discussions
3. Contact with others
4. Importance of being exact or accurate.
Read more on work contexts here: brainly.com/question/22826220
Answer:
Combination circuit; The basic strategy for the analysis of combination circuits involves using the meaning of equivalent resistance for parallel branches to transform the combination circuit into a series circuit.
Example:
The use of both series and parallel connections within the same circuit. In this case, light bulbs A and B are connected by parallel connections and light bulbs C and D are connected by series connections. This is an example of a combination circuit.
Answer:
%Reduction in area = 73.41%
%Reduction in elongation = 42.20%
Explanation:
Given
Original diameter = 12.8 mm
Gauge length = 50.80mm
Diameter at the point of fracture = 6.60 mm (0.260 in.)
Fractured gauge length = 72.14 mm.
%Reduction in Area is given as:
((do/2)² - (d1/2)²)/(do/2)²
Calculating percent reduction in area
do = 12.8mm, d1 = 6.6mm
So,
%RA = ((12.8/2)² - 6.6/2)²)/(12.8/2)²
%RA = 0.734130859375
%RA = 73.41%
Calculating percent reduction in elongation
%Reduction in elongation is given as:
((do) - (d1))/(d1)
do = 72.14mm, d1 = 50.80mm
So,
%RA = ((72.24) - (50.80))/(50.80)
%RA = 0.422047244094488
%RA = 42.20%
