Answer:
Viscosity is notated using the common classification “XW-XX”. The number preceding the “W” (winter) rates the oil's flow (viscosity) at zero degrees Fahrenheit (-17.8 degrees Celsius). The lower the number, the less the oil thickens in cold weather.
Answer:
Unit testing is a software development process in which the smallest testable parts of an application, called units, are individually and independently scrutinized for proper operation. This testing methodology is done during the development process by the software developers and sometimes QA staff. The main objective of unit testing is to isolate written code to test and determine if it works as intended.
Unit testing is an important step in the development process, because if done correctly, it can help detect early flaws in code which may be more difficult to find in later testing stages.
Unit testing is a component of test-driven development (TDD), a pragmatic methodology that takes a meticulous approach to building a product by means of continual testing and revision. This testing method is also the first level of software testing, which is performed before other testing methods such as integration testing. Unit tests are typically isolated to ensure a unit does not rely on any external code or functions. Testing can be done manually but is often automated. It might be helpful
Answer:
A sole proprietorship, also known as the sole trader, individual entrepreneurship or proprietorship, is a type of enterprise that is owned and run by one person and in which there is no legal distinction between the owner and the business entity.
Explanation:
Answer: resistor
Explanation: Not quite sure. Need more research
Answer:
a) -8 lb / ft^3
b) -70.4 lb / ft^3
c) 54.4 lb / ft^3
Explanation:
Given:
- Diameter of pipe D = 12 in
- Shear stress t = 2.0 lb/ft^2
- y = 62.4 lb / ft^3
Find pressure gradient dP / dx when:
a) x is in horizontal flow direction
b) Vertical flow up
c) vertical flow down
Solution:
- dP / dx as function of shear stress and radial distance r:
(dP - y*L*sin(Q))/ L = 2*t / r
dP / L - y*sin(Q) = 2*t / r
Where dP / L = - dP/dx,
dP / dx = -2*t / r - y*sin(Q)
Where r = D /2 ,
dP / dx = -4*t / D - y*sin(Q)
a) Horizontal Pipe Q = 0
Hence, dP / dx = -4*2 / 1 - 62.4*sin(0)
dP / dx = -8 + 0
dP/dx = -8 lb / ft^3
b) Vertical pipe flow up Q = pi/2
Hence, dP / dx = -4*2 / 1 - 62.4*sin(pi/2)
dP / dx = 8 - 62.4
dP/dx = -70.4 lb / ft^3
c) Vertical flow down Q = -pi/2
Hence, dP / dx = -4*2 / 1 - 62.4*sin(-pi/2)
dP / dx = -8 + 62.4
dP/dx = 54.4 lb / ft^3
