Answer:
6.9
Explanation:
I had the same question lol your welcomr if itd not right in sorry
Answer:
The essence including its problem is listed throughout the clarification section following.
Explanation:
Projects build deliverable that seem to be the products of the venture or indeed the implementation of the project. This ensures that perhaps the agile methodology may be as broad as either the goal of the study itself as well as the coverage that would be part of a much larger venture.
For every other production to have been marked as "deliverable" within the same project, this should satisfy a few eligibility requirements:
- It should be within the development of the work.
- The interested parties-external or internal-must consent to the above. This is perhaps the product of hard effort.
So that the above seems to be the right answer.
Answer:
true
Explanation:
True, there are several types of polymers, thermoplastics, thermosets and elastomers.
Thermosets are characterized by having a reticulated structure, so they have low elasticity and cannot be stretched when heated.
Because of the above, thermosetting polymers burn when heated.
Answer:
6.6 kilo volts = 6.6 k volts
Explanation:
A prefix is a word, number or a letter that is added before another word. In physics we have different prefixes for the exponential powers of 10, that are placed before units in place of those powers. Some examples are:
deci (d) ------ 10⁻¹
centi (c) ------ 10⁻²
milli (m) ------ 10⁻³
kilo (k) ------ 10³
mega (M) ----- 10⁶
giga (G) ------ 10⁹
We have:
6600 volts
converting to exponential form:
=> 6.6 x 10³ volts
Thus, we know that the prefix of kilo (k) is used for 10³.
Hence,
=> <u>6.6 kilo volts = 6.6 k volts</u>
Answer:
The pressure drop across the pipe also reduces by half of its initial value if the viscosity of the fluid reduces by half of its original value.
Explanation:
For a fully developed laminar flow in a circular pipe, the flowrate (volumetric) is given by the Hagen-Poiseulle's equation.
Q = π(ΔPR⁴/8μL)
where Q = volumetric flowrate
ΔP = Pressure drop across the pipe
μ = fluid viscosity
L = pipe length
If all the other parameters are kept constant, the pressure drop across the circular pipe is directly proportional to the viscosity of the fluid flowing in the pipe
ΔP = μ(8QL/πR⁴)
ΔP = Kμ
K = (8QL/πR⁴) = constant (for this question)
ΔP = Kμ
K = (ΔP/μ)
So, if the viscosity is halved, the new viscosity (μ₁) will be half of the original viscosity (μ).
μ₁ = (μ/2)
The new pressure drop (ΔP₁) is then
ΔP₁ = Kμ₁ = K(μ/2)
Recall,
K = (ΔP/μ)
ΔP₁ = K(μ/2) = (ΔP/μ) × (μ/2) = (ΔP/2)
Hence, the pressure drop across the pipe also reduces by half of its initial value if the viscosity of the fluid reduces by half of its value.
Hope this Helps!!!
