Answer:
The fluid property responsible for the development of velocity boundary layer is majorly the fluid's viscosity.
For non-viscous fluids (in theory, because no fluid is entirely non-viscous), there will be no velocity boundary layer.
Explanation:
The velocity boundary layer is the thin layer of viscous fluid that is in direct contact with the pipe surface. The velocity of fluid in this layer is 0 as fluid doesn't move in this layer.
This phenomenon is due to the viscosity of the fluid. Viscosity of the fluid refers to the internal friction that exists between fluid layers, so, the layer of fluid in contact with non-moving, static surface of the pipe experiences friction that causes this layer to not move, causing the fluid velocity to vary from 0 at this surface to the maximum value at the centre of the pipe, before the velocity begins to drop again until it reaches 0 at the other end of the circular pipe.
Since viscosity is the primary cause of this, non-viscous or inviscid fluids are saved from this phenomenon as their flows do not have the velocity boundary layer.
Although, a completely non-viscous or inciscid fluid is an idealized concept because all fluids will experience some sort of viscosity (no matter how small) between their fluid layers. Hence, a velocity boundary layer, no matter how thin (or of minute thickness), will exist in the flow of real fluids.
But, an idealized non-viscous or inviscid fluid will not have a velocity boundary layer.
Hope this Helps!!!
Answer:
d) phase sequence of the voltage applied to the motor stator windings
Explanation:
The direction of rotation of a three-phase squirrel-cage induction motor depends on the phase sequence of the voltage applied to the motor stator windings.
The dealer would accept any offer that is at least 500 dollars over the dealer's cost The algorithms for your query are 1) Declare list Price as a double READ list Price• Declare and initialize most-0.85*list Price+500.0.
<h3>What is an easy definition of a set of rules?</h3>
A set of rules is a method used for fixing a hassle or acting a computation. Algorithms act as a specific listing of commands that behavior certain moves grade by grade in both hardware- or software-primarily based totally routines. Algorithms are broadly used during all regions of IT.
- Function calculateAvg(Scores) Declare and initialize SUM as double identical to 0 • For i=1 to length(Scores) SUM=SUM+Scores.get(i).
- EndLoop- Return SUM/length(Scores)
- EndFunction- Function printBelowAvg(Names, Scores)
- Define and claim AVG=calculateAvg(Scores), For i=1 to length(Scores)
- If Scores.get(i)MAX then MAX=Scores.get(i)
- Endif, EndLoop.
- RETURN MAX-EndFunction- Function blended Perform(Names, Scores).
- PRINT calculateAvg(Scores), print BelowAvg(Names, Scores)
- PRINT highestScore(Scores), EndFunction
- Call the characteristics as combinedPerform(Names, Scores)
- Kindly revert for any queries.
Read more about the algorithm :
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Answer:
component of acceleration are a = 3.37 m/s² and ar = 22.74 m/s²
magnitude of acceleration is 22.98 m/s²
Explanation:
given data
velocity = 10 m/s
initial time to = 0
distance s = 400 m
time t = 14 s
to find out
components and magnitude of acceleration after the car has travelled 200 m
solution
first we find the radius of circular track that is
we know distance S = 2πR
400 = 2πR
R = 63.66 m
and tangential acceleration is
S = ut + 0.5 ×at²
here u is initial speed and t is time and S is distance
400 = 10 × 14 + 0.5 ×a (14)²
a = 3.37 m/s²
and here tangential acceleration is constant
so velocity at distance 200 m
v² - u² = 2 a S
v² = 10² + 2 ( 3.37) 200
v = 38.05 m/s
so radial acceleration at distance 200 m
ar = ![\frac{v^2}{R}](https://tex.z-dn.net/?f=%5Cfrac%7Bv%5E2%7D%7BR%7D)
ar = ![\frac{38.05^2}{63.66}](https://tex.z-dn.net/?f=%5Cfrac%7B38.05%5E2%7D%7B63.66%7D)
ar = 22.74 m/s²
so magnitude of total acceleration is
A = ![\sqrt{a^2 + ar^2}](https://tex.z-dn.net/?f=%5Csqrt%7Ba%5E2%20%2B%20ar%5E2%7D)
A = ![\sqrt{3.37^2 + 22.74^2}](https://tex.z-dn.net/?f=%5Csqrt%7B3.37%5E2%20%2B%2022.74%5E2%7D)
A = 22.98 m/s²
so magnitude of acceleration is 22.98 m/s²
Answer:
The change of entropy is 1.229 kJ/(kg K)
Explanation:
Data
(Individual Gas Constant for air)
For variable specific heats
where
is evaluated from table attached
Replacing in equation