Answer:
Just answered this to confirm my profile.
Explanation:
I dont have a clue, this is just to confirm my profile.
Answer:
B. The thickness of the heated region near the plate is increasing.
Explanation:
First we know that, a boundary layer is the layer of fluid in the immediate vicinity of a bounding surface where the effects of viscosity are significant. The fluid is often slower due to the effects of viscosity. Advection i.e the transfer of heat by the flow of liquid becomes less since the flow is slower, thereby the local heat transfer coefficient decreases.
From law of conduction, we observe that heat transfer rate will decrease based on a smaller rate of temperature, the thickness therefore increases while the local heat transfer coefficient decreases with distance.
Answer:
modulus of elasticity for the nonporous material is 340.74 GPa
Explanation:
given data
porosity = 303 GPa
modulus of elasticity = 6.0
solution
we get here modulus of elasticity for the nonporous material Eo that is
E = Eo (1 - 1.9P + 0.9P²) ...............1
put here value and we get Eo
303 = Eo ( 1 - 1.9(0.06) + 0.9(0.06)² )
solve it we get
Eo = 340.74 GPa
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 = 
ar = 
ar = 22.74 m/s²
so magnitude of total acceleration is
A = 
A = 
A = 22.98 m/s²
so magnitude of acceleration is 22.98 m/s²
Answer:
Feedforward basically configured and used mainly to avoid errors in a control system entering or disrupting a control loop
Explanation:
Feedforward basically configured and used mainly to avoid errors in a control system entering or disrupting a control loop. Although Feedforward control seems to be a very attractive idea, it imposes a high responsibility on both the system developer and the operator to examine and consider mathematically the effect of disruptions on the process concerned.
example of feedforward is
Shower
which consist of following control points
Hear toilet flush (measurement)
Customize water to compensate
feedback refers to that point when water turns hot before the configuration changes
