Answer:
Qx = 9.10
m³/s
Explanation:
given data
diameter = 85 mm
length = 2 m
depth = 9mm
N = 60 rev/min
pressure p = 11 × 
Pa
viscosity n = 100 Pas
angle = 18°
so Qd will be
Qd = 0.5 × π² ×D²×dc × sinA × cosA ..............1
put here value and we get
Qd = 0.5 × π² × ( 85 
)²× 9 × sin18 × cos18
Qd = 94.305 × 
m³/s
and
Qb = p × π × D × dc³ × sin²A ÷ 12 × n × L ............2
Qb = 11 × 
× π × 85 × ( 9 )³ × sin²18 ÷ 12 × 100 × 2
Qb = 85.2 × m³/s
so here
volume flow rate Qx = Qd - Qb ..............3
Qx = 94.305 × - 85.2 ×
Qx = 9.10 m³/s
Reflection helps designers to learn from their experiences, to integrate and co-ordinate different aspects of a design situation, to judge the progress of the design process, to evaluate interactions with the design context, and to plan suitable future design activities.
Answer:
battery life in year = 9 years and 48 days
Explanation:
given data
Battery Ampere-hours = 1.5
Pulse voltage = 2 V
Pulse width = 1.5 m sec
Pulse time period = 1 sec
Electrode heart resistance = 150 Ω
Current drain on the battery = 1.25 µA
to find out
battery life in years
solution
we get first here duty cycle that is express as
duty cycle = 
...............1
duty cycle = 1.5 × 
and applied voltage will be
applied voltage = duty energy × voltage ...........2
applied voltage = 1.5 × × 2
applied voltage = 3 mV
so current will be
current = 
................3
current = 
current = 20 µA
so net current will be
net current = 20 - 1.25
net current = 18.75 µA
so battery life will be
battery life = 
battery life = 80000 hours
battery life in year = 
battery life in year = 9.13 years
battery life in year = 9 years and 48 days
Answer:
X_cp = c/2
Explanation:
We are given;
Chord = c
Angle of attack = α
p u (s) = c 1
p1(s)=c2,
and c2 > c1
First of all, we need to find the resultant normal force on the plate and the total moment about leading edge.
I've attached the solution
