Answer: the constant angular velocity of the arms is 86.1883 rad/sec
Explanation:
First we calculate the linear velocity of the single sprinkler;
Area of the nozzle = π/4 × d²
given that d = 8mm = 8 × 10⁻³
Area of the nozzle = π/4 × (8 × 10⁻³)²
A = 5.024 × 10⁻⁵ m²
Now total discharge is dived into 4 jets so discharge for single jet will be;
Q_single = Q / n = 0.006 / 4 = 1.5 × 10⁻³ m³/sec
So using continuity equation ;
Q_single = A × V_single
V_single = Q_single/A
we substitute
V_single = (1.5 × 10⁻³) / (5.024 × 10⁻⁵)
V_single = 29.8566 m/s
Now resolving the forces as shown in the second image,
Vt = Vcos30°
Vt = 29.8566 × cos30°
Vt = 25.8565 m/s
Finally we calculate the angular velocity;
Vt = rω
ω_single = Vt / r
from the given diagram, radius is 300mm = 0.3m
so we substitute
ω_single = 25.8565 / 0.3
ω_single = 86.1883 rad/sec
Therefore the constant angular velocity of the arms is 86.1883 rad/sec
The answer to the problem b.
Answer:
Explanation:
a )
Time to reach the speed of 20 m/s with an acceleration of 2 m/s² can be calculated as follows .
v = u + a t
20 = 0 + 2 t
t = 20 /2 = 10 s .
Total time = 10 s + 20 s + 5 s = 35 s .
b) Average velocity = Total distance travelled / total time
Distance travelled in first 10 s
S₁ = ut + 1/2 a t²
= 0 + .5 x 2 x 10²
= 100 m
Distance travelled in next 20 s
S₂= 20s x 20 m/s = 400 m
Distance travelled in last 5 s .
deceleration in last 5 s
v = u + at
0 = 20 m/s + a x 5
a = - 4 m/s²
v² = u² - 2 a s
0 = (20 m/s)² - 2 x 4 m/s² x s
s = 50 m
S₃ = 50 m
Total distance = S₁ + S₂ + S₃
= 100 m + 400 m + 50 m
= 550 m .
Average velocity = 550 m / 35 s
= 15.71 m /s .
Answer:
Thrust due to fuel consumption must overcome gravitational force from the Earth to send the rocket up into space.
Explanation:
From the concept of Escape Velocity, derived from Newton's Law of Gravitation, definition of Work, Work-Energy Theorem and Principle of Energy Conservation, which is the minimum speed such that rocket can overcome gravitational forces exerted by the Earth, and according to the Tsiolkovski's Rocket Equation, which states that thrust done by the rocket is equal to the change in linear momentum of the rocket itself, we conclude that thrust due to fuel consumption must overcome gravitational force from the Earth to send the rocket up into space.
Answer:
The answer is "
"
Explanation:
Please find the complete question in the attached file.
pi = pressure only at two liquids' devices
PA = pressure atmosphere.
1 = oil density
2 = uncertain fluid density
The pressures would be proportional to the quantity 
cm from below the surface at the interface between both the oil and the liquid.
