Answer:

Explanation:
Previous concepts
Angular momentum. If we consider a particle of mass m, with velocity v, moving under the influence of a force F. The angular momentum about point O is defined as the “moment” of the particle’s linear momentum, L, about O. And the correct formula is:

Applying Newton’s second law to the right hand side of the above equation, we have that r ×ma = r ×F =
MO, where MO is the moment of the force F about point O. The equation expressing the rate of change of angular momentum is this one:
MO = H˙ O
Principle of Angular Impulse and Momentum
The equation MO = H˙ O gives us the instantaneous relation between the moment and the time rate of change of angular momentum. Imagine now that the force considered acts on a particle between time t1 and time t2. The equation MO = H˙ O can then be integrated in time to obtain this:

Solution to the problem
For this case we can use the principle of angular impulse and momentum that states "The mass moment of inertia of a gear about its mass center is
".
If we analyze the staritning point we see that the initial velocity can be founded like this:

And if we look the figure attached we can use the point A as a reference to calculate the angular impulse and momentum equation, like this:

](https://tex.z-dn.net/?f=0%2B%5Csum%20%5Cint_%7B0%7D%5E%7B4%7D%2020t%20%280.15m%29%20dt%20%3D0.46875%20%5Comega%20%2B%2030kg%5B%5Comega%280.15m%29%5D%280.15m%29)
And if we integrate the left part and we simplify the right part we have

And if we solve for
we got:

Answer:
The minimum mass flow rate will be "330 kg/s".
Explanation:
Given:
For steam,


For water,


They add energy efficiency as condenser becomes adiabatic, with total mass flow rate of minimal vapor,
⇒ 
On putting the estimated values, we get
⇒ 
⇒ 
⇒ 
Answer: Coefficient= 0.35 per day
Explanation:
To find the bio degradation reaction rate coefficient, we have
k= 
Here, the C lagoon= 20 mg/L
Q in= Q out= 8640 m³/d
C in= 100 mg/L
C out= 20 mg/L
V= 10 ha* 1* 10
V= 10⁵ m³
So, k= 
k= 0.35 per day
Question
Determine the average water exit velocity
Answer:
53.05 m/s
Explanation:
Given information
Volume flow rate, 
Diameter d= 8cm= 0.08 m
Assumptions
- The flow is jet flow hence momentum-flux correction factor is unity
- Gravitational force is not considered
- The flow is steady, frictionless and incompressible
- Water is discharged to the atmosphere hence pressure is ignored
We know that Q=AV and making v the subject then
where V is the exit velocity and A is area
Area,
where d is the diameter
By substitution

To convert v to m/s from m/s, we simply divide it by 60 hence
