Answer:
<em>His angular velocity will increase.</em>
Explanation:
According to the conservation of rotational momentum, the initial angular momentum of a system must be equal to the final angular momentum of the system.
The angular momentum of a system =
'ω'
where
' is the initial rotational inertia
ω' is the initial angular velocity
the rotational inertia = 
where m is the mass of the system
and r' is the initial radius of rotation
Note that the professor does not change his position about the axis of rotation, so we are working relative to the dumbbells.
we can see that with the mass of the dumbbells remaining constant, if we reduce the radius of rotation of the dumbbells to r, the rotational inertia will reduce to
.
From
'ω' =
ω
since
is now reduced, ω will be greater than ω'
therefore, the angular velocity increases.
Answer:
Angular acceleration of the disk will be 
Explanation:
We have given mass of the disk m = 5 kg
Diameter of the disk d = 30 cm = 0.3 m
So radius 
Moment of inertia of disk is given by 
Force is given by F=4 N
Torque is given as 
We also know that torque is given by 


Answer:
Hindi ko alam pasensya ka ha godbless
Answer:
I believe it's the first option
Answer:
Explanation:
We have here values from SI and English Units. I will convert the units to English Units.
We hace for the power P,


we have other values such
and
(specific weight of the water), and 0.85 for \eta
We need to figure the flow rate of the water (V) out, that is,

Where
is the turbine efficiency, at which is,

Replacing,


With this value (the target of this question) we can also calculate the mass flow rate of the waters,
through the density and the flow rate,

converting the slugs to lbm, 1slug = 32.174lbm, we have that the mass flow rate of the water is,
