Answer:
Explanation:
We can use the conservation of the angular momentum.


Now the Inertia is I(professor_stool) plus mR², that is the momentum inertia of a hoop about central axis.
So we will have:

Now, we just need to solve it for ω.

I hope it helps you!
Answer:
The maximum energy stored in the combination is 0.0466Joules
Explanation:
The question is incomplete. Here is the complete question.
Three capacitors C1-11.7 μF, C2 21.0 μF, and C3 = 28.8 μF are connected in series. To avoid breakdown of the capacitors, the maximum potential difference to which any of them can be individually charged is 125 V. Determine the maximum energy stored in the series combination.
Energy stored in a capacitor is expressed as E = 1/2CtV² where
Ct is the total effective capacitance
V is the supply voltage
Since the capacitors are connected in series.
1/Ct = 1/C1+1/C2+1/C3
Given C1 = 11.7 μF, C2 = 21.0 μF, and C3 = 28.8 μF
1/Ct = 1/11.7 + 1/21.0 + 1/28.8
1/Ct = 0.0855+0.0476+0.0347
1/Ct = 0.1678
Ct = 1/0.1678
Ct = 5.96μF
Ct = 5.96×10^-6F
Since V = 125V
E = 1/2(5.96×10^-6)(125)²
E = 0.0466Joules
Answer:
776.6 w
1.04 hp
Explanation:
given:
Mass, m = 190kg
height change, h = 25m
time elapsed, t = 60 s
acceleration due to gravity, g = 9.81 m/s²
Potential energy required raising 190 kg of water to a height of 25m
= mgh
= 190 x 9.81 x 25
= 46,597.5 J
Power required in 60 s
= Energy required ÷ time elapsed
= 46,597.5 ÷ 60
= 776.6 Watts (Use conversion 1 W = 0.00134102 hp)
= 776.6 w x 0.00134102 hp/w
= 1.04 hp