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:
Mass, m = 125 kg
Explanation:
Let us assume that the question says, "What is the mass of an object whose velocity is 400 m/s and the kinetic energy of 10⁷ J.
The kinetic energy of an object is :
So, the mass of the object is 125 kg.
Probably it whould separate becase hot water departs things! Hope this helps
Answer:
V = (v1 + v2) / 2 = (8 + 6.5) / 2 = 7.25 m/s average speed
t = 7.2 / 7.25 = .993 sec time to cross patch
a = (v2 - v1) / t = (6.5 - 8) / .993 = -1.51 m/s^2 or 1.5 m/s^2
