The question is incomplete. The complete question is :
A dielectric-filled parallel-plate capacitor has plate area A = 10.0 cm2 , plate separation d = 10.0 mm and dielectric constant k = 3.00. The capacitor is connected to a battery that creates a constant voltage V = 15.0 V . Throughout the problem, use ϵ0 = 8.85×10−12 C2/N⋅m2 .
Find the energy U1 of the dielectric-filled capacitor. I got U1=2.99*10^-10 J which I know is correct. Now I need these:
1. The dielectric plate is now slowly pulled out of the capacitor, which remains connected to the battery. Find the energy U2 of the capacitor at the moment when the capacitor is half-filled with the dielectric.
2. The capacitor is now disconnected from the battery, and the dielectric plate is slowly removed the rest of the way out of the capacitor. Find the new energy of the capacitor, U3.
Solution :
Given :
d = 10 mm
= 0.010 m
Then, Capacitance,
Now,
And
In parallel combination,
Then energy,
b). Now the charge on the is :
Now when the capacitor gets disconnected from battery and the is slowly of the way out of the is :
Without the dielectric,
Answer:
0.22m/s
Explanation:
The total momentum of the System is conserved. Total momentum of the system before the collision is equal to the total momentum of the system after collision. The total momentum is the sum of individual momentum of all the objects in that system.
momentum of an object = mass* velocity
Total Momentum before collision = 0.2*0.3 + 0.1*0.1= 0.07 kg⋅m/s;
Total momentum after collision = 0.1*0.26 + 0.2*x = 0.07;
Solve for x.
If you think more specifically about energy,such as thermal energy,the amount of that energy a substance has is determined by the movement of the molecules.hope this helps u
Answer:
hydraulic lift pump moves the lift piston out in a series of movements. and a hydraulic brake moves it out in one short break.