Pressured is measured by the Force per unit area
It's just in the name! Accurate data is helpful, and correct, but reproducible data is all of that, and is able to be given to other people through different sources! At least, that's what my understanding of them are. Hope it helps!
Let north and east be posetive
Y=20sin20
=6.840402867 south
X=20cos20
18.79385243 west
Net horizontal components=60+(-18.79)
=41.20614758m
Net vertical=-6.84040267
X^2=41.20614758^2+(-6.684040267)^2
√x^2 =√1744.73771m
x=41.77m
Answer:
I wish I learned what I could do in the real world with the information I learned
Explanation:
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,