No. If there are forces on an object, they need to have a counterforce, otherwise the object won't be zero. Since there are only 5 forces, the 5th force has no counterforce, resulting in movement or reforming the object.
Answer:
D
Explanation:
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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,
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I believe that is the correct answer.
I hope this helps you and have a great day! :-)
Answer: Option (c) is the correct answer.
Explanation:
It is known that the relation between resistance, length and cross-sectional area is as follows.
R =
Let the resistance of resistor A is denoted by R and the resistance of resistor B is denoted by R'.
Hence, for resistor A the expression for resistance according to the given data is as follows.
R =
On cancelling the common terms we get the expression as follows.
R =
Now, the resistance for resistor B is as follows.
R' =
Thus, we can conclude that the statement, Wire A has the same resistance as wire B, accurately compares the resistances of wire resistors A and B.