Answer:
a) before immersion
C = εA/d = (8.85e-12)(25e-4)/(1.31e-2) = 1.68e-12 F
q = CV = (1.68e-12)(255) = 4.28e-10 C
b) after immersion
q = 4.28e-10 C
Because the capacitor was disconnected before it was immersed, the charge remains the same.
c)*at 20° C
C = κεA/d = (80.4*)(8.85e-12)(25e-4)/(1.31e-2) = 5.62e-10 F
V = q/C = 4.28e-10 C/5.62e-10 C = 0.76 V
e)
U(i) = (1/2)CV^2 = (1/2)(1.68e-12)(255)^2 = 5.46e-8 J
U(f) = (1/2)(5.62e-10)(0.76)^2 = 1.62e-10 J
ΔU = 1.62e-10 J - 5.46e-8 J = -3.84e-8 J
Good question. if it's a physical change what you would see is a change in the physical object/appearance like let's say I Tear a piece of paper, that would be a Physical change. I believe your answer would be a chemical change.
Answer : The de-Broglie wavelength of this electron, 
Explanation :
The formula used for kinetic energy is,
..........(1)
According to de-Broglie, the expression for wavelength is,
or,
...........(2)
Now put the equation (2) in equation (1), we get:
...........(3)
where,
= wavelength = ?
h = Planck's constant = 
m = mass of electron = 
K.E = kinetic energy = 
Now put all the given values in the above formula (3), we get:
conversion used : 
Therefore, the de-Broglie wavelength of this electron,
Fish swimming forward in the water, the water gets pushed backward because the fish moving forward is forcing the water to move backward, the motion forward and backward are the same, they are opposite and equal.
I think the answer for the question above its b 1.2
