Answer:
(a) Since net charge remains same,after immersion Q is same
(b) I. 14.56pF ii. 3.05V
(c) ΔU = 5.204nJ
Explanation:
a)
C = kεA/d
k=1 for air
ε is 8.85x10-12F/m
A = .0025m2
d = .125m
C = 8.85x10-12x.0025/.125 = 1.77x10-13F = 0.177pF
Q = CV = .177pF * 244V = 43.188pC
Since net charge remains same,after immersion Q is same
b)
C = kεA/d, for distilled water k is approx. 80
Cwater = Cair x k
=0.177pF x 80 = 14.16pF
Q is same and C is changed V=Q/c holds. where Q is still 43.188pC and C is now 14.16pF, so V = 43.188pC/14.16pF = 3.05V
c) Change in energy: ΔU = Uwater - Uair
Uwater = Q2/2C = (43.188)2/2x.177pF = 5.27nJ
Uair = Q2/2C = (43.188)2/2x14.16pF = 0.066nJ
ΔU = 5.204nJ
Answer:
59.4 N
Explanation:
The force exerted on a current-carrying wire due to a magnetic field perpendicular to the wire is given by

where
I is the current in the wire
L is the length of the wire
B is the strength of the magnetic field
Here in this problem, we model the strike as a current-carrying wire, so we have:
I = 49,500 A is the current
L = 1 m is the length (we want to find the force per each meter of length)
is the strength of the magnetic field
Therefore, the force on each meter of the current due to the magnetic field is:

Answer:
vi = 14.610
Explanation:
initial velocity of the first object (vi) = 0 m/s, because it was dropped
distance (y) = -20 m
initial position (y0) = 0 m
acceleration due to gravity (g) = -9.8 m/s^2
y = 1/2 gt^2 + vi*t + y0
-20 = 1/2(-9.8)t^2 + 0 + 0
-20 = -4.9t^2
4.081 = t^2
+√4.081 = t
t = 2.020
time of second object = 2.020 - 1 = 1.020
Now we can plug in the new time to solve for vi of the second object.
y = 1/2 gt^2 + vi*t + y0
-20 = 1/2(-9.8)1.020^2 + vi*1.020 + 0
-20 = -5.098 + 1.020vi
14.902 = 1.020vi
vi = 14.610 m/s
Complete Question
The complete question iws shown on the first uploaded image
Answer:
a

b

Explanation:
Now looking at the diagram let take that the magnetic field is moving in the x-axis
Now the magnetic force is mathematically represented as
x B
Note (The x is showing cross product )
Note the force(y-axis) is perpendicular to the field direction (x-axis)
Now when the loop is swinging forward
The motion of the loop is from y to z to to x to y
Now since the force is perpendicular to the motion(velocity) of the loop
Hence the force would be from z to y and back to z
and from lenze law the induce current opposes the force so the direction will be from y to z to x
Now when the loop is swinging backward
The motion of the induced current will now be x to z to y