Answer:
Incomplete question, check attachment for the graph needed to solve problem.
A 8.1nm........
Explanation:
Electric Field is given as
E=V/d
Where V is voltage
And d is the distance apart
E is the electric field
The voltage V just before action of potential is -70mV,
The value d=8.1nm
d=8.1×10^-9m
E=V/d
E=-70×10^-3/8.1×10^-9
E=-8.6×10^6 N/C
Then the magnitude of the electric field is 8.6×10^6N/C
F=nmv
where;
n=no. of bullets = 1
m=mass of bullets=2g *10^-3
V=velocity of bullets200m/sec
F=1
loss in Kinetic energy=gain in heat energy
1/2MV^2=MS∆t
let M council M
=1/2V^2=S∆t
M=2g
K.E=MV^2/2
=(2*10^-3)(200)^2/2
2 councils 2
2*10^-3*4*10/2
K.E=40Js
H=mv∆t
(40/4.2)
40Js=40/4.2=mc∆t
40/4.2=2*0.03*∆t
=158.73°C
The new natural frequency would be ω/2.
we know that,
= ω. -> equation 1
now, when capacitance is quadrupled,
. -> equation 2
substituting value of equation 1 in equation 2 , we get,
Hence, the new natural frequency of the circuit is ω/2.
what do you mean by frequency ?
The resonant frequency for a particular circuit is the frequency at which this equality stands true. Where L is the inductance in henries and C is the capacitance in farads, this is the LC circuit's resonant frequency.
Learn more about frequency here:-
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