Answer:

Explanation:
Given that:
- Area of the plate of capacitor 1= Area of the plate of capacitor 2=A
- separation distance of capacitor 2,

- separation distance of capacitor 1,

- quantity of charge on capacitor 2,

- quantity of charge on capacitor 1,

We know that the Capacitance of a parallel plate capacitor is directly proportional to the area and inversely proportional to the distance of separation.
Mathematically given as:
.....................................(1)
where:
k = relative permittivity of the dielectric material between the plates= 1 for air

From eq. (1)
For capacitor 2:

For capacitor 1:

![C_1=\frac{1}{2} [ \frac{k.\epsilon_0.A}{d}]](https://tex.z-dn.net/?f=C_1%3D%5Cfrac%7B1%7D%7B2%7D%20%5B%20%5Cfrac%7Bk.%5Cepsilon_0.A%7D%7Bd%7D%5D)
We know, potential differences across a capacitor is given by:
..........................................(2)
where, Q = charge on the capacitor plates.
for capacitor 2:


& for capacitor 1:


![V_1=8\times [\frac{Q.d}{k.\epsilon_0.A}]](https://tex.z-dn.net/?f=V_1%3D8%5Ctimes%20%5B%5Cfrac%7BQ.d%7D%7Bk.%5Cepsilon_0.A%7D%5D)

Answer: 14. 49 m
Explanation:
We can solve this problem with the following equations:
(1)
(2)
Where:
is the horizontal distance between the cannon and the ball
is the cannonball initial velocity
since the cannonball was shoot horizontally
is the time
is the final height of the cannonball
is the initial height of the cannonball
is the acceleration due gravity
Isolating
from (2):
(3)
(4)
(5)
Substituting (5) in (1):
(6)
Finally:
We first calculate the acceleration on the ball using:
2as = v² - u²; u = 0 because ball is initially at rest
a = (36)²/(2 x 0.35)
a = 1850 m/s²
F = ma
F = 0.058 x 1850
= 107.3 Newtons
Answer:
6.02 radian
Explanation:
At 12 noon position is zero radian on Monday
at 12 midnight the position is π/2
at 11 am position on Tuesday is 2π/24 x 23 ( after 23 hours ) = 6..02 radian.