Solution:
With reference to Fig. 1
Let 'x' be the distance from the wall
Then for 
DAC:
⇒ 
Now for the BAC:
⇒ 
Now, differentiating w.r.t x:
![\frac{d\theta }{dx} = \frac{d}{dx}[tan^{-1} \frac{d + h}{x} - tan^{-1} \frac{d}{x}]](https://tex.z-dn.net/?f=%5Cfrac%7Bd%5Ctheta%20%7D%7Bdx%7D%20%3D%20%5Cfrac%7Bd%7D%7Bdx%7D%5Btan%5E%7B-1%7D%20%5Cfrac%7Bd%20%2B%20h%7D%7Bx%7D%20-%20%20tan%5E%7B-1%7D%20%5Cfrac%7Bd%7D%7Bx%7D%5D)
For maximum angle, 
= 0
Now,
0 = [/tex]\frac{d}{dx}[tan^{-1} \frac{d + h}{x} - tan^{-1} \frac{d}{x}][/tex]
0 = 
After solving the above eqn, we get
x = 
The observer should stand at a distance equal to x =
Answer:
it shows the products of a chemical reaction to the right of the reaction arrow
Answer:
U₁ = (ϵAV²)/6d
This means that the new energy of the capacitor is (1/3) of the initial energy before the increased separation.
Explanation:
The energy stored in a capacitor is given by (1/2) (CV²)
Energy in the capacitor initially
U = CV²/2
V = voltage across the plates of the capacitor
C = capacitance of the capacitor
But the capacitance of a capacitor depends on the geometry of the capacitor is given by
C = ϵA/d
ϵ = Absolute permissivity of the dielectric material
A = Cross sectional Area of the capacitor
d = separation between the capacitor
So,
U = CV²/2
Substituting for C
U = ϵAV²/2d
Now, for U₁, the new distance between plates, d₁ = 3d
U₁ = ϵAV²/2d₁
U₁ = ϵAV²/(2(3d))
U₁ = (ϵAV²)/6d
This means that the new energy of the capacitor is (1/3) of the initial energy before the increased separation.
The solution for this problem is:
500 revolution per
minute = 8.33rev /s = 2π*8.33 rad /s = 52.36 rad /s
Angular velocity ω = 2π N
Angular acceleration α= (ω2 - ω1) /t
ω2 = 0
α = - ω1/t = -2π N /t
N = 500 rpm = 8.33 r p s.
α = -2π 8.33 /2.6 =- 20 rad/s^2
