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 = 
It is important to maintain septic systems in order to <span>prevent untreated sewage from contaminating groundwater</span>
Answer:
This is because the 11 positive protons and 10 negative electrons end up with an overall charge of +1.
Explanation:
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Answer:
the reason for the acceleration month that the coefficient of kinetic friction is less than the coefficient of satic frictionExplanation:
This exercise uses Newton's second law with the condition that the acceleration is zero, by the time the body begins to slide. At this point the balance of forces is
fr- w || = 0
The expression for friction force is that it is proportional to the coefficient of friction by normal.
fr = μ N
When the system is immobile, the coefficient of friction is called static coefficient and has a value, this is due to the union between the surface, when the movement begins some joints are broken giving rise to coefficient of kinetic friction less than static.
In consequence a lower friction force, which is why the system comes out of balance and begins to accelerate.
μ kinetic <μ static
In all this movement the normal with changed that the angle of the table remains fixed.
Consequently, the reason for the acceleration month that the coefficient of kinetic friction is less than the coefficient of satic friction