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:
(a) 1.414 km
(b) 1.06 m/s
Explanation:
(a) For John:
Distance = 1 km north and then 1 km east
Speed = 1.5 m/s
total distance traveled = 1 + 1 = 2 km = 2000 m
Time taken to travel = Distance / speed
t = 2000 / 1.5 = 1333.3 seconds
Displacement = 
(b) For jane :
Time is same as john = 1333.33 second
Distance = 1.414 km = 1414 m
Speed = distance / time = 1414 / 1333.33 = 1.06 m/s
All of the orbitals in a given subshell have the same value of the "<span>magnetic and principal" quantum number
Hope this helps!</span>
The gravitational attraction between two planets is 4905.95 N
<h3>What is gravitational attraction?</h3>
When two objects with masses are placed at a distance, there will an attractive force acting between them.
According to the Newton's law of gravitation, gravitational force is
F = Gm₁m₂ /r²
where r is the distance between the masses m₁ and m₂ and G is the gravitational constant G = 6.67 x 10⁻¹¹ N-m²/kg²
Substitute the values into the expression, we get
F = 6.67 x 10⁻¹¹ x 2.25 x 10²⁰ x 6.20 x 10¹⁸ / (435,500 x 1000)²
F= 4905.95 N
Thus, the gravitational attraction between two planets is 4905.95 N.
Learn more about gravitational attraction.
brainly.com/question/19822389
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No one knows yet, but Alpha Centauri is an obvious target. It's the nearest star system to our sun at 4.3 light-years away. That's about 25 trillion miles (40 trillion km) away from Eart
