Answer:
<em>T</em><em>h</em><em>e</em><em>r</em><em>e</em><em> </em><em>are</em><em> </em><em>t</em><em>wo hydrogen </em><em>atom</em><em> </em><em>in</em><em> </em><em>all</em><em> </em><em>the</em><em> </em><em>reactants</em><em>.</em>
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) 152.85 Nm
(b) 1528.5 Nm
Explanation:
According to the formula of power
P = τ ω
ω = 2 π f
(a) f = 2500 rpm = 2500 / 60 = 41.67 rps
So, 40 x 1000 = τ x 2 x 3.14 x 41.67
τ = 152.85 Nm
(b) f = 250 rpm = 250 / 60 = 4.167 rps
So, 40 x 1000 = τ x 2 x 3.14 x 4.167
τ = 1528.5 Nm
Cars 'A' and 'C' look like they're moving at the same speed. If their tracks are parallel, then they're also moving with the same velocity.
#8 positive kinetic energy