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:
5.38 m/s^2
Explanation:
NET force causing the object to accelerate = 50 -10 = 40 N
Mass of the object = 73 N / 9.81 m/s^2 = 7.44 kg
F = ma
40 = 7.44 * a a = 5.38 m/s^2
Answer:
mechanical advantage!
Explanation:
The Mechanical advantage of a machine is the factor by which the machine changes the input force.
When a a machine multiplies an input force, that's called a mechanical advantage.
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Defenition of Mechanical Advantage
brainly.com/question/16617083?referrer=searchResults
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Hope this helps! <3
Answer:
V = 0.0806 m/s
Explanation:
given data
mass quarterback = 80 kg
mass football = 0.43 kg
velocity = 15 m/s
solution
we consider here momentum conservation is in horizontal direction.
so that here no initial momentum of the quarterback
so that final momentum of the system will be 0
so we can say
M(quarterback) × V = m(football) × v (football) ........................1
put here value we get
80 × V = 0.43 × 15
V = 0.0806 m/s
